Download to read offline
![CHAPTER 3
OSCILLATIONS
3.1 ( ) [ ]
1
0.002sin 2 512
x s t m
π −
=
( )( )( )
max 0.002 2 512 6.43
m m
x
s s
π
= =
( )( ) ( )
2 2 4
max 2 2
0.002 2 512 2.07 10
m m
x
s s
π
= =
×
3.2 0.1sin
x t
ω
= [m] 0.1 cos
m
x t
s
ω ω
=
When t = 0, x = 0 and 0.5 0.1
m
x
s
ω
= =
1
5s
ω −
=
2
1.26
T s
π
ω
= =
3.3 ( ) cos sin
x
x t x t t
ω ω
ω
= + and 2 f
ω π
=
( ) ( )[ ]
0.25cos 20 0.00159sin 20
x t t
π π
= + m
3.4 ( )
cos cos cos sin sin
α β α β α
− = + β
( )
cos cos cos sin sin
x A t A t A t
ω φ φ ω φ
= − = + ω
cos sin
x t t
ω ω
= Α + Β , cos
A φ
Α = , sin
A φ
Β =
3.5 2 2 2 2
1 1 2
1 1 1 1
2 2 2 2
mx kx mx kx
+ = + 2
( ) ( )
2 2 2 2
1 2 2 1
k x x m x x
− = −
1
2 2 2
2 1
2 2
1 2
k x x
m x x
ω
−
= =
−
2 2
1 1
1 1 1
2 2 2
kA mx kx
= + 2
2 2 2 2
2 2 2 1 1 2 1
1 1 2 2
2 1
m x x x x
A x x
k x x
−
= + = +
−
2
1
x
1
2 2 2 2 2
1 2 2 1
2 2
2 1
x x x x
A
x x
−
=
−
1](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-27-2048.jpg)
![( ) 2 2
1
sin cos cos 0
2
z t gt v t v t
α ω α λ
′ ′
= − + + =
D D
or
2 sin 2 sin
2 cos cos
v v
t
g v g
α α
ω α λ
′ ′
= ≈
′
−
D D
D
We have ignored the second term in the denominator—since v′
D would have to be
impossibly large for the value of that term to approach the magnitude g
For example, for 41 45
o o
and
λ α
= = 2 cos cos
g v g v
ω α λ ω
′ ′
− ≈ −
D D
or 144
g k
s
ω
′ ≈ ≈
D
m
v !
Substituting t into equation 5.4.15b to find the lateral deflection gives
( ) [ ]
3
2 2
2
4
cos sin sin sin cos
v
y t v t
g
ω
ω α λ λ α
′
= − = − D
D
α
5.18 Let … aD
G
= acceleration of object relative to Earth
G
ωD = k̂
ωD
AD
= its angular speed
R
G
0
R
G
r
G
y x G
= acceleration of satellite
k̂
ω ω
=
G
= its angular speed
( )
2
a a v r A
ω ω ω
= + × + × × +
D D
G
G G G G G G G
(Equation 5.2.14)
G
G G G G G G G
2
a a A v r
ω ω ω
= − − × − × ×
D D
As in prob em 5.7 Evaluate the term
G
G G G
l …
( )
a A r R R R R
ω ω ω ω ω ω ω ω
= − − × × = × × − × × − × × −
D D D D D D
G G
G G G G G G G G G G
( )
2 2
a R
ω ω
∆ = −
D D
G G
a
∆
… given the condition that 2 3 2 3
R R
ω ω
=
D D
3
2
3
1
a
R
R
R
ω
∆ = − −
D
D
G G
G G G G
but ( ) ( ) 2 2
2 cos
R R R r R r R r rR θ
⋅ = + ⋅ + = + +
D D
G G
Letting x cosθ
= 2 2 2 2
2 1
x
R R R r Rx R
R
⋅ = + + ≈ +
D D
G G
for small r
and
3
2
3 3 2
1
x
R R
R
≈ +
D
or
3
3
2
3
2
1
R x
R R
−
= +
D
3
2
2 2
2 ˆ
1 1 3 3
a
R
x 2
R x
R R
ω ω
−
∆ = − − + ≈ − ≈ −
D
D xi
ω
G
G G
for small r
9](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-64-2048.jpg)
![b) ( )
( )
( ) ( )
2 2
1 1
2 2
2 2
2 2
1
,
2
1
x y
y
x y x y
α α
α α
−
V x
+
= − − −
− + + − +
(7.4.13)
( )( )
[ ]
( )
[ ]
3 3
2 2
1 1
x x
V
x
x
α α α α
− − + −
∂
= +
∂
−
Now x 0.5
α
= − at and
4
L 5
L
also, each bracket term in the denominator equals 1 at ,
4
L 5
L
( )( ) ( ) (
1 0.5 0.5 1 0.5
V
x
α α α α α α α
∂
= − − − + − + − − −
∂
)
0.5 0.5 0.5 0.5 0
α α α
= − + + − + ≡
( )
[ ] [ ]
3 3
2 2
1 y
V y
y
y
α α
−
∂
= +
∂
−
Again, the denominator in brackets equals 1 @ ,
4
L 5
L
So, ( )
3 3
1
2 2
V
y
α α
= − ± + ± − ±
∂
3
2
∂
3 3 3 3
0
2 2 2 2
α α
= ± ± ≡
∓ ∓
Thus ( ) ˆ ˆ
, 0
V V
V x y i j
x y
∂ ∂
+
∂ ∂
∇ = ≡
n of moment :
at L , .
4 5
L
7.14 Conservatio um
4
p p p p
m v m v m vα
′ ′
= +
45 o
v’
φ
v’p
cos45 4 cos
p
v v vα φ
′ ′
= +
4 cos
2
p
v
v v
α φ
′
′ = −
vo
0 sin 45 4 sin
p
v vα φ
′ ′
= −
4 sin
2
p
v
vα φ
′
′ =
2 2
16 2 p p
v v v v v
α
′ ′
= − + 2
′
Conservation of energy:
2 2
1 1 1
4
2 2 2
p p p
m v m v m v 2
p α
′ ′
= +
16 2 2
4 4 p
v v
α
′ ′
= − v
Subtracting: 2 2
2 5
p p
v v v
0 3 v
′ ′
= − − +
6](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-95-2048.jpg)
![( )
3 3 3 1 2 3 2 1
0 I I I
ω ω ω ω ω
= + −
Adding, 0 1 1 1 2 2 2 3 3 3
I I I
ω ω ω ω ω
= + +
ω
( )
2 2 2
1 1 2 2 3 3
1
0
2
d
I I I
dt
ω ω ω
= + +
From equation 9.2.5, [ ]
rot
d
T
dt
=
0
rot
T constant
=
Multiplying equations 9.3.5 by 1 1
I ω , 2 2
I ω , and 3 3
I ω , respectively:
( )
2
1 1 1 1 1 2 3 3 2
0 I I I
ω ω ω ω ω
= + −
I
( )
2
2 2 2 2 1 2 3 1 3
0 I I I
ω ω ω ω ω
= + −
I
( )
2
3 3 3 3 1 2 3 2 1
0 I I I
ω ω ω ω ω
= + −
I
Adding, 0 2 2 2
1 1 1 2 2 2 3 3 3
I I I
ω ω ω ω ω
= + +
ω
( )
2 2 2 2 2 2
1 1 2 2 3 3
1
0
2
d
I I I
dt
ω ω ω
= + +
From equation 9.2.4, ( )
2
d
L
dt
=
0
2
L constant
=
9.8 From equations 9.3.5 for zero torque …
( )
1 1 2 3 3 2
0 I I I
ω ω ω
= + −
( )
2 2 1 3 1 3
0 I I I
ω ω ω
= + −
From the perpendicular axis theorem, 3 1 2
I I I
= +
1 1 2 3 1
0 I I
ω ω ω
= +
2 2 1 3 2
0 I I
ω ω ω
= −
Multiplying by 1
1
I
ω
and 2
2
I
ω
, respectively:
1 1 1 2 3
0 ω ω ω ω ω
= +
2 2 1 2 3
0 ω ω ω ω ω
= −
Adding, ( )
2 2
1 1 2 2 1 2
1
2
d
dt
0 ω ω ω ω ω ω
= + = +
2 2
1 2 constant
ω ω
+ =
If 1 2
I I
= , from the third of Euler’s equations … ( )
3 3 0
I ω =
3 constant
ω =
9.9 (a)From symmetry … 3
s
I I
= and 1 2
I I I
= =
From the perpendicular axis theorem, 3 1 2
I I I
= + or 2
s
I I
=
From eqn 9.5.8, ( )
2 1 cos
ω α
−
Ω =
5](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-130-2048.jpg)
![( )
( )
( )
( )
2 2
2 2
1
2 2 2 2
2 2
2
a b c
m
L b
a b c
c a b
ω
+
= +
+ +
+
G
a c
From equation 9.1.32,
1
2
ω
= ⋅
T L
G
G
( )
[ ]
( )
( )
( )
2 2
2
2 2
2 2 2
2 2
1 2
2
a b c
m
T a b c b
a b c
c a b
ω
a c
+
= +
+ +
+
( ) ( ) (
2
2 2 2 2 2 2 2 2 2
2 2 2
m
T a b c b a c c a b
a b c
ω
= + + + +
+ +
)
+
( )
2
2 2 2 2 2 2
2 2 2
2m
T a b a c
a b c
ω
= +
+ +
b c
+
9.21 For Problem 9.1:
2
1 1
0
3 2
1 4
0
2 3
5
0 0
3
I ma
−
= −
I
,
2
1
5
0
ω
ω
=
G
2
1 1
0
3 2 2
1 4
0 1
2 3
5
0
5
0 0
3
ma
L I
ω
ω
−
= = −
G I G
2 2
2 1 1
3 2 6 1
4 1
1 2
3 3
5 5 6
0
0 0
ma ma ma
ω ω
−
= − + = =
2
5
ω
( )
2
1
1 1
2 1 0 2
2 2 5 6 5
0
ma
T L
ω ω
ω
= ⋅ =
G
G
( )
2 2 2 2
2 2 0
60 15
ma ma
ω ω
= + + =
14](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-139-2048.jpg)
![( )
[ ]
( )
( )
( )
2 2
2
2 2
2 2 2
2 2
1
2 3
a b c
m
T a b c b
a b c
c a b
ω
a c
+
= +
+ +
+
( )
( ) ( ) (
2
2 2 2 2 2 2 2 2 2
2 2 2
6
m
T a b c b a c c a b
a b c
ω
= + + + +
+ +
)
+
( )
( )
2
2 2 2 2 2 2
2 2 2
3
m
T a b a c
a b c
ω
= +
+ +
b c
+
With the origin at one corner, from the parallel axis theorem:
( )
( ) ( )
2 2
2 2 2 2
4
3 3
xx
m b c m
I m b c b c
+
= + + = +
( )
2 2
4
3
yy
m
I a c
= + , ( )
2 2
4
3
zz
m
I a b
= +
xy
I xydm xy dV
ρ
= − = −
∫ ∫
2 2 2
2 2
0 0 0
8
x a y b z c
xy
x y z
I xydxdydz a b c
ρ ρ
= = =
= = =
= − = −
∫ ∫ ∫
, so
( )( )( )
2 2 2 8
m a b c abc
ρ ρ
= = xy
I mab
= −
xy
I mac
= − , yz
I mbc
= −
( )
( )
( )
2 2
2 2
2 2
4
3
4
3
4
3
b c ab ac
I m ab a c bc
ac bc a b
+ − −
= − + −
− − +
I
9.23 (See Figure 9.7.1)
( ) ( ) ( )
z x y z
z z
z
L L L L
′ ′
= + + ′
( ) ( )
sin cos sin sin cos
z y z s
L L L I I S
θ θ φ θ θ
′ ′
= + = +
θ
9.24 (See Figure 9.7.1) If the top precesses without nutation, it must do so at θ θ
= D
whereV ( )
θD is a minimum of ( )
V θ …
( ) 0
dV
d θ θ
θ
θ =
=
D
( )
( )
2
2
cos
cos
2 sin
z z
L L
V
I
θ
mgl
θ θ
θ
′
−
= + (See equation 9.8.7)
16](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-141-2048.jpg)
![CHAPTER 10
LAGRANGIAN MECHANICS
10.1 Solution … ( ) ( ) ( )
0,
x t x t t
αη
= +
( ) ( ) ( )
0,
x t x t t
αη
= +
where ( )
0, sin
x t t
ω
= and ( )
0, cos
x t t
ω ω
=
2
1
2
T m
=
x 2 2
1 1
2 2
V kx mω
= = 2
x
so: ( ) ( )
2
1
2 2 2
2
t
t
m
J x
α ω
= −
∫ x dt
( ) (
2
1
2 2
2
cos sin
2
t
t
m
t t
ω ω αη ω ω αη
= + − +
∫ ) dt
( ) ( ) ( ) ( )
2 2
1 1
2
2 2 2 2 2 2
cos sin cos sin
2 2
t t t
t t
m m
J t t dt m t t dt dt
α
α ω ω ω αω η ω ωη ω η ω η
= − + − + −
∫ ∫ ∫
2
1
t
Examine the term linear in α :
( ) ( )
2 2 2
2
1
1 1 1
cos sin cos sin sin 0
t t
t
t
t t
t t dt t t tdt tdt
η ω ωη ω η ω ωη ω ωη ω
− = + −
∫ ∫
t
t
≡
∫
(1st
term vanishes at both endpoints: ( ) ( )
2 1 0
t t
η η
= = )
so ( ) ( )
2 2
1 1
2 2 2
1 1
cos2
2 2
t t
t t
J m tdt m
α ω ω α η ω η
= + −
∫ ∫ 2 2
dt
[ ] ( )
2
1
2 2 2 2
2 1
1 1
sin 2 sin 2
4 2
t
t
m t t m
ω ω ω α η ω η
= − + −
∫ dt
which is a minimum at 0
α =
10.2 V mgz
=
( )
2 2 2
1
2
T m x y z
= + +
( )
2 2 2
1
2
L T V m x y z mgz
= − = + + −
L
mx
x
∂
=
∂
,
L
my
y
∂
=
∂
,
L
mz
z
∂
=
∂
d L
mx
dt x
∂
=
∂
,
d L
my
dt y
∂
=
∂
,
d L
mz
dt z
∂
=
∂
1](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-144-2048.jpg)
![( )
2
F ma m v m r
ω ω ω
′ ′
= + × + × ×
G G G G G G
′
G
G
( ) ( ) ( )
ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
2
F m ix jy kz m k ix jy kz m k k ix jy kz
ω ω ω ˆ
= + + + × + + + × × + +
( )
2
2
x
F m x y x
ω ω
= − −
( )
2
2
y
F m y x y
ω ω
= − −
z
F m
=
z
10.10
( )
2 2 2
1
2
T m r r θ
= +
θ
r
( )
2
1
cos
2
V k r l mgr θ
= − −
D
( ) ( )
2
2 2 2
cos
2 2
m k
L T V r r r l mgr
θ θ
= − = + − − +
D
L
mr
r
∂
=
∂
,
d L
mr
dt r
∂
=
∂
( )
2
cos
L
mr k r l mg
r
θ θ
∂
= − − +
∂
D
( )
2
cos
mr mr k r l mg θ
= − − +
D
2
L
mr θ
θ
∂
=
∂
sin
L
mgr θ
θ
∂
= −
∂
( )
2
sin
d
mr mgr
dt
θ θ
= −
10.11 (See Example 4.6.2) ( )
2 sin 2
4
a
x θ θ
= + ( )
1 cos2
4
a
y θ
= −
at 0
θ = 0
x y =
( )
2 2
1
2
T m x y
= +
V mgy
=
( ) ( )
cos2 s2
x 1 co
2 2
a aθ
θ θ θ
= +
θ
= +
sin 2
2
a
y θ θ
=
( ) ( )
2 2
2 2
1 cos2 sin 2 1 cos2
8 4
ma mga
L T V
θ
θ θ θ
= − = + + − −
[ ] ( )
2 2
1 2cos2 1 1 cos2
8 4
ma mga
θ
θ θ
+ − −
= +
2 2
2 2
cos sin
2 2
ma mga
θ
θ θ
= −
where we used the trigonometric identies … 2
2cos 1 cos2
θ θ
= + and 2
2sin 1 cos2
θ θ
= −
8](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-151-2048.jpg)
![(b) Atwood’s machine …
( )
1 2
V m gx m g l x
= − − −
2
2 2
1 2 2
1 1 1
2 2 2
x
T m x m x I
a
= + +
[Includes the pulley]
1 2 2
T I
p m m x
x a
∂
= = + +
∂
,
1 2 2
p
x
I
m m
a
=
+ +
( )
2
1 2 2
1 2 2
2
p
H T V m m gx m g
I
m m
a
= + = − − −
+ +
l
1 2 2
H p
x
I
p
m m
a
∂
= =
∂
+ +
( )
1 2
H
m m g
x
∂
= − − = −
∂
p , ( )
1 2
p m m g
= −
(c) Particle sliding down a smooth inclined plane …
sin
V mgx θ
= − 2
1
2
T m
=
x
T
p mx
x
∂
= =
∂
,
p
x
m
=
2
sin
2
p
H T V mgx
m
θ
= + = −
H p
x
p m
∂
= =
∂
sin
H
mg p
x
θ
∂
= − = −
∂
, sin
p mg θ
=
10.28
(a)
r
F
( , ) ( , )
i i i
L T q q T q t
= −
…note, potential energy is time dependent.
V
F
r
∂
= −
∂
so t
k
V Fdr e
r
β
−
= = −
∫
( )
2 2 2
1 1
2 2
T mv v m r r θ
= ⋅ = +
G G
22](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-165-2048.jpg)
![4
[ ]
sin cos
V mg a b
θ θ θ
′= − +
[ ]
cos sin cos
V mg b b a
θ θ θ θ
′′= − −
( )
0
V mg b a
′′ = −
Equilibrium …
Stable Unstable
a b
a b
11.6
2 4
cos 1 ...
2! 4!
θ θ
θ = − + −
3 5
sin ...
3! 5!
θ θ
θ θ
= − + −
( )
2 4 3 5
1 ... ...
2! 4! 3! 5!
V mg a b b
θ θ θ θ
θ θ
= + − + − + − + −
2 4
3
...
2 24
a b a b
V mg a b θ θ
− −
= + − + −
For a = b, 4
2 ...
12
a
g a θ
= −
V m +
3
...
3
mga
V θ
′=− + terms in higher order of θ
2
...
V mgaθ
′′=− +
2 ...
V mgaθ
′′′=− +
2 0
V mga
′′′′=− ∴ Equilibrium is unstable
11.7
The center of mass (CM) of the hemisphere is
3
8
a from the flat side (see Equation 8.1.8).
The height of CM) above the point of contact
between the two hemispheres is designated
by h2 in the figure. h1 is the height of the
point of contact above the ground.
a
φ θ
h1 θ b
h2
( ) (
1 2
3
cos cos cos
8
V mg h h mg b a a )
θ θ θ
= + = + − +
ϕ and a b
ϕ θ
=](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-172-2048.jpg)
![8
Equilibrium @ 0
φ =
2
3
0
cos2
e
GM a
V
r
φ
′′≈ and
2
0 3
0
e
GM a
r
′ ≈
V
2
1
3
M I ma
= =
0
3
0
3 e
V G
M r
ω
′′
= =
M
3
0
0
2
2
3 e
r
T
GM
π
π
ω
= =
11.13
The amplitude of the symmetric component is A1 and the amplitude of the anti-symmetric
component is A2 (See Equations 11.3.19a through 11.3.20b).
[ ]
2
2
1 1 2
1
(0) (0)
4
A x x
= +
[ ] 0
1 1 2
1
(0) (0)
2 2
A
A x x
= + =
[ ]
2
2
2 2 1
1
(0) (0)
4
A x x
= −
[ ] 0
2 2 1
1
(0) (0)
2 2
A
A x x
= − = 1 2
A A
∴ =
From Equation 11.3.18 the solution for x1 is …
( )
0
1 1
( ) cos cos
2
A
2
x t t t
ω ω
= + (The phase 2
δ is 180o
, which insures that 1 0
(0)
x A
= )
( ) ( )
0 1 2 1 2
1( ) 2cos cos
2 2 2
A t t
x t
ω ω ω ω
+ −
=
Letting … 1 2 1
2 2
and 2
ω ω ω
ω
+ −
= ∆ =
ω
( )
1 0
( ) cos cos
x t A t t
ω
= ∆
From Equation 11.3.18, the solution for x2 is …
( )
0
2 1
( ) cos cos
2
A
2
x t t t
ω ω
= −
( ) ( )
0 1 2 1 2
2 ( ) 2sin sin
2 2 2
A t t
x t
ω ω ω ω
+ −
=
( )
2 0
( ) sin sin
x t A t t
ω
= ∆](https://image.slidesharecdn.com/engineeringphysics-compilationofequationsandcorollaries-250324175850-5d09bf08/75/Engineering-Physics-Compilation-of-Equations-and-Corollaries-pdf-176-2048.jpg)
Uploaded byDharmsinh Desai of University
Engineering Physics- Compilation of Equations and Corollaries.pdf
FUNDAMENTAL CONCEPTS: VECTORS NEWTONIAN MECHANICS; RECTILINEAR MOTION; OSCILLATIONS; MOTION OF A PARTICLE IN 3D; NONINERTIAL REFERENCE ; SYSTEMS; GRAVITATIONAL AND CENTRAL FORCE DYNAMICS OF SYSTEMS OF PARTICLE MECHANICS OF RIGID BODIES PLANAR MOTION; Movement of rigid bodies in 3D; LAGRANGIAN MECHANICS; EQUILIBRIUM MOTION; Engineering Physics Mechanics
More Related Content
![Cálculo ii howard anton - capítulo 16 [tópicos do cálculo vetorial]](https://cdn.slidesharecdn.com/ss_thumbnails/clculoii-howardanton-captulo16tpicosdoclculovetorial-141009135024-conversion-gate02-thumbnail.jpg?width=640&height=640&fit=bounds)
Similar to Engineering Physics- Compilation of Equations and Corollaries.pdf
More from Dharmsinh Desai of University
Engineering Physics- Compilation of Equations and Corollaries.pdf
- 2. Engineering Physics Compilation ofEquations and Corollaries Prof. Samirsinh P Parmar Department of Civil Engineering, Dharmsinh Desai University, Nadiad-387001 Gujarat, India
- 3. Index 1. FUNDAMENTAL CONCEPTS:VECTORS 2. NEWTONIAN MECHANICS: RECTILINEAR MOTION OF A PARTICLE 3. OSCILLATIONS 4. GENERAL MOTION OF A PARTICLE IN THREE DIMENSIONS 5. NONINERTIAL REFERENCE SYSTEMS 6. GRAVITATIONAL AND CENTRAL FORCES 7. DYNAMICS OF SYSTEMS OF PARTICLES 8. MECHANICS OF RIGID BODIES: PLANAR MOTION 9. MOTION OF RIGID BODIES IN THREE DIMENSIONS 10.LAGRANGIAN MECHANICS 11. EQUILIBRIUM MOTION
- 4. ABSTRACT This book presentsa comprehensive collection of equations and corollaries in Engineering Mechanics and Engineering Physics, covering fundamental and advanced topics essential for understanding the principles of motion and equilibrium. Beginning with vector analysis, the foundation of mechanics, and the book systematically progresses through Newtonian mechanics, addressing rectilinear motion, oscillations, and three-dimensional particle motion. It further explores the complexities of no inertial reference systems, gravitational and central forces, and the dynamics of particle systems. For rigid body mechanics, the book delves into planar and three-dimensional motion, providing key formulations for analysing forces and torques. Advanced theoretical mechanics is introduced through Lagrangian mechanics, offering a powerful mathematical framework for deriving equations of motion. The final chapter on equilibrium motion consolidates the principles necessary for static and dynamic stability analysis. Designed as a reference guide, this book is valuable for students, researchers, and professionals seeking a structured and equation-driven approach to engineering mechanics and physics. With its focus on concise formulations and fundamental corollaries, it serves as a vital resource for problem-solving and theoretical analysis in mechanical and structural systems.
- 5. ABOUT THE AUTHOR Prof.Samirsinh P. Parmar is an esteemed academician and researcher in the field of Civil Engineering, specializing in Geotechnical Engineering and Nanotechnology applications. He has been serving as an Assistant Professor in the Department of Civil Engineering at Dharmsinh Desai University, Nadiad, since July 2006, contributing significantly to both teaching and research. Prof. Parmar completed his B.Tech in Civil Engineering from Lukhdhirji Engineering College, Morbi, affiliated with Saurashtra University, Rajkot, in 2003. He pursued his M.Tech in Geotechnical Engineering from Dharmsinh Desai University, Nadiad, in 2005 and later joined IIT Kanpur as a QIP Research Scholar from 2014 to 2016. Currently, he is pursuing a Ph.D. in Nanotechnology applications in Geotechnical Engineering, exploring innovative materials and techniques to enhance soil and foundation performance. With a strong passion for research and knowledge dissemination, Prof. Parmar has published more than 40 research papers in renowned journals and conferences, covering diverse topics such as Geotechnical Engineering, Ancient Temple Architecture, Nanotechnology applications in Geotechnical Engineering, and Social Sciences. His dedication to academic outreach is further reflected in his contribution of over 200 PPT articles on Slideshare.com, making technical knowledge accessible to a broader audience. Through his extensive academic journey and research endeavours, Prof. Parmar continues to shape the future of engineering education and research, bridging the gap between traditional engineering principles and emerging technologies.
- 6. CHAPTER 1 FUNDAMENTAL CONCEPTS:VECTORS 1.1 (a) A B ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) 2 i j j k i j k + = + + + = + + K K 1 2 (1 4 1) 6 A B + = + + = K K (b) 3 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ 3( ) 2( ) 3 2 A B i j j k i j − = + − + = + − K K k (c) (1)(0) (1)(1) (0)(1) 1 A B ⋅ = + + = K K (d) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1 1 0 (1 0) (0 1) (1 0) 0 1 1 i j k A B i j k i j k × = = − + − + − = − + K K 1 2 (1 1 1) 3 A B × = + + = K K K K 1.2 (a) ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ 2 4 (2)(1) (1)(4) (0)(1) 6 A B C i j i j k ⋅ + = + ⋅ + + = + + = K ( ) ( ) ˆ ˆ ˆ ˆ 3 4 (3)(0) (1)(4) (1)(0) 4 A B C i j k j + ⋅ = + + ⋅ = + + = K K K (b) ( ) 2 1 0 1 0 1 8 0 4 0 A B C ⋅ × = = − K K K ( ) ( ) 8 A B C A B C × ⋅ = ⋅ × = − K K K K K K (c) ( ) ( ) ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ 4 2 4 4 8 A B C A C B A B C i k j i j k × × = ⋅ − ⋅ = + − = − + K K K K K K K K K 4 ( ) ( ) ( ) ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ 0 2 4 4 4 A B C C A B C B A C A B i j i k i k × × = − × × = − ⋅ − ⋅ = − + − + = + K K K K K K K K K K K K 1
- 7. 1.3 2 2 2 2 ( )() (2 )(2 ) (0)(3 ) 5 cos 5 14 5 14 A B a a a a a a AB a a a θ ⋅ + + = = = K K 1 5 cos 53 14 θ − = ≈ ° 1.4 (a) ˆ ˆ ˆ A i j k = + + K : body diagonal K K ˆ ˆ ˆ ˆ ˆ ˆ 3 A A A i i j j k k = ⋅ = ⋅ + ⋅ + ⋅ = (b) ˆ ˆ B i j = + K : face diagonal K 2 B B B = ⋅ = K (c) ˆ ˆ ˆ 1 1 1 1 1 0 i j k B = × = K K K C A (d) 1 1 0 3 2 A B AB θ ⋅ − = = = K K cos 90 θ ∴ = D 1.5 sin B B A C AC θ = = × = K K K sin y B C C A θ ∴ = = cos A C AC u θ ⋅ = = K K cos x u C C A θ ∴ = = 2 x y A B A u B A B C C C A A A A B A × × = + = + × K K K B A K K K K K K 2 2 1 u A B A A = + × A K K K 1.6 ( ) ( ) ( ) 2 3 ˆ ˆ ˆ ˆ ˆ ˆ2 3 d d d i t j t k t i j t k dt dt dt dt 2 t dA α β γ α β + = + + K γ = + 2 2 ˆ ˆ2 6 d A j k t dt β γ = + K 2
- 8. 1.7 ( )() ( )( ) ( )( ) 2 1 2 A B q q q q q = ⋅ = + − + = − + K K 0 3 3 2 0 ( ) , q ( ) 2 1 q q − − = 1 = or 2 1.8 ( ) ( ) 2 2 2 2 A B A B A B A B + = + ⋅ + = + + ⋅ K K K K K K A B K K 2 2 2 2 A B A B AB + = + + K K Since , cos A B AB AB θ ⋅ = ≤ K K A B A B + ≤ + K K K K K K cos cos A B AB A B A B θ θ ⋅ = = ≤ K K K K cos B B θ ≤ 1.9 Show ( ) ( ) ( ) C A C B A B C × × = ⋅ − ⋅ K K K K K K K A B K K or ( ) ( ˆ ˆ ˆ x y z x x y y z z x x y y z z x y z i j k B B A C A C A C B A B A B A B C C C C × = + + − + + K K K ( )ˆ x x y x y z x z x x x y y x z z x C A B C A B C A B C A B C A B C i = + + − − − ( ) ˆ x y y y z y z x x y y y y z z y C A B C A B C A B C A B C A B C j + + + − − − ( ) ˆ x y z y z z z x x z y y z z z z C A B C A B C A B C A B C A B C k + + + − − − ) A B x A B x y A B x z A B ( ) ( ) ( ) ˆ ˆ ˆ y x y z x z y y x z z x x y x z y z x x y z z y x z x y z y x x z y y z A B C A B C A B C A B C i A B C A B C A B C A B C j A B C A B C A B C A B C k + − − = + + − − + + − − 3
- 9. ˆ ˆ ˆ x y yz z y z x x z x y y x i j k A A Az B C B C B C B C B C B C − − − ( ) ( ) ( ) ˆ ˆ ˆ y x y y y x z z x z x z z y z z z y x x y x y x x z x x x z y y z y z y i A B C A B C A B C A B C j A B C A B C A B C A B C k A B C A B C A B C A B C − − + = + − − + + − − + 1.10 y Asinθ = ( ) 1 2 xy y B x xy yB xy AB 2 sinθ = + − = + − = Α A Α = B × K K 1.11 ( ) ˆ ˆ ˆ x y x y z x y x y z x y i j k A A A C A B B B B B B C C C C C C ⋅ × = ⋅ = K K K K z z z A B ( ) x y z x y z x y z B B B A A A B A C C C C = − = − ⋅ K K K × 1.12 x z u B x G C G A G G Let = (Ax,Ay,Az), A K B K = (0,By,0) and C G = (0,Cy,Cz) K Cz is the perpendicular distance between the plane A , B K and its opposite. u B is directed along the x-axis since the vectors xC = G G G B G , C G are in the y,z plane. x = u B = B xC G G yCz is the area of the parallelogram formed by the vectors B G , C G . Multiply that area times the height of plane K , A B K = Ax to get the volume of the parallopiped ( ) x x x y z u A B C A B xC = = = • G G V A G 4
- 10. 1.13 For rotationabout the z axis: i i ˆ ˆ ˆ ˆ cos , j j φ ′ ′ ⋅ = = ⋅ ˆ ˆ 1 k k′ ⋅ = i j ˆ ˆ sinφ ′ ⋅ = − ˆ ˆ sin j i φ ′ ⋅ = For rotation about the axis: y′ i i ˆ ˆ ˆ ˆ cos , k k θ ′ ′ ⋅ = = ⋅ ˆ ˆ 1 j j′ ⋅ = i k̂ ˆ sinθ ′ ⋅ = k i ˆ ˆ sinθ ′ ⋅ = − T cos 0 sin cos sin 0 cos cos cos sin sin 0 1 0 sin cos 0 sin cos 0 sin 0 cos 0 0 1 sin cos sin sin cos θ θ φ φ θ φ θ φ φ φ φ φ θ θ θ θ φ θ − − = − = − I φ θ 1.14 3 ˆ ˆ cos30 2 1 ˆ ˆ sin30 2 ˆ ˆ 0 i i i k ′ ⋅ = = ′ ⋅ = − = − ′ ⋅ = D D i j 1 ˆ ˆ sin30 2 3 ˆ ˆ cos30 2 ˆ ˆ 0 j i j j j k ′ ⋅ = = ′ ⋅ = = ′ ⋅ = D D k j ˆ ˆ 0 ˆ ˆ 0 ˆ ˆ 1 k i k k ′ ⋅ = ′ ⋅ = ′ ⋅ = 3 1 3 0 3 2 2 2 2 1 3 3 0 3 3 1 2 2 2 1 0 0 1 1 x y z A A A ′ ′ ′ + = − = − − − K ˆ ˆ ˆ 3.232 1.598 A i j k ′ ′ ′ = + − 5
- 11. 1.15 1. Rotatethru φ about z-axis 45 φ = D Rφ 2. Rotate thru θ about x’-axis 45 θ = D Rθ 3. Rotate thru ψ about z’-axis 45 ψ = D Rψ 1 1 0 2 2 1 1 0 Rφ = − 2 2 0 0 1 1 0 0 1 1 0 2 2 1 1 0 2 2 Rθ = − 1 1 0 2 2 1 1 0 2 2 0 0 Rψ = − 1 ( ) 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 , , 2 2 2 2 2 2 1 1 2 2 2 R R ψ θ φ ψ θ φ − + = = − − − + − 2 2 1 R R or ( ) 1 0 , , 0 R α ψ θ φ β γ = Condition is: x Rx ′ = K K where x 1 0 0 ′ K = and x α β γ = K Since 1 x x ⋅ = K K we have: 2 2 2 1 ψ β α + + = After a lot of algebra: 1 2 , 2 4 α = − 1 2 , 2 4 β = + 1 2 γ = 1.16 ˆ ˆ v v ct τ τ = = K 2 2 ˆ ˆ ˆ v c a v n c n b τ τ ρ = + = + K 2 ˆ t 6
- 12. at b t = , c ˆ vτ = K bc and ˆ ˆ a c cn τ = + K 2 1 cos 2 2 v a c bc va bc c θ ⋅ = = = K K 45 θ = D 1.17 ( ) ( ) ( ) ˆ ˆ sin 2 cos v t ib t j b t ω ω ω = − + K ω ( ) ( ) 1 1 2 2 2 2 2 2 2 2 2 sin 4 cos 1 3cos v b t b t b t ω ω ω ω ω ω = + = + K K ( ) 2 2 ˆ ˆ cos 2 sin a t ib t j b t ω ω ω = − − ω ( ) 1 2 2 2 1 3sin a b t ω ω = + K at , 0 t = 2 v bω = K ; at 2 t π ω = , v bω = K 1.18 ( ) ˆ ˆ ˆ cos sin 2 v t ib t jb t k ct ω ω ω ω = − + K K ( ) 2 2 ˆ ˆ ˆ sin cos 2 a t ib t jb t k c ω ω ω ω = − − + ( ) ( ) 1 1 2 4 2 2 4 2 2 2 4 2 2 2 sin cos 4 4 a b t b t c b c ω ω ω ω ω = + + = + K 1.19 ˆ ˆ ˆ kt kt r r v re r e bke e bce ê θ θ θ = + = + K K ( ) ( ) ( ) 2 2 2 ˆ ˆ ˆ 2 2 kt kt r r a r r e r r e b k c e e bcke ê θ θ θ θ θ = − + + = − + ( ) ( ) ( ) 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 cos 4 kt kt kt kt b k k c e b c ke v a va be k c be k c c k φ − + ⋅ = = + − + K K ( ) ( ) ( ) ( ) 2 2 1 1 2 2 2 2 2 2 2 2 cos k k c k k c k c k c φ + = = + + + , a constant 1.20 ( ) ( ) ˆ ˆ 2 R z a R R e R R e ze φ φ φ φ = − + + + K K ˆ e 2 ˆ ˆ 2 R z a b e c ω = − + ( ) 1 2 4 2 2 4 a b c ω = + K 7
- 13. 1.21 ( )( ) ˆ ˆ 1 kt kt r t i e je − = − + K K ( ) ˆ ˆ kt kt r t ike jke − = + K ( ) 2 2 ˆ ˆ kt kt r t ik e jk e − = − + 1.22 ˆ ˆ ˆ sin r v e r e r e r φ θ φ θ θ = + + K ( ) ( ) 1 ˆ ˆ sin 1 cos 4 sin 4 2 4 2 v e b t e b t φ θ π π ω ω ω = + − K ω ( ) ( ) ˆ ˆ cos cos 4 sin 4 8 2 v e b t e b t φ θ π π ω ω ω = − K ω 1 2 2 2 2 cos cos4 sin 4 8 4 v b t t π π ω ω = + K ω Path is sinusoidal oscillation about the equator. 1.23 2 v v v ⋅ = K K 2 dv dv v v vv dt dt ⋅ + ⋅ = K K K K 2 2 v a vv ⋅ = K K v a vv ⋅ = K K 8
- 14. 1.24 ( )( ) ( ) d dr d r v a v a r v a dt dt dt ⋅ × = ⋅ × + ⋅ × K K K K K K K K K ( ) dv da v v a r a v dt dt = ⋅ × + ⋅ × + × K K K K K K K K ( ) 0 0 r v a = + ⋅ + × K K K ( ) ( ) d r v a r v a dt ⋅ × = ⋅ × K K K K K K 1.25 ˆ v vτ = K and ˆ ˆ n a a a n ττ = + K v a vaτ ⋅ = K K , so v a a v τ ⋅ = K K 2 2 n a a a τ = + 2 , so ( ) 1 2 2 2 n a a aτ = − 1.26 For 1.14, ( ) 2 3 2 3 2 1 2 2 2 2 2 2 2 2 2 cos sin sin cos 4 cos sin 4 b t t b t t a b t b t c t τ ω ω ω ω ω ω ω ω ω ω − ⋅ + ⋅ + = + + c t ( ) 2 1 2 2 2 2 2 4 4 c t a b c t τ ω = + 1 4 2 2 2 2 2 2 2 2 2 16 4 4 n c t a b c b c t ω ω = + − + For 1.15, ( ) ( ) ( ) 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 kt kt kt kt b k k c e b c ke a b be k c τ − + = = + ke k c + ( ) ( ) ( ) 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 kt kt kt n a b e k c b k e k c bce k c = + − + = + 1.27 ˆ v vτ = K , 2 ˆ ˆ v n τ a v ρ = + K 2 3 n v v v a v a v ρ ρ × = ⋅ = = K K 9
- 15. 1.28 ˆ ˆ sin cos P rib jb θ θ = + D K K ˆ ˆ cos sin rel v ib jb θ θ θ = − K θ ( ) ( ) 2 2 ˆ ˆ cos sin sin cos rel a ib jb θ θ θ θ θ θ θ θ = − − + at the point 2 π θ = , rel v v = − K K So, rel θ = = K v b v v b θ = v a b b θ = = D Now, 2 2 ˆ ˆ ˆ rel rel rel v v n a b τ τ ρ = + = + D K n̂ a v 1 4 2 2 2 rel v a a b = + D K K K K P rel v v v = + K K and P rel a a a = + D K 2 2 2 2 ˆ ˆ cos sin sin cos P a v a v a i a b jb b b b b θ θ θ = + − − + D D D K θ 1 4 2 2 2 2 2 2 2cos sin P v v a a a b a b θ θ = + + − D D D K is a maximum at P a K 0 θ = , i.e., at the top of the wheel. 2 2 2sin cos 0 v a b θ θ − − D = 2 1 tan v a b θ − = − D 1.29 Therefore, 2 2 0 2 0 0 0 0 2 0 0 1 0 0 1 x -x 0 x x x = x x 0 x x x 0 0 1 − = RR 0 1 2 x = The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2) 10
- 16. 1.30 (a) ˆ ˆ cossin θ θ + a = i j φ θ b a ˆ ˆ cos sin ϕ ϕ = + b i j ( ) ( ) ( ) ˆ ˆ ˆ ˆ cos cos sin cos sin θ ϕ θ θ ϕ ⋅ = − = + ⋅ + a b i j i j ϕ ( ) cos cos cos sin sin θ ϕ θ ϕ θ − = + ϕ (b) ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ sin cos sin cos sin θ ϕ θ θ ϕ × = − = + × + b a k i j i j ϕ ( ) sin sin cos cos sin θ ϕ θ ϕ θ − = − ϕ -------------------------------------------------------------------------------------------------------- 11
- 17. CHAPTER 2 NEWTONIAN MECHANICS: RECTILINEARMOTION OF A PARTICLE 2.1 (a) ( ) 1 x F ct m = + ( ) 2 0 1 2 t F c x F ct dt t t m m = + = + ∫ m 2 2 0 2 6 t F F c c 3 x t t dt t m m m m = + = + ∫ t (b) sin F x ct m = ( ) 0 0 sin cos 1 cos t t F F F x ct dt ct ct m cm cm = = − = − ∫ ( ) 0 1 1 cos 1 sin t F F x ct dt ct cm cm c = − = − ∫ (c) ct F x e = m ( ) 0 1 t ct ct F F x e e cm cm = = − ( ) 2 1 1 1 ct ct F F x e t e cm c c c m = − − = − ct − 2.2 (a) dx dx dx dx x x dt dx dt dx ⋅ = = = ( ) 1 dx x F cx dx m = + ( ) 1 xdx F cx dx m = + 2 2 1 1 2 2 cx x F x m = + ( ) 1 2 2 x x F cx m = + (b) 1 cx dx x x F dx m − e = = 1
- 18. 1 cx xdx Fe dx m − = ( ) ( ) 2 1 1 1 2 cx cx F F x e cm cm − − = − − = − e ( ) 1 2 2 1 cx F x e cm − = − (c) ( ) 1 cos dx x x F dx m cx = = cos F xdx cx dx m = 2 1 sin 2 F x cx cm = 1 2 2 sin F x cx cm = 2.3 (a) ( ) ( ) 2 2 x x cx F cx dx F x C = − + = − − + ∫ V x (b) ( ) x cx cx x F F e dx e C c − − = − = + ∫ V x (c) ( ) cos sin x x F F cx dx cx C c = − = − + ∫ V x 2.4 (a) ( ) ( ) dV x kx dx = − = − F x ( ) 2 0 1 2 x kx dx kx = = ∫ V x (b) T T ( ) ( ) x V x = + ( ) ( ) ( ) 2 1 2 T V x k A x = − = − T x (c) 2 1 2 E T kA = = (d) turning points @ T x ( ) 1 0 → 1 x A ∴ = ± 2.5 (a) ( ) 3 2 kx F x kx A − + so ( ) 3 4 2 2 2 0 1 1 2 4 x kx kx kx dx kx A A = − = − ∫ V x (b) ( ) ( ) 4 2 2 1 1 2 4 kx T V x T kx A = − = − + T x (c) E T = 2
- 19. (d) V hasmaximum at ( ) x ( ) 0 m F x → 3 2 0 m m kx A − = kx m x A = ± ( ) 4 2 2 2 1 1 1 2 4 4 m kA x kA kA A = − = V If turning points exist. ( m E V x ) Turning points @ T x let u ( ) 1 0 → 2 1 x = 2 2 1 1 0 2 4 ku E ku A − + = solving for u , we obtain 1 2 2 2 4 1 1 E kA = ± − u A or 1 2 1 2 4 1 1 E x A kA = ± − − 2.6 ( ) x v x x α = = 2 2 3 x x x x α α = − = − ( ) 2 3 m F x mx x α = = − 2.7 sin F Mg θ ≥ 2.8 dx F mx mx dx = = 3 x bx− = 4 3 dx bx dx − = − ( )( 3 4 3 F m bx bx − − = − ) 2 7 3 F mb x− = − 2.9 (a) ( ) ( ) 2 .145 9.8 1250 .3048 541 m m gx kg ft J s ft = = = V m 3
- 20. (b) 2 2 2 2 2 1 1 11 2 2 2 2 .22 t mg m g v mv m c D = = = = T m ( ) ( )( ) ( )( ) 2 2 2 .145 9.8 87 2 .22 2 .0366 m kg s kg m = = T J 3 2 3 tanh t t Fdx cv dx c v dt c v dt τ = − = − = − − ∫ ∫ ∫ ∫ 3 2 1 tanh tanh 2 t t t cv d τ t τ τ τ + ∫ = − 3 2 1 tanh ln cosh 2 t t t cv τ τ τ + = − Now 2 tanh 1 t τ ≅ for t τ Meanwhile tanh ln cosh t t t t x vdt v dt vτ τ τ = = − = ∫ ∫ ln cosh t t x v τ τ = ( ) 1250 .3048 381 m x ft m ft = = ( ) ( )( ) 1 2 1 2 2 2 2 .145 9.8 34.72 .22 .0732 t m kg mg m s v kg c s m = = = ( ) ( )( ) 1 2 1 2 2 2 2 .145 3.543 .22 .0732 9.8 kg m s kg m c g m s τ = = = ( )( ) ( ) ( ) ( )( ) 2 3 3.81 .22 .0732 34.72 3.543 .5 454 34.72 3.54 Fdx J = − + ∫ = 541 87 454 V T J J J − = − = 2.10 For 0 : 1 t t ≤ ≤ F v t m = , 2 1 2 F x t m = For 1 1 2 : t t t ≤ ≤ 1 F v t m = , 2 1 2 F x t m = , 1 t t = 4
- 21. ( ) () 2 2 1 1 1 1 2 2 2 F F F 1 x t t t t t t m m m = + − + − At t 1 2 : t = 2 2 2 1 1 1 5 2 2 F F F F 2 1 x t t t m m m m = + + = t 2.11 3 2 dv dv dx dv c a v dt dx dt dx m = = ⋅ = ⋅ = − v 1 2 c v dx m − = − v d max 1 2 0 v x v c v dv dx m − = − ∫ ∫ 1 2 max 2 c v x m − = − 1 2 max 2mv x c = 2.12 Going up: sin30 cos30 x F mg mg µ = − − ( ) 2 sin30 0.1cos30 5.749 m x g s = − + = − v v at = + at the highest point so 0 v = 0.174 up v a = − = t v s 2 2 2 1 0.174 .087 0.087 2 up up up 2 x v t at v v v m = + = − = Going down: 2 0.087 x v ′ = , v 0 ′ = , ( ) 9.8 0.5 0.0866 ′ = − − a 2 2 1 0 0.087 4.0513 2 down down x v t − = = 0.207 down t v = s s 0.381 total up down t t t v = + = 2.13 At the top so 0 v = max 2 2 kx g k e g v k − = + Coming down max x x = and at the bottom 0 x = ( ) 2 2 2 2 2 1 1 g v g g k v g g k k v v k k = − = + + 5
- 22. ( ) 1 2 22 t t v v v v v = + , 2 t g m v k c = = g 2.14 Going up: 2 2 x F mg c v = − − 2 dv a v g kv dx = = − − , 2 c m = k 2 0 v x v vdv dx g kv = − − ∫ ∫ ( ) 2 1 ln 2 v v g kv x k − − − = 2 2 2 kx g kv e g kv − + = + 2 2 2kx g g v v e k k − = + − Going down: 2 2 x F mg c = − + v 2 dv v g dx = − + kv 2 0 0 v x vdv dx g kv = − + ∫ ∫ ( ) 2 0 1 ln 2 v g kv x x k − + = − 2 2 2 kx kx k v e e g − − = 1 2 2 2 kx kx g g v e k k − = − e 2.15 2 1 2 dv m mg c v c dt = − − v 2 0 0 1 2 t v dt dv m mg c v c v = − − ∫ ∫ Using 2 2 2 2 1 2 4 ln 4 2 4 dx cx b b ac a bx cx b ac cx b b a + − − = + + − + + − ∫ c , 6
- 23. 2 2 1 12 2 2 1 2 2 1 1 0 2 4 1 ln 4 2 4 v c v c c mgc t m c mgc c v c c mgc − − − + = + − − + + 2 ( ) ( )( ) ( )( ) 2 2 1 2 1 1 2 1 1 2 2 2 1 2 2 2 2 1 1 2 1 1 2 2 4 4 4 ln 2 4 4 c v c c mgc c c mgc t c mgc m c v c c mgc c c mgc + + + − + + = + − + + + as t , → ∞ 2 2 1 1 2 t c v c c mgc + − + = 2 4 0 1 2 2 1 1 2 2 2 2 2 t c c mg v c c c = − + + Alternatively, when v , t v = 2 1 2 0 t t dv m mg c v dt = = − − c v 1 2 2 1 1 2 2 2 2 2 t c c mg v c c c = − + + 2.16 2 dv k a v x dx m − = = − 2 0 v x b kdx vdv mx = − ∫ ∫ 2 1 1 2 k v m x b = − 1 1 1 2 2 2 1 1 2 dx k k b x v dt m x b mb x − = = − = 1 2 1 1 3 2 2 0 0 0 1 2 2 1 t b x mb x mb x b dt dx d x k b x k b b = = − − ∫ ∫ ∫ Since x b ≤ , say 2 sin x b θ = ( ) 1 1 3 3 2 2 0 0 2 2 2 sin 2sin cos 2 sin 2 cos d mb mb t d k k π π θ θ θ θ θ θ θ − − = = ∫ ∫ 1 3 2 8 mb t k π = 7
- 24. 2.17 ( )( ) dv dv m mv f x g dt dx = = i v ( ) ( ) mvdv f x dx g v = By integration, get ( ) dx v v x dt = = If F x : ( ) ( ) ( ,t f x g t = i ) ( ) ( ) 2 2 d x d dx m m f x g dt dt dt = = i t This cannot, in general, be solved by integration. If ( ) ( ) ( ) ,t f v g t = i F v : ( ) ( ) dv m f v g dt = i t ( ) ( ) mdv g t dt f v = Integration gives v v( ) t = ( ) dx v t dt = ( ) dx v t dt = A second integration gives ( ) x x t = 2.18 ( )( ) 4 2 6 1 1.55 10 10 1.55 10 kg c s − − − = × = × ( )( ) 2 2 5 2 0.22 10 2.2 10 kg c s − − = = × ( )( ) 1 2 7 2 6 6 5 5 10 9.8 1.55 10 1.55 10 2 2.2 10 2 2.2 10 2.2 10 t v − − − − − × × = − + + × × × × × 5 − 0.179 t m v s = Using equation 2.29, ( )( ) 7 5 10 9.8 0.211 2.2 10 t m v s − − = = × 2.19 or ( ) x F x Ae mx α = − = ( ) v F v Ae mv α = − = v dv A dt e m α = − Let u v eα = v du e dv α α = v du du dv e u α α α = = 2 du A dt u m α ∴ = − 8
- 25. Integrating 1 1 A t uu m α − = and substituting v e u α = (a) 1 ln 1 v A e t m α v v α α = − + (b) t T @ 0 v = = ln 1 v A v e m α T α α = + 1 v v A e e m α α T α = + 1 v m T e A α α − = − (c) v dv A e dx m v v α = = − v vdv A dx e m α = − again, let u v eα = du udv α = or du dv u α = 1 ln v u α = 1 ln du u A u dx u m α α = − Integrating and solving ( ) 2 1 1 v m x v A α α α − = − + e 2.20 ( ) d mv F mv vm dt = = + = mg but 3 4 3 m r ρ π = 2 1 m r ρ π = v so (1) 3 2 2 2 1 4 4 3 3 r v r v r g πρ πρ π πρ + = = 3 4 3 Now 3 1 10 ρ ρ − ≈ so, second term is negligible-small hence v and g ≈ v gt ≈ speed t ∝ but or 2 2 1 4 m r r ρ π ρ π = = r v 1 1 4 r v ρ ρ ≅ Hence 1 1 4 r ρ ρ ≈ gt and rate of growth t ∝ The exact differential equation from (1) above is: 2 1 1 1 4 4 4 4 3 3 r r r r ρ ρ g πρ πρ π ρ ρ + = ρ 9
- 26. which reduces to: 2 1 3 4 r rg r ρ ρ + = Using Mathcad, solve the above non-linear d.e. letting 3 1 10 ρ ρ − ≈ and 0.01 R mm ≈ (small raindrop). Graphs show that and v r ∝ ∝ t 2 r t ∝ 10
- 27. CHAPTER 3 OSCILLATIONS 3.1 () [ ] 1 0.002sin 2 512 x s t m π − = ( )( )( ) max 0.002 2 512 6.43 m m x s s π = = ( )( ) ( ) 2 2 4 max 2 2 0.002 2 512 2.07 10 m m x s s π = = × 3.2 0.1sin x t ω = [m] 0.1 cos m x t s ω ω = When t = 0, x = 0 and 0.5 0.1 m x s ω = = 1 5s ω − = 2 1.26 T s π ω = = 3.3 ( ) cos sin x x t x t t ω ω ω = + and 2 f ω π = ( ) ( )[ ] 0.25cos 20 0.00159sin 20 x t t π π = + m 3.4 ( ) cos cos cos sin sin α β α β α − = + β ( ) cos cos cos sin sin x A t A t A t ω φ φ ω φ = − = + ω cos sin x t t ω ω = Α + Β , cos A φ Α = , sin A φ Β = 3.5 2 2 2 2 1 1 2 1 1 1 1 2 2 2 2 mx kx mx kx + = + 2 ( ) ( ) 2 2 2 2 1 2 2 1 k x x m x x − = − 1 2 2 2 2 1 2 2 1 2 k x x m x x ω − = = − 2 2 1 1 1 1 1 2 2 2 kA mx kx = + 2 2 2 2 2 2 2 2 1 1 2 1 1 1 2 2 2 1 m x x x x A x x k x x − = + = + − 2 1 x 1 2 2 2 2 2 1 2 2 1 2 2 2 1 x x x x A x x − = − 1
- 28. 3.6 1 1 2.5 9.8 2 6 l T s g ππ = = ≈ s 3.7 For springs tied in parallel: ( ) ( ) 1 2 1 2 s F x k x k x k k x = − − = − + ( ) 1 2 1 2 k k m ω + = For springs tied in series: The upward force is m eq k x . Therefore, the downward force on spring is 2 k eq k x . The upward force on the spring is 2 k 1 k x′ where x′ is the displacement of P, the point at which the springs are tied. Since the spring is in equilibrium, 2 k 1 eq k x k ′ = x . Meanwhile, The upward force at P is 1 k x′ . The downward force at P is ( ) 2 k x x′ − . Therefore, ( ) 1 2 k x k x x ′ ′ = − 2 1 2 k x x k k ′ = + And eq k 2 1 1 2 k x x k k k = + ( ) 1 2 1 2 1 2 eq k k k m k k m ω = = + 3.8 For the system ( ) M m + , ( ) kX M m X − = + The position and acceleration of m are the same as for ( ) M m + : m m k x x M m = − + cos cos m k k x A t d t M m M δ = + = + + m r The total force on m, m m F mx mg F = = − 2
- 29. cos r m mk mkdk F mg x mg t M m M m M = + = + + + m + For the block to just begin to leave the bottom of the box at the top of the vertical oscillations, 0 r F = at m x d = − : 0 mkd mg M m = − + ( ) g M m d k + = 3.9 ( ) cos t d x e A t γ ω φ − = − ( ) ( ) sin cos t t d d d dx e A t e A t dt γ γ ω ω φ γ ω φ − − = − − − − maxima at ( ) ( ) 0 sin cos d d d t t dt dx ω ω φ γ ω φ − + − = = ( ) tan d d t γ ω φ ω − = − thus the condition of relative maximum occurs every time that t increases by 2 d π ω : 1 2 i i d t t π ω + = + For the i th maximum: ( ) cos i t i d x e A t γ i ω φ − = − ( ) 1 2 1 1 cos i d t i d i i x e A t e π γ γ ω ω φ + − − + + = − x = 2 1 d d T i i x e e x π γ ω γ − + = = 3.10 (a) 1 3 2 c s m γ − = = 2 2 25 k s m ω − = = 2 2 2 16 d s ω ω γ 2 − = − = 2 2 2 7 r d s ω ω γ 2 − = − = 1 7 r s ω − ∴ = (b) max 48 0.2 60.4 d F Cω = = = A m m (c) ( ) 2 2 2 2 2 n 2 3 r r r r γω γω ω φ γ γ ω ω = = = − 7 = ta 41.4 φ ∴ ≈ 3
- 30. 3.11 (a) 2 17 30 2 mx mx mx β β + + = 3 2 γ β = and 2 2 17 2 ω β = 2 2 2 2 4 r 2 ω ω γ β = − = 2 r ω β ∴ = (b) max 2 d F m A γω = 2 2 2 25 4 d 2 ω ω γ β = − = 5 2 d ω β ∴ = 2 2 15 A β = 3.12 1 2 d T e γ − = 1 ln 2 ln 2 d d f T γ = = (a) 1 2 2 2 ( ) d ω ω γ = − So, 1 2 2 2 ( ) d ω ω γ = + 1 1 2 2 2 2 2 ln 2 1 2 2 d d f f f γ π π = + = + 100.6 f Hz = (b) ( ) 1 2 2 2 r d ω ω γ = − 1 1 2 2 2 2 2 ln 2 1 2 2 r d d f f f γ π π = − = − 99.4 r f Hz = 3.13 Since the amplitude diminishes by e d T γ − in each complete period, ( ) 1 1 d n T e e e γ − − = = 1 d T n γ = 1 2 d d T n n ω γ π = = Now ( ) 1 2 2 2 d ω ω γ = − So ( ) 1 1 2 2 2 2 2 2 1 1 4 d d n γ ω π = + = + ω ω 4
- 31. 1 2 2 2 2 1 1 2 4 dd d T T n π ω ω π ω π ω = = = + For large n, 2 2 1 1 8 d T n π T ≈ + 3.14 (a) ( ) 1 2 1 2 2 2 2 2 2 2 0.707 2 ω r ω ω γ ω = − = − = ω (b) ( ) 1 2 1 2 2 2 1 1 4 0.866 2 2 2 2 d ω ω γ ω ω γ γ − − = = Q = = (c) ( ) 2 2 2 2 2 2 2 2 2 n 4 3 ω ω γω ω ω ω ω = − − ta φ = = − 1 2 tan 146.3 3 φ − = − = (d) ( ) ( ) ( ) 1 2 2 2 2 2 2 4 4 4 3.606 2 ω 2 D ω ω ω ω ω = − + = ( ) ( ) 2 0.277 F F m A D m ω ω ω = = 3.15 ( ) ( ) max 1 2 2 2 A A γ ω ω ω γ ≈ − + for ( ) max 1 2 ω = A A , ( ) 1 2 2 2 1 2 γ ω ω γ = − + ( ) 2 2 2 4 ω ω γ γ − + = 3 ω ω γ − = ± 3 ω ω γ = ± 5
- 32. 3.16 (b) 2 2 22 d Q ω γ ω γ γ − = = 2 1 LC ω = , 2 R L γ = 2 2 2 1 4 1 4 2 2 R LC L L R R C L − = = Q − (c) 2 L R C Q R R ω γ = = = 3.17 sin Im i t ext F F t F eω ω = = and ( ) x t is the imaginary part of the solution to: i t mx cx kx F eω + + = i.e. x t ( ) ( ) ( ) Im sin i t Ae A t ω φ ω φ − = = − where, as derived in the text, ( ) 1 2 2 2 2 2 F A k m c ω ω = − + and 2 2 2 tan γω φ ω ω = − 3.18 Using the hint, , where ( ) Re t ext F F β = e i β α ω = − + , and x(t) is the real part of the solution to: mx t cx kx F eβ + + = . Assuming a solution of the form: t i x Aeβ φ − = ( ) 2 i F m c k x xe A φ β β + + = ( ) 2 2 2 c F m im m c ic k i A os sin α αω ω α ω φ φ − − − + + = + ( ) 2 2 cos F m c k A α ω α − − + = φ ( ) 2 s F m c A in ω α φ − + = 6
- 33. ( ) ( ) 1 22 2 tan c m m c ω α φ α ω α − − = − − + k Using sin , 2 2 cos 1 φ φ + = ( ) ( ) 2 2 2 2 2 2 2 2 F m c k c A m α ω α ω α = − − + + − ( ) ( ) { } 1 2 2 2 2 2 2 2 F A m c k c m α ω α ω α = − − + + − and x t ( ) ( ) cos t Ae t α ω φ − = − + the transient term. 3.19 (a) 1 2 2 2 1 8 l A T g π − ≈ − for 4 A π = , 2 1.041 l T g π ≈ × (b) 2 2 4 1.084 l g T π = × Using 2 l g π = T gives 2 2 4 l g T π = , approximately 8% too small. (c) 3 2 32 A B λ ω = − and 2 6 ω λ = 2 192 B A A = for 4 A π = , 0.0032 B A = 3.20 ( ) in t n n f t c e ω = ∑ . . . 0, 1, 2, n = ± ± ( ) cos sin n n n n f t c n t c i n t ω ω = + ∑ ∑ , n 0, 1, 2, = ± ± . . . and ( ) 2 2 1 T in t T n t e T ω − − = ∫ c f , dt 0, 1, 2, n = ± ± . . . ( ) ( ) ( ) ( ) 2 2 2 2 1 cos sin T T T T n i c f t n t dt f t n T T ω ω − − = − ∫ ∫ t dt The first term on is the same for and n c n n − ; the second term changes sign for n vs. . The same holds true for the trigonometric terms in n − ( ) f t . Therefore, when terms that cancel in the summations are discarded: 7
- 34. ( ) () ( ) 2 2 1 cos cos T T n f t c f t n t dt n T t ω ω − = + ∑ ∫ ( ) ( ) 2 2 1 sin sin T T n f t n t dt n T t ω ω − ∑ ∫ + , . . ., and 1, n = ± ±2, ( ) 2 2 1 T T T − = ∫ c f t dt Now, due to the equality of terms in n ± : ( ) ( ) ( ) 2 2 2 cos cos T T n f t c f t n t dt n t T ω ω − = + ∑ ∫ ( ) ( ) 2 2 2 sin sin T T n f t n t dt n T t ω ω − ∑ ∫ + , . . . 1,2 n = ,3, Equations 3.9.9 and 3.9.10 follow directly. 3.21 ( ) in t n n f t c e ω = ∑ , ( ) 2 2 1 T in t T n c f t e T ω − − = ∫ dt , and 0, 1, 2, n = ± ± … 2 T π ω = so ( ) 2 in t n c f t e π ω ω π ω ω π − − = ∫ dt ( ) 0 0 2 in t in t e dt e d π ω ω ω π ω ω π − − − t = − + ∫ ∫ 0 0 1 1 2 in t in t e e in in π ω ω ω π ω ω π ω ω − − − = − 1 1 1 2 in in e e in π π π + − = − − + For n even, e e and the term in brackets is zero. 1 in in π π − = = For n odd, 1 in in e e π π − = = − 4 2 n c in π = , . . . 1, 3, n = ± ± ( ) 4 2 in t n f t e in ω π = ∑ , . . . 1, 3, n = ± ± ( ) 4 1 1 2 in t in t n e e n i ω ω π − = − ∑ , 1,3,5, n = . . . ( ) 4 1 sin n n t n ω π = ∑ , 1,3,5, n = . . . ( ) 4 1 1 sin sin3 sin5 3 5 f t t t t ω ω ω π = + + + … 8
- 35. 3.22 In steadystate, ( ) ( ) n i n t n n A e x t ω φ − = ∑ ( ) 1 2 2 2 2 2 2 2 2 4 n n F m A n n ω ω γ ω = − + Now 4 n F n F π = , . . . and 1,3,5, n = 3 ω ω = 100 2 Q ω γ = ≈ so 2 2 9 40,000 ω γ ≈ ( ) 1 1 2 2 2 2 2 2 4 1 9 9 4 40000 F A mπ ω ω ω ω = ⋅ ⋅ − + 1 2 2 F A mπω ≈ ( ) 3 1 2 2 2 2 2 2 4 1 3 9 9 9 4 200 F A mπ ω ω ω = ⋅ − + 3 2 400 27 F A mπω ≈ ( ) ( ) 5 1 2 2 2 2 2 2 4 1 5 3 9 25 4 5 200 F A mπ ω ω ω ω = ⋅ − + 5 2 20 F A mπω ≈ i.e., A : A : 1 : 29.6 : 0.1 1 3 5 A = 3.23 (a) Thus 2 0 x x ω + = y x = 2 y x ω = − x y = divide these two equations: 2 y dy x x dx y ω = = − (b) Solving 2 0 ydy xdx ω + = and Integrating 2 2 2 2 2 y x C ω + = Let 2 2 C A = 2 2 2 2 2 1 y x A A ω + = an ellipse → 9
- 36. 3.24 The equationof motion is ( ) 3 F x x x mx = − = . For simplicity, let m=1. Then 3 x x x = − . This is equivalent to the two first order equations … x y = and 3 y x x = − (a) The equilibrium points are defined by ( )( ) 3 1 1 x x x x x − = − + = 0 Thus, the points are: (-1,0), (0,0) and (+1,0). We can tell whether or not the points represent stable or unstable points of equilibrium by examining the phase space plots in the neighborhood of the equilibrium points. We’ll do this in part (c). (b) The energy can be found by integrating 3 dy y x x dx x y − = = or or ( ) 3 y dy x x dx C = − + ∫ ∫ 2 2 4 2 2 4 y x x C = − + ` In other words … 2 4 2 2 4 2 y x x E T V C = + = + − = . The total energy C is constant. (c) The phase space trajectories are given by solutions to the above equation 1 4 2 2 2 2 x y x C = ± − + . The upper right quadrant of the trajectories is shown in the figure below. The trajectories are symmetrically disposed about the x and y axes. They form closed paths for energies C0 about the two points (-1,0) and (+1,0). Thus, these are points of stable equilibrium for small excursions away from these points. The trajectory passes thru the point (0,0) for C=0 and is a saddle point. Trajectories never pass thru the point (0,0) for positive energies C0. Thus, (0,0) is a point of unstable equilibrium. 0 0.5 1 1.5 2 0 0.5 1 10
- 37. 3.25 sin 0 θθ + = ⇒ 2 cos 0 2 d dt θ θ − = Integrating: 2 0 cos 2 θ θ θ θ θ = or ( ) 2 2 cos cos θ θ θ = − ( ) 1 0 2 4 2 cos cos d T θ θ θ θ ∴ = − ∫ Time for pendulum to swing from 0 θ = to θ θ = is 4 T Now—substitute sin 2 sin 2 sin θ φ θ = so 2 π φ = at θ θ = and use the identity 2 cos 1 2sin 2 θ θ = − 1 0 2 2 2 4 4 sin sin 2 2 d T θ θ θ θ ∴ = − ∫ and after some algebra … 1 1 2 2 2 2 2 1 sin 4 sin sin 2 2 2 d d φ θ θ θ θ = − − or (a) 2 1 2 0 2 4 1 sin d T π φ α φ = − ∫ where 2 sin 2 θ α = (b) ( ) 1 2 2 2 2 1 3 1 sin 1 sin sin 2 8 α φ α φ α φ − − ≈ + + … 4 + 2 2 2 4 0 1 3 4 1 sin sin 2 8 T d π φ α φ α φ = + + + ∫ … 2 9 2 1 4 64 T α π α = + + + … (c) 2 3 2 2 sin 2 2 48 4 θ θ θ θ α = ≈ − + … ∼ . . . 2 2 1 16 T θ π = + + … ------------------------------------------------------------------------------------------------------ 11
- 38. CHAPTER 4 GENERAL MOTIONOF A PARTICLE IN THREE DIMENSIONS --------------------------------------------------------------------------------------------------------- Note to instructors … there is a typo in equation 4.3.14. The range of the projectile is … 2 0 sin 2 v R x g α = = … NOT 2 2 0 sin 2 v g ... α --------------------------------------------------------------------------------------------------------- 4.1 (a) ˆ ˆ ˆ V V F V i j k V x y z ∂ ∂ ∂ = −∇ = − − − ∂ ∂ ∂ ( ) ˆ ˆ ˆ F c iyz jxz kxy = − + + (b) ˆ ˆ ˆ 2 2 2 F V i x j y k z α β γ = −∇ = − − − (c) ( ) ( ) ˆ ˆ ˆ x y z F V ce i j k α β γ α β γ − + + = −∇ = + + (d) 1 1 ˆ ˆ ˆ sin r V V F V e e e r r r θ φ V θ θ φ ∂ ∂ = −∇ = − − − ∂ ∂ ∂ ∂ 1 ˆ n r F e cnr − = − 4.2 (a) ˆ ˆ ˆ 0 i j k F x y z x y z ∂ ∂ ∂ ∇× = = ∂ ∂ ∂ conservative (b) ( ) 2 ˆ ˆ ˆ ˆ 1 1 0 i j k F k x y z y x z ∂ ∂ ∂ ∇× = = − − ≠ ∂ ∂ ∂ − conservative (c) ( ) 3 ˆ ˆ ˆ ˆ 1 1 0 i j k F k x y z y x z ∂ ∂ ∂ ∇× = = − = ∂ ∂ ∂ conservative (d) 1
- 39. 2 ˆ ˆ ˆsin 1 0 sin 0 0 r n e e r e r F r r kr θ φ θ θ θ φ − ∂ ∂ ∂ ∇× = = ∂ ∂ ∂ − conservative 4.3 (a) ( ) 2 3 ˆ ˆ ˆ 2 i j k F k x y z xy cx z ∂ ∂ ∂ ∇× = = − ∂ ∂ ∂ cx x 2 0 cx x − = 1 2 c = (b) 2 ˆ ˆ ˆ i j k F x y z z cxz x y y y ∂ ∂ ∂ = ∂ ∂ ∂ ∇× 2 2 2 2 1 1 ˆ ˆ ˆ x cx cz z i j k y y y y y y − + − + + = − 2 2 0 x cx y y = − − c 1 = − also 2 2 0 z y y = cz + implies that 1 c = − as it must 4.4 (a) constant E = ( ) 2 1 , , 2 V x y z mv = + at the origin 2 1 0 2 = + E mv at ( ) 1,1,1 2 2 1 1 2 2 E mv α β γ = + + + = mv 2
- 40. ( ) 2 22 m v v α β γ = − + + ( ) 1 2 2 2 m α β γ v v = − + + (b) ( ) 2 2 0 m α β γ + = v − + ( ) 1 2 2 m α β γ = + + v (c) x V mx F x ∂ = = − ∂ mx α = − 2 V y y my β ∂ = − = − ∂ 2 3 V z z mz γ ∂ = − = − ∂ 4.5 (a) ˆ ˆ F ix jy = + on the path x y = : ˆ ˆ dr idx jdy = + ( ) ( ) 1,1 1 1 1 1 0,0 0 0 0 0 1 x y F dr F dx F dy xdx ydy ⋅ = + = + = ∫ ∫ ∫ ∫ ∫ on the path along the x-axis: ˆ dr idx = and on the line 1 x = : ˆ dr jdy = ( ) ( ) 1,1 1 1 0,0 0 0 1 x y F dr F dx F dy ⋅ = + = ∫ ∫ ∫ is conservative. F (b) ˆ ˆ F iy jx = − on the path x y = : ( ) ( ) 1,1 1 1 1 1 0,0 0 0 0 0 x y F dr F dx F dy ydx xdy ⋅ = + = − ∫ ∫ ∫ ∫ ∫ and, with x y = ( ) ( ) 1,1 1 1 0,0 0 0 0 F dr xdx ydy ⋅ = − ∫ ∫ ∫ = on the x-axis: ( ) ( ) 1,0 1 1 0,0 0 0 x F dr F dx ydx ⋅ = = ∫ ∫ ∫ and, with on the x-axis 0 y = ( ) ( ) 1,0 0,0 0 F dr ⋅ = ∫ on the line 1 x = : ( ) ( ) 1,1 1 1 1,0 0 0 y F dr F dy xdy ⋅ = = ∫ ∫ ∫ 3
- 41. and, with 1 x= ( ) ( ) 1,1 1 1,0 0 1 F dr dy ⋅ = = ∫ ∫ on this path ( ) ( ) 1,1 0,0 0 1 1 F dr ⋅ = + = ∫ is not conservative. F 4.6 From Example 2.3.2, ( ) ( ) 2 e e r mg r z = − V z + ( ) 1 1 e e z mgr r V z − = − + From Appendix D, ( ) 1 2 1 1 x x x − + = − + +… ( ) 2 2 1 e e e z z mgr r r = − − + + … V z ( ) 2 e e mgz mgr mgz r = − + − +… V z With − an additive constant, e mgr ( ) 1 e z mgz r ≈ − V z ( ) ˆ F V k V z ∂ = −∇ = − ∂ z 1 ˆ 1 e e z kmg z r r = − − + − 2 ˆ 1 e z F kmg r = − − , 0 x mx F = = 0 y my F = = 2 1 e z mz mg r = − − 2 1 e dz z mz mg dz r = − − 0 0 2 1 z h v e z zdz g dz r = − − ∫ ∫ 2 2 1 2 z e h v g h r − = − − 2 2 0 2 e z e r v h g − + = h r 4
- 42. 2 2 2 1 2 2 ee e r r h r g = − − z v ( ) , e h z r 2 2 1 2 2 e e z e r r v gr = − − h From Appendix D, ( ) 2 1 2 1 1 2 8 x x x + = + − +… 2 4 2 2 2 4 e e z z e r r v v h g gr = − + + +… 2 2 1 2 2 z z e v v h g gr ≈ + From Example 2.3.2, 2 2g v h = 1 2 1 2 e v gr − − And with ( ) 1 1 1 x x − − ≈ + , 2 2 1 2 2 e v v h g gr ≈ + 4.7 For a point on the rim measured from the center of the wheel: ˆ ˆ cos sin r ib jb θ θ = − v t t b θ ω = = , so r i ˆ ˆ sin cos v jv θ θ = − − Relative to the ground, ( ) ˆ ˆ 1 sin cos v jv v i θ θ = − − For a particle of mud leaving the rim: sin y b θ = − and v v cos y θ = − So v v cos y y gt v gt θ = − = − − and 2 1 sin cos 2 v t gt θ θ = − − − y b At maximum height, 0 vy = : cos v g t θ = − 2 cos 1 cos sin cos 2 v v v g g g h b θ θ θ θ = − − − − − 2 2 cos sin 2 v h b g θ θ = − + Maximum h occurs for 2 2 cos sin 0 cos 2 dh v b d g θ θ θ θ = = − − 5
- 43. 2 sin gb v θ = − 42 2 2 4 cos 1 sin v g b v θ θ − = − = 2 2 4 2 2 2 2 max 2 2 2 2 2 gb v g b gb v h v gv v − = + = + 2g Measured from the ground, 2 2 max 2 2 2 gb v h b v g ′ = + + The mud leaves the wheel at 1 2 sin gb v θ − = − 4.8 cos x R φ = and ( ) cos x x v t v t α = = so co R t s cos v φ α = sin y R φ = and ( ) 2 2 1 1 sin 2 2 y v t gt v t g α = − = − y t α φ ( ) 2 cos 1 cos sin sin cos 2 cos R R R v g v v φ φ φ α α α = − 2 2 2 cos sin tan cos 2 cos gR v φ φ α φ α = − ( ) ( ) 2 2 2 2 2 2 cos 2 cos tan cos sin sin cos cos sin cos cos v v R g g α α α φ φ α φ α φ φ = − = − φ From Appendix B, sin( ) sin cos cos sin θ φ θ φ θ + = + φ ( ) 2 2 2 cos sin cos v R g α α φ φ = − R is a maximum for ( ) ( 2 2 2 0 sin sin cos cos cos dR v d g ) α α φ α α φ α φ = = − − + − Implies that ( ) ( ) cos sin sin 0 α α φ α α φ − − − = cos From appendix B, ( ) cos cos cos sin sin θ φ θ φ θ + = − φ so ( ) cos 2 0 α φ − = 2 2 π α φ − = 4 2 π φ α = + 2 max 2 2 cos sin cos 4 2 4 2 v R g π φ π φ = + φ − 6
- 44. Now sin coscos 4 2 2 4 2 4 2 π φ π π φ π − = − − = + φ 2 2 max 2 2 cos cos 4 2 v R g π φ φ = + Again using Appendix B, 2 2 2 cos2 cos sin 2cos 1 θ θ θ θ = − = − 2 max 2 2 1 1 cos cos 2 2 2 v R g π φ φ = + + = 2 2 cos 1 cos 2 v g π φ φ + + Using cos sin 2 π θ θ + = − , ( ) ( ) 2 max 2 1 sin 1 sin v R g φ φ = − − ( ) 2 max 1 sin v R g φ = + 4.9 (a) Here we note that the projectile is launched “downhill” towards the target, which is located a distance h below the cannon along a line at an angle φ below the horizon. α is the angle of projection that yields maximum range, Rmax. We can use the results from problem 4.8 for this problem. We simply have to replace the angle φ in the above result with the angle -φ, to account for the downhill slope. Thus, we get for the downhill range … φ α Rmax h ( ) 2 0 2 cos sin 2 cos v R g α α ϕ ϕ + = The maximum range and the angle is α are obtained from the problem above again by replacing φ with the angle -φ … ( ) 2 0 max 2 1 sin cos v R g ϕ ϕ + = and … 2 2 π α ϕ = − . We can now calculate α … ( ) ( ) 2 2 0 0 max 2 1 sin sin cos 1 sin v v h R g g ϕ ϕ ϕ ϕ + = = = − Solving for sinϕ … 2 2 0 0 1 gh gh v v ϕ = + sin But, from the above … 2 in sin 2 cos2 1 2sin 2 π s ϕ α α = − = = − α Thus … 2 2 2 0 0 sin 1 gh gh v v α − = + 1 2 7
- 45. 2 2 2 2 00 2 0 1 2 2sin 1 1 csc 1 gh gh gh v v v α α = = − + = + Finally … 2 2 0 2 1 gh v α = + csc (b) Solving for Rmax … max 2 2 sin 1 2sin 1 2 csc h h h R ϕ α α = = = − − Substituting for 2 csc α and solving … 2 0 max 2 0 1 v gh R g v = + 4.10 We can again use the results of problem 4.8. The maximum slope range from problem 4.8 is given by … ( ) 2 0 max 1 sin sin v h R g ϕ ϕ = = + Solving for sinϕ … 2 2 0 0 sin 1 gh gh v v ϕ = − Thus … max max cos cos sin x R h ϕ ϕ ϕ = = We can calculate cosϕ from the above relation for sinϕ … ( ) 1 1 2 2 2 2 2 0 0 cos 1 sin 1 2 1 gh gh v v ϕ ϕ = − = − − Inserting the results for sinϕ and cosϕ into the above … 1 2 2 0 max 2 0 cos 1 2 sin v gh x h g v ϕ ϕ = = − 4.11 We can simplify this problems somewhat by noting that the trajectory is symmetric about a vertical line that passes through the highest point of the trajectory. Thus we have the following picture … 8
- 46. α v0 x z zmax h1 h0 x0 δ x1 R We have “reversedthe trajectory so that h0 (= 9.8 ft), and x0 , the height and range within which Mickey can catch the ball represent the starting point of the trajectory. h1 (=3.28 ft) is the height of the ball when Mickey strikes it at home plate. δ is the distance behind home plate where the ball would be hypothetically launched at some angle α to achieve the total range R. x1 (=328 ft) is the distance the ball actually would travel from home plate if not caught by Mickey. (Note, because of the symmetry, v0 is the speed of the ball when it strikes the ground … also at the same angle α at which was launched. We will calculate the value of x0 assuming a time-reversed trajectory!) (1) The range of the ball … 2 2 0 0 sin 2 2 sin cos v v R g g α α α = = (2) The maximum height … 2 max 2 2 0 tan 2 2 cos 2 R g R z v α α = − (3) The height at x1 … ( ) 2 1 1 1 2 2 0 tan 2 cos g h x x v α α = − From (1) … 2 2 0 tan 2 cos g v R α α = and inserting this into (2) gives … max tan tan tan 2 4 4 R R R z α α α = − = Thus, max 4 tan z R α = and inserting this expression and the first previously derived into (3) (4) ( ) 2 1 1 1 max tan tan 4 x h x z α α = − Let u x1 tanα = and we obtain the following quadratic … and solving for u … 2 max max 1 4 4 u z u z h − + 0 = ( ) 1 2 max 1 max 2 1 1 u z h z = ± − and letting 1 max h z ε = , we get … 9
- 47. max 1 u zh ε ≈ = ax or . This result is the correct one … Thus, ( ) ( ) max max max 2 2 2 2 .0475 3.9 u z z z ε ≈ − = − = 0 m 1 3.9 tan 0.821 39.4 z x α α = = ∴ = Now solve for x0 using a relation identical to (4) … 0 0 max tan 4 h x z α = − ( ) 2 0 tan x α A es gain we obtain a quadratic expr sion for 0 tan u x α = which we solve as before. This time, though, the first result for u is the correct one to use … max 0 u z h ε = ≈ and we obtain … 0 11.9 h x ft = = 0 tanα .12 The 4 x and positions of the ball vs. time are z 2 2 1 cos x v t θ = 2 1 1 cos sin z v t gt θ θ = − Since x v 2 1 cos 2 v θ = The ho i rizontal range s 2 2 1 cos sin 2 2 v R g θ θ = The maximum range occurs @ 0 dR dθ = 2 1 2cos cos v dR θ = 2 1 1 2 cos sin sin 2 0 2 2 2 d g θ θ θ θ θ − Thus, 2 = 2 1 1 1 cos cos2 cos sin sin 2 2 2 2 θ θ θ θ = We get θ Using the identities: 2 2cos 1 cos2 θ θ = + and sin 2 2sin cos θ θ θ = ( )( 1 cos 2 cos θ θ θ + − : ) ( ) 2 2 cos 1 sin sin 1 cos cos θ θ θ = = − θ − = or (1 cos + )( ) 2 3cos cos 1 0 θ θ θ − Thus cos 1 θ = − , ( ) 1 cos 1 13 6 θ = ± Only the s po itive root applies for the θ -range: 0 2 π θ ≤ ≤ ( ) cos 1 13 0.7676 6 θ = + = θ 1 39 51′ = 1 25 v m s− = max 55.4 @ 39 51 R m θ ′ = = Thus (b) for 10
- 48. (a) The maximumheight occurs at 0 dz dt = cos sin 2 v gT θ θ or at 1 1 cos sin 2 v T g θ = dH θ = or 2 2 2 1 cos sin 2 2 v H g θ θ = maximum at fixed θ 1 dH The maximum possible height occurs @ 0 dα = Using the above trigonometric identities, 2 2 2 1 1 2cos sin cos cos sin sin 0 2 2 2 2 v d g θ θ θ θ θ θ α = − = we get ( ) ( ) 2 2 1 1 1 cos sin cos sin sin sin 1 cos 2 2 θ θ θ θ θ θ θ + = = There ar − e 3-roots: or ( )( ) sin 1 cos 3cos 1 0 θ θ θ + − = sin 0 θ = , cos 1 θ = − , 1 cos 3 θ = Thus, max 18.9 os H The first two roots give minimum heights; the last gives the maximum 1 1 @ c 70 32 3 m θ − ′ = = = z r s 4 The trajectory of the shell is given by Eq. 4.3.11 with r replacing .13 x 2 2 2 z g r r = − co r v θ = sin z v θ = r where Thus, 2 2 tan sec g r z r θ θ = − 2 2v Since 2 2 sec 1 tan θ θ = + We have: 2 2 2 2 2 tan tan 0 2 2 g r g r r z v v θ θ − + + = rdinates. The above equation yields two possible roots: (r,z) are target coo ( ) 1 2 4 2 2 2 2 1 tan 2 v v gzv g r θ = ± − − gr ≥ The critical The roots are only real if 4 2 2 2 2 0 v gzv g r − − 4 2 2 2 surface is therefore: 0 v gzv g r − − = 2 4.14 If the velocity vector, of magnitude , makes an angle s θ with the z-axis, and its 11
- 49. projection on thexy-plane make an angle φ cos , and cos z r 2 2 2 2 with the x-axis: φ sin cos x s θ = , and x sin cos x r F F m θ φ = = Since ( F x z + sin sin y s θ φ = , and y sin sin y r F F m θ φ = = separabl z s θ = F mg F mz θ = − + = ) 2 2 r c s c y = − = − + 2 c sx 2 dx c s , the differential equations of motion are not e. θ φ y x z 2 2 2 sin cos mx c s c sx θ φ = − = − = dx d dx m m ms dt ds dt ds = ⋅ = = − x ds ds ds x m γ = − = − 2 c m γ , where ln ln ln x x x s x γ − = = − s x x e γ − = 4.15 From eqn 4.3.1 Similarly s y y e γ − = 6, max max 2 2 ln 1 0 x x z g g x x γ γ γ γ γ + + − max x γ γ = From Appendix D: ( ) 2 3 ln 1 2 3 u u u u − = − − − −… for 1 u max max m z x gx gx 2 2 2 3 3 max max max 2 3 ln 1 2 3 x x x x x x x γ γ − = − − − 4 γ + terms in 2 3 ax max max 2 3 2 3 gx g x x x x x x γ γ γ + − − − 0 γ = + terms in 2 2 max max 3 3 0 2 x x z x x g γ γ + − ≈ 1 2 2 2 max 2 3 9 3 4 16 x x x z x g γ γ γ ≈ − ± + 1 2 max 3 3 16 1 4 4 3 x x z x g γ γ γ ≈ − ± + γ Since , the + sign is used. From Appendix D: max 0 x 8 g g 1 2 2 16 8 16 1 1 1 3 3 3 z z z g γ γ + = + − 3 γ + terms in 3 3 x x x = − + + 2 max 2 2 8 4 4 3 x z x z g g γ γ γ − + terms in 2 γ 12
- 50. 2 max 2 2 8 3 xz x z x g g γ = − +… For sin z v α = and 2 2 2 sin x z v α = ) α y x : 2 3 max 2 sin 2 4 sin 2 sin 3 v v x g g α α α γ = − … + 4.16 ( cos x A t ω = + , ( ) sin x A t ω ω α = − + my from 0 x = , 0 α = 2 from x A = , cos x A t ω = with , ω ( ) cos y B t ω β = + , ( ) sin y B t ω ω β = − + 2 2 1 1 1 2 2 kB ky = + 2 4 y A = 3 y A ω = and k m = : 2 2 2 1 16 B ω = + 5 B A = Then 4 5 cos A A β = ( ) 2 2 9 25 A A ω = 2 A and 3 5 sin A A ω ω β = − Since m respectiv in 1 1 4 3 cos sin 36.9 5 5 β − − = = − = − rectangl aximum x and y displacements are ( ) 5 cos 36.9 y A t ω = − A ± and 5A ± 2 2 A B − , ely, the motion takes place entirely with a e of dime and . From eqn 4.4.15, nsion 2A 10A ( )( ) ( ) 2 5 cos 6.9 A A 36.9 0 36.9 β α ∆ = − = − − = − tan 2 4 ψ = = 2 cos tan 2 AB ψ ∆ = ψ = .17 ( ) 2 2 10 3 1 5 2 3 5 A A − − − 4 4 = − 1 1 1 tan 9.2 2 3 − − = − co x A + 2 x V mx F kx mx x π ∂ = = − = − = − ∂ ( ) s cos k t A t m α π α + = = 13
- 51. 2 4 V my mx y π ∂ = −= − ∂ ( ) cos 2 y B t π β = + S t = 2 9 V mz mz z π ∂ = − = − ∂ ( ) cos 3 z C t π γ = + ince at , 0 x y z = = = 0 2 π α β γ = = = − cos sin 2 x A t A π t π π = − = cos x A t π π = Since 2 and 2 2 2 v x y z = + + x y z = = , 3 v x Aπ = = t 3 v A π = sin 3 v x π π = t sin 2 y B π = , t 2 cos2 y Bπ π = 2 3 v y B π = = π 2 3 v B π = sin 2 2 3 v y t π = sin3 z C t π = , t 3 cos3 z Cπ π = 3 3 v z Cπ = = π 3 3 v C π = Since x ω π = , sin3 3 3 v z t π = min t = 2 y ω π = , and 3 z ω π = the ball does retrace its path. 3 1 2 2 2 2 x y n n n π π π ω ω ω = = z ime occurs at 1 1 n = , 2 2 n = , 3 3 n = . The minimum t 14
- 52. min 2 2 t π π = = 4.18 Equation4.4.15 is 2 2 2 cos tan 2 AB A B ψ ∆ = − the z-axis through a Transforming the coordinate axes xyz to the new axes x y z ′ ′ ′ ψ given, from or, cos sin x x y ψ ψ ′ ′ = − , and sin co y x y ψ ′ ′ = + From eqn. 4.4.10: 2 2 2cos sin x y y ∆ − + = ∆ by a rotation about n angle Section 1.8: cos sin x x y ψ ψ ′ = + , sin cos y x y ψ ψ ′ = − + sψ Substitu 2 2 x A AB B 2 A 2 ting: ( ) 2 2 2 2 1 cos 2 cos sin sin x x y y ψ ψ ψ ψ ′ ′ ′ ′ − + 2 2 2cos x ′ − ( ) 2 2 cos sin cos sin cos sin x y y AB ψ ψ ψ ψ ψ ∆ ′ ′ ′ + − − For x′ t ψ ( ) 2 2 2 2 2 2 1 sin 2 cos sin cos sin x x y y B ψ ψ ψ ψ ′ ′ ′ ′ + + + = 2cos − ∆ o be a major or minor axis of the ellipse, the coefficient of x y ′ ′ A AB B 2 2 must vanish. ( ) 2 2 2 2 sin 2cos 2cos sin cos sin 0 ψ ψ ψ ψ ψ ψ ∆ − − + B, 2cos sin sin 2 ψ ψ ψ = and cos sin cos2 ψ ψ ψ − = 1 tan 2ψ − = = From Appendix 2 2 sin 2 2cos cos2 sin 2 0 A AB B ψ ψ ψ ∆ − − + = 2 2 1 2cos B A AB ∆ 2 2 2 cos tan 2 AB A B ψ ∆ = − within a 4.19 Shown below is a face-centered cubic lattice. Each atom in the lattice is centered cube on whose 6 faces lies another adjacent atom. Thus each atom is surrounded by 6 nearest neighbors at a distance d. We neglect the influence of atoms that lie at further distances. Thus, the potential energy of the central atom 2d can be approximated as 6 1 i i V cr α − = = ∑ 15
- 53. ( ) 1 2 22 2 1 r d x y z = − + + ( ) 2 2 2 2 2 2 2 2 2 1 2 2 2 1 x x y z r d dx x y z d d d α α α α − − − − + + = − + + + − + 2 2 = From Appendix D, ( ) ( ) 2 1 1 1 1 2 n x nx n n x + = + + − +… 2 2 d 2 1 2 1 1 1 2 2 x x y z r d d α α α α − − + + = − − + + − − − 2 2 2 2 2 2 2 2 2 2 2 2 2 x x x z x y z d d d d + + + + − − + + term y s in 3 3 x d 2 α ( ) 2 2 2 2 1 2 2 4 1 1 2 4 2 x x r d x y z d d d α α α α α α − − = + − + + + + + terms in 3 3 x d ( ) 2 2 2 2 1 2 2 1 1 2 2 x r d x y z x d d d α α α α α α − − ≈ + − + + + + ( ) 1 1 2 2 2 2 2 2 2 2 2 2 2 r d x y z d dx x y z = − − + + = + + + + 2 2 2 2 2 2 2 1 x x y z r d d d α α α − − − + + = + + 2 ( ) 2 2 2 2 2 2 2 1 1 2 2 x r d x y z x d d d α α α α α α − − ≈ − − + + + + ( ) ( ) 2 2 2 1 2 2 2 2 2 r r d x y z x d d α α α α α α − − − + ≈ − + + + + Similarly: ( ) ( ) 2 2 2 2 3 4 2 2 2 2 r r d x y z y d d α α α α α α − − − + ≈ − + + + + 5 6 2 r r d ( ) ( ) 2 2 2 2 2 2 x y z z d d α α α α α − − + + + + α − − 2 + ≈ ( ) ( ) 2 2 2 2 2 2 2 2 2 2 3 2 6 V cd x y z x y z d d d α α α α − ≈ − + + + + + + ( )( ) 2 2 2 2 2 6cd cd x y z α α α α − − − ≈ + − + + 4.20 ( ) 2 2 2 V A B x y z ≈ + + + 16
- 54. h x 17 ( ) F qE v B = + × ( ) ˆ ˆ ˆ ˆ ˆ ˆ v B ix jy kz kB iyB jxB × = + + × = − ( ) ˆ ˆ F iqyB jq E xB = + − x mx F qyB = = qB x x y m − = y y qB my F qE qxB qE qB x m = = − = − + y k̂B ĵE x z 0 ˆ iv 2 2 qE qBx qB eE eBx eB y y m m m m m m = − − = − + − y 2 eE y y m x ω ω + = − + , eB m ω = ( ) 2 1 cos eE y x A m t ω ω θ ω = − + + + ( ) sin y A t ω ω θ = − + 0 y = , so 0 θ = 0 y = , so 2 1 eE A x m ω ω = − − + ( ) 1 cos y a t ω = − , 2 1 eE a x m ω ω = − + ( ) 1 cos qB x x y x y x a m t ω ω ω = + = − = − − ( ) cos x x a a t ω ω ω = − + ( ) sin x x a t a t ω ω = − + sin x a t bt ω = + , b x a ω = − 0 z mz F = = 4 0 z z t z = + = .21 2 1 2 2 b mv mqh mg + = sθ + y b ( ) 2 2 v g b h = − 2 co r mv F mg b = − = − R
- 55. cos h b θ = ( )( ) 2 2 3 h mv mg mg R mg h b h h b b b b b = − = − − = − the particle leaves the side of the sphere when 0 R = 3 3 b h = , i.e., b above the central plane 4.22 2 1 0 2 mv mgh + = h b = − 2 at the bottom of the loop, so 2 1 mv mgb = v = 18 , 2g + b 2 r mv F mg R b = − = 2 2 3 mv R mg mg mg mg b = + = + = the equation for the energy as a function of s in Example 4.6.2, 4.23 From 1 1 mg v h b 2 2 E ms = + s is und 2 2 4A s , ergoing harmonic motion with: 1 k g ω = = = 4 2 m A A g Since 4 sin s A φ = , φ increases by 2π radians during the time interval: 2 2 g π 2 A T π ω ′ = = For cycl z oidal motion, x 2φ cycloidal motion is one-half the period for s . and are functions of so they undergo a complete cycle every time φ changes by π 1 A π ′ = = . Therefore, the period for the 2 2 T T g ------------------------------------------------------------------------------------------------------
- 56. CHAPTER 5 NONINERTIAL REFERENCESYSTEMS 5.1 (a) The non-inertial observer believes that he is in equilibrium and that the net force acting on him is zero. The scale exerts an upward force, N G , whose value is equal to the scale reading --- the “weight,” W’, of the observer in the accelerated frame. Thus 0 0 N mg mA + − = G G G 0 A G N G mg G (b) (a) 0 5 0 4 4 g N mg mA N mg m N mg − − = − − = − = 5 5 4 4 W N mg W ′ = = = 150 . W lb ′ = (b) The acceleration is downward, in the same direction as g G 0 4 g N mg m − + = 3 4 4 W W ′ = − = W W 90 . W l ′ = b r 5.2 (a) ( ) cent F mω ω ′ = − × × G G G G G G G For r ω ′ ⊥ , 2 ˆ cent r F m r ω e ′ ′ = 1 1 500 1000 s s ω π − − = = ( ) 2 6 2 ˆ 10 1000 5 5 cent r F e π π − = × × = G dynes outward (b) ( ) 2 2 4 1000 5 5.04 10 980 cent g F m r F mg π ω ′ = = = × 5.3 (See Figure 5.1.2) 0 mg T mA + − = D G G G ˆ ˆ ˆ ˆ cos sin 0 10 g mg j T j T i m i θ θ − + + − = cos T mg θ = , and sin 10 mg θ = T 1 tan 10 θ = , 5.71 θ = D 1.005 cos mg T m θ = = g 5.4 The non-inertial observer thinks that g′ G points downward in the direction of the hanging plumb bob… Thus 1
- 57. ˆ ˆ 10 g g gA g j ′ = − = − D G G G i For small oscillations of a simple pendulum: 1 2 g π = ′ T 2 2 1.005 10 g g g ′ = + = g 1 1 2 1.995 1.005g g π π = = T 5.5 (a) f mg µ = − is the frictional force acting on the b x, so G G o 0 f mA ma′ − = G ( a′ G is the acceleration of the box relative to the truck. See Equation 5.1.4b.) Now, f G the only real force acting horizontally, so the acceration relative to the road is 0 A G f G (b) 3 f mg g a g m m µ µ = = − = − = − (For + in the direction of the moving truck, the – indicates that friction opposes the forward sliding of the box.) 2 g A = − D (The truck is decelerating.) from above, 0 ma mA ma′ − = so (a) 3 2 6 g g g a a A ′ = − = − + = D 5.6 (a) ( ) ˆ ˆ cos sin r i x R t jR = + Ω + D K K t Ω t 2 ˆ ˆ sin cos r i R t j R = − Ω Ω + Ω Ω K K 2 2 r r v R ⋅ = = Ω v R ∴ = Ω circular motion of radius R (b) r r where r ω ′ ′ = − × K K K K ˆ ˆ r ix jy ′ ′ = + K ′ ( ) ˆ ˆ ˆ ˆ sin cos i R t j R t k ix jy ω ˆ ′ ′ = − Ω Ω + Ω Ω − × + 2
- 58. ˆ ˆ ˆ sincos i R t j R t j x i y ˆ ω ω ′ ′ = − Ω Ω + Ω Ω − + sin x y R t ω ′ ′ = − Ω Ω cos y x R t ω ′ ′ = − + Ω Ω (c)Let here u x iy ′ ′ = + ′ 1 i = − ! sin cos u x iy y R t i x i R t ω ω ′ ′ ′ ′ ′ = + = −Ω Ω − + Ω Ω i u ω ′ = y ω ′ = − i x ω ′ + sin cos Rei t u i u R t i R t i ω Ω ′ ′ ∴ + = −Ω Ω + Ω Ω = Ω Try a solution of the form i t i t u Ae Be ω − Ω ′ = + i t i t u i Ae i Be ω ω − Ω ′ = − + Ω i t i t i u i Ae i Be ω ω ω ω − Ω ′ = + ( ) i t u i u i Beω ω ω ′ ′ ∴ + = + Ω so R B ω Ω = + Ω Also at t the coordinate systems coincide so 0 = ( ) ( ) 0 0 u A B x iy x R ′ ′ ′ = + = + = + D R A x R B x R ω Ω ∴ = + − = + − + Ω D D so, R A x ω ω = + + Ω D Thus, i t i t R R u x e e ω ω ω ω − Ω Ω ′ = + + + Ω + Ω D 5.7 The x, y frame of reference is attached to the Earth, but the x-axis always points away from the Sun. Thus, it rotates once every year relative to the fixed stars. The X,Y frame of reference is fixed inertial frame attached to the Sun. (a) In the x, y rotating frame of reference ( ) ( ) cos x t R t Rε ω = Ω − − ( ) ( ) sin y t R t ω = − Ω − where R is the radius of the asteroid’s orbit and RE is the radius of the Earth’s orbit. Ω is the angular frequency of the Earth’s revolution about the Sun and ω is the angular frequency of the asteroid’s orbit. (b) ( ) ( ) ( ) sin 0 x t R t ω ω = − Ω − Ω − → at 0 t = ( ) ( ) ( ) ( ) cos y t R t R ω ω = − Ω − Ω − Ω − → − ω at 0 t = (c) 2 a A A r r r ε = − −Ω× − Ω× − Ω×Ω× K K K K K K K K K K K Where is the acceleration of the asteroid in the x, y frame of reference, a 3
- 59. , A Aε K K arethe accelerations of the asteroid and the Earth in the fixed, inertial frame of reference. 1st : examine: A A r ε − − Ω×Ω× K K K K K K K K K K ( ) R R R R ε ε ω ω = × × −Ω×Ω× −Ω×Ω× − K K K K K K K K K K K =( ) ( ) 2 2 R R ω ω ω × − Ω×Ω × = − − Ω note: k̂ ω ω = K , k̂ Ω = Ω K Thus: ( ) 2 2 2 a R ω = Ω − − Ω× K K K K v Therefore: ( ) ( ) ( ) ( ) 2 2 ˆ ˆ ˆ ˆ cos sin ix jy iR t jR t ω ω + = Ω − Ω − − Ω − ω ˆ ˆ 2 2 j x i y − Ω + Ω Thus: ( ) ( ) 2 2 cos 2 x R t ω ω = Ω − Ω − + Ω y x ) ( ) ( ) 2 2 sin 2 y R t ω ω = − Ω − Ω − − Ω Let ( x y ω = Ω − and ( ) y x ω = Ω − Then, we have ( ) ( ) ( ) 2 cos y R t ω ω ω y Ω = Ω − Ω − + Ω − which reduces to ( ) ( ) cos y R t ω ω = − Ω − Ω − Integrating … ( ) sin 0 y R t ω = − Ω − → at t 0 = Also, ( ) ( ) ( ) 2 2 sin 2 x R t ω ω ω − Ω − = − Ω − Ω − − Ω x or ( ) ( ) sin 2 x R t ω ω = Ω + Ω − + Ω x ) ( ) ( sin x R t ω ω = − Ω − Ω − Integrating … ( ) cos x R t const ω = Ω − + ( ) cos x R t R R R ε ε ω − − → − = Ω at 0 t = 5.8 Relative to a reference frame fixed to the turntable the cockroach travels at a constant speed v’ in a circle. Thus 2 ˆr v a e b ′ ′ ′ = − G . Since the center of the turntable is fixed. G 0 A = D The angular velocity, ω, of the turntable is constant, so 4 ω G x′ y′ b
- 60. k̂ ω ω ′ = G ,with 0 ω = G G G G r , so ˆr be ′ ′ = ( ) 2 ˆr r b ω ω ω e ′ ′ × × = − G G G G v v êθ′ ′ ′ = , so ˆr v v e ω ω ′ ′ ′ × = − G G G From eqn 5.2.14, 2 ( ) a a v r ω ω ω ′ ′ = + × + × × G G G G and putting in terms from above ′ 2 2 2 r v a v b b ω ω ′ ′ ′ = − − − For no slipping s F m µ ≤ G g , so s a g µ ≤ G 2 2 2 s v v b g b ω ω µ ′ ′ + + ≤ v b 2 2 2 2 0 m m s v b b g ω ω µ ′ ′ + + − = 2 2 2 2 m s v b b b b g ω ω ω µ ′ = − ± − + Since v was defined positive, the +square root is used. ′ m s b v b g ω µ ′ = − + (b) v v êθ′ ′ ′ = − G G G ˆr v v e ω ω ′ ′ ′ × = + 2 2 2 r v b a v b ω ω ′ ′ ′ = − + − 2 2 2 s v v b g b ω ω µ ′ ′ − + ≤ m s b v b g ω µ ′ = + 5.9 As in Example 5.2.2, ˆ V j ω ρ ′ = D G and 2 ˆ V A i ρ ′ = D D G For the point at the front of the wheel: 2 ˆ V b ′ ′ = D G G r j and v V k̂ ′ ′ = − D G 0 ω = ( ) ˆ ˆ ˆ V V r k bj b i ω ρ ρ ′ ′ ′ × − = D D ′ × = G G ( ) 2 ˆ ˆ ˆ V V b V b r k i ω ω ρ ρ ρ ′ ′ ′ × × = × = D D D G G G j′ ( ) ˆ ˆ 0 V v k V k ω ρ ′ ′ ′ × = × − = D D G G ( ) 2 2 2 2 ˆ ˆ V V V b r A i j b ω ω ρ ρ ′ ′ ′ = + × × + = + + D D D D a r ′ G G G G G G 5
- 61. 5.10 (See Example5.3.3) 2 m x mx ω ′ ′ = ( ) t t x t Ae Be ω ω − ′ = + ( ) t t x t Ae Be ω ω ω ω − ′ = − Boundary Conditions: ( ) 2 0 2 l x A B = = + ( ) ( ) 0 0 x A B ω ′ = = − 4 l A ∴ = 4 l B = (a) ( ) cosh 2 l x t t ω ′ = ( ) sinh 2 l x t t ω ω ′ = (b) ( ) cosh 2 2 2 l l l x T T ω ′ = + = when the bead reaches the end of the rod cosh 2 T ω ∴ = or 1 1 1 cosh 2 T .317 ω ω − = = (c) ( ) sinh 2 l x T T ω ω ′ = ( ) 1 sinh cosh 2 1.732 0.866 2 2 l l l ω ω ω − = = = or 1 2 2 cosh 1 3 0.866 2 2 T l ω l l ω ω ω − = = 5.11 mph ˆ 400 v j ′ ′ = G ˆ 586.67 j′ = ft⋅s 1 − G ( ) 5 ˆ ˆ 7.27 10 cos41 sin 41 j k ω − ′ ′ = × + D D s 1 − G G ( )( )( ) 5 ˆ 7.27 10 586.67 sin 41 v i ω − ′ ′ × = − × D ft⋅s 2 − G G 2 cor grav m v F F mg ω ′ − × = ( )( )( ) 5 2 7.27 10 586.67 sin 41 32 − × = D 0.0017 cor grav F F = The Coriolis force is in the v ω ′ − × G G direction, i.e., ˆ i′ + or east. 5.12 (See Figure 5.4.3) ˆ ˆ y z j k ω ω ω ′ ′ ′ ′ = + G G ˆ ˆ x y v v i v j ′ ′ ′ ′ = + ′ 6
- 62. ˆ ˆ ˆ z yz x y x v v i v j v ω ω ω ω k ′ ′ ′ ′ ′ ′ ′ ′ × = − + − G G G G ′ ′ v j′ ( ) ˆ ˆ z y z x horiz v v i ω ω ω ′ ′ ′ ′ ′ ′ ′ × = − + ( ) ( ) ( ) 1 1 2 2 2 2 2 2 2 2 z y z x z x y z horiz v v v v v ω ω ω ω ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ × = + = + = G G G G G v ω v 2 cor F mω ′ = − × G ( ) ( ) 2 cor z horiz horiz F m v m 2 v ω ω ′ = × = G G , independent of the direction of v . ′ G ′ ′ 5.13 From Example 5.4.1 … 1 3 2 1 8 cos 3 h h x g ω λ ′ = and 0 h y′ = . ( ) 1 3 3 2 5 1 2 1 8 1250 7.27 10 cos41 3 32 h ft x s ft s − − − × ′ = × ⋅ D 0.404 h x ft ′ = to the east. 5.14 From Example 5.4.2: 2 sin H v ω λ ∆ ≈ D is the deflection of the baseball towards the south since it was struck due East at Yankee Stadium at latitude (problem 5.13). v 41 N λ = D 0 is the initial speed of the baseball whose range is H. From eqn 4.3.18b, without air resistance in an inertial reference frame, the horizontal range is … 2 sin 2 v H g α = D Solving for v0 … 1 2 sin 2 gH v α = D 1 2 2 1 32 200 113 sin30 ft s ft − v ft s− ⋅ × = = D D ⋅ ( )( ) 5 1 2 2 1 7.27 10 200 sin 41 0.0169 0.2 113 s ft ft i ft s − − − × n ∴ ∆ ≈ = = ⋅ D A deflection of 0.2 inches should not cause the outfielder any difficulty. 5.15 Equation 5.2.10 gives the relationship between the time derivative of any vector in a fixed and rotating frame of reference. Thus … fixed rot da da r a dt dt ω = = + G G G G × G G G G G G G G ( ) 2 a r r r r ω ω ω ω ′ ′ ′ = + × + × + × × G G ′ 7
- 63. 2 2 rot da r rr r dt ω ω ω ω ′ ′ ′ ′ = + × + × + × + × G G G G G G G G G r′ G ( ) ( ) ( ) r r r ω ω ω ω ω ω ′ ′ ′ × + × × + × × G G G G G G G G G + × ( ) ( ) ( ) 2 a r r r r ω ω ω ω ω ω ω ω ω ′ ′ ′ × = × + × × + × × + × × × ′ G G G G G G G G G G G G G G G G G G Now ( ) is ⊥ to r ω ′ × ω and r′ . Let this define a direction : n̂ G G G G ˆ r r ω ω ′ ′ × = × G G n Since n̂ ω ⊥ , is in the plane defined by ( r ω ω ′ × × G ) ω G and r′ G and G G G G ˆ ( ) r n r ω ω ω ω ω ω ′ ′ × × = × × = × G G G G G G G G . r′ Since … ( ) r ω ω ω ′ ⊥ × × G G G G ( ) 2 r r ω ω ω ω ω ′ ′ × × × = × G G G G G G And is in the direction ( r ω ω ω ′ × × × ) n̂ − G G G G Thus ( ) ( ) 2 r r ω ω ω ω ω ′ ′ × × × = − × G G G G G G G G G ( ) ( ) ( ) 2 2 a r r r ω ω ω ω ω ω ω ω ′ ′ ′ × = × + × × + × × − × G G G G G G G G G G G G G G G r′ ( ) 3 3 r r r r r r ω ω ω ω ω ′ ′ ′ ′ = + × + × + × + × × G G G G G G G G ′ ( ) ( ) ( ) 2 2 3 r r ω ω ω ω ω ω ′ ′ + × × + × × − × r′ 5.16 With x y , and 0 z x y ′ ′ ′ ′ ′ = = = = = D D D D D z v ′ ′ = D D Equations 5.4.15a – 5.4.15c become: ( ) 3 2 1 cos cos 3 x t gt t v ω λ ω λ ′ ′ = − D ( ) 0 y t ′ = ( ) 2 1 2 z t gt v t ′ ′ = − + D When the bullet hits the ground ( ) 0 z t ′ = , so 2v t g ′ = D 3 2 3 2 8 4 1 cos cos 3 v v x g v g g ω λ ω λ ′ ′ ′ ′ = − D D D 3 2 4 cos 3 v x g ω λ ′ ′ = − D x′ is negative and therefore is the distance the bullet lands to the west. 5.17 With x y and 0 z ′ ′ ′ = = = D D D cos x v α ′ = D D 0 y′ = D sin z v α ′ = D D we can solve equation 5.4.15c to find the time it takes the projectile to strike the ground … 8
- 64. ( ) 22 1 sin cos cos 0 2 z t gt v t v t α ω α λ ′ ′ = − + + = D D or 2 sin 2 sin 2 cos cos v v t g v g α α ω α λ ′ ′ = ≈ ′ − D D D We have ignored the second term in the denominator—since v′ D would have to be impossibly large for the value of that term to approach the magnitude g For example, for 41 45 o o and λ α = = 2 cos cos g v g v ω α λ ω ′ ′ − ≈ − D D or 144 g k s ω ′ ≈ ≈ D m v ! Substituting t into equation 5.4.15b to find the lateral deflection gives ( ) [ ] 3 2 2 2 4 cos sin sin sin cos v y t v t g ω ω α λ λ α ′ = − = − D D α 5.18 Let … aD G = acceleration of object relative to Earth G ωD = k̂ ωD AD = its angular speed R G 0 R G r G y x G = acceleration of satellite k̂ ω ω = G = its angular speed ( ) 2 a a v r A ω ω ω = + × + × × + D D G G G G G G G G (Equation 5.2.14) G G G G G G G G 2 a a A v r ω ω ω = − − × − × × D D As in prob em 5.7 Evaluate the term G G G G l … ( ) a A r R R R R ω ω ω ω ω ω ω ω = − − × × = × × − × × − × × − D D D D D D G G G G G G G G G G G G ( ) 2 2 a R ω ω ∆ = − D D G G a ∆ … given the condition that 2 3 2 3 R R ω ω = D D 3 2 3 1 a R R R ω ∆ = − − D D G G G G G G but ( ) ( ) 2 2 2 cos R R R r R r R r rR θ ⋅ = + ⋅ + = + + D D G G Letting x cosθ = 2 2 2 2 2 1 x R R R r Rx R R ⋅ = + + ≈ + D D G G for small r and 3 2 3 3 2 1 x R R R ≈ + D or 3 3 2 3 2 1 R x R R − = + D 3 2 2 2 2 ˆ 1 1 3 3 a R x 2 R x R R ω ω − ∆ = − − + ≈ − ≈ − D D xi ω G G G for small r 9
- 65. Hence: ( ) 2ˆ ˆ ˆ 2 3 2 a xi k ix ω ω ω = ∆ − × = − − × + ĵy a v G G G G G a i 2 ˆ ˆ ˆ ˆ 3 2 2 x jy xi yi xĵ ω ω ω = + = − + − So x y 2 2 3 x ω ω − − = 0 2 0 y x ω + = 5.19 ( ) mr qE q v B = + × G G G G Equation 5.2.14 ( ) 2 r v r ω ω ω ω r r′ ′ ′ = + × + × + × × G G G G G G G G G ′ G G G G Equation 5.2.13 v v r ω ′ ′ = + × 2 q B m ω = − G G so 0 ω = G ( ) ( ) ( ) 2 q mr q B v B r qE q v r B ω ω ′ ′ ′ ′ ′ − × − × × = + + × × G G G G G G G G G G G ( ) ( ) ( ) ( ) 2 q mr q v B r B qE q v B q r B ω ω ′ ′ ′ ′ ′ + × + × × = + × + × × G G G G G G G G G G G G ( ) 2 q mr qE r B ω ′ ′ = + × × G G G G G ( ) ( )( )( ) 2 sin 2 2 2 q q qB r B r B B m ω θ ′ ′ × × = ∝ G G G G G Neglecting terms in 2 B , mr qE ′ = 5.20 For cos sin x x t y t ω ω ′ ′ = + ′ t sin cos y x t y ω ω ′ ′ = − + ′ cos sin sin cos x x t x t y t y t ω ω ω ω ω ω ′ ′ ′ ′ ′ ′ = − + + ′ t sin cos cos sin y x t x t y t y ω ω ω ω ω ω ′ ′ ′ ′ ′ ′ = − − + − ′ cos sin x x t y t y ω ω ω ′ ′ ′ = + + ′ ′ x sin cos y x t y t ω ω ω ′ ′ ′ = − + − ′ ′ ′ ′ ′ ′ cos sin sin cos x x t x t y t y t y ω ω ω ω ω ω ω ′ ′ ′ = − + + + ′ ′ ′ ′ ′ ′ ′ ′ ′ x sin cos cos sin y x t x t y t y t ω ω ω ω ω ω ω = − − + − − 2 ′ ′ cos sin 2 x x t y t y x ω ω ω ω ′ ′ ′ ′ ′ = + + + 2 ′ ′ y ′ ′ sin cos 2 y x t y t x ω ω ω ω ′ ′ ′ ′ ′ = − + − + Substituting into Eqns 5.6.3: 2 cos sin 2 x t y t y x ω ω ω ω ′ ′ ′ ′ + + + ′ ′ cos sin 2 g g x t y t l l y ω ω ω ′ ′ = − − + ′ ′ 2 sin cos 2 x t y t x y ω ω ω ω ′ ′ ′ ′ + − + ′ ′ sin cos 2 g g x t y t l l x ω ω ω ′ ′ = + − − ′ ′ 10
- 66. Collecting terms andneglecting terms in 2 ω′ : cos sin 0 g g x x t y y t l l ω ω ′ ′ + + + = sin cos 0 g g x x t y y t l l ω ω ′ ′ + − + = 5.21 24 sin T λ = hours 24 73.7 sin19 T = = D hours 5.22 Choose a coordinate system with the origin at the center of the wheel, the x′ and axes pointing toward fixed points on the rim of the wheel, and the axis pointing toward the center of curvature of the track. Take the initial position of the y′ z′ x′ axis to be horizontal in the V − D G direction, so the initial position of the y′ axis is vertical. The bicycle wheel is rotating with angular velocity V about its axis, so … b D ˆ l V k b ω ′ = D G A unit vector in the vertical direction is: ˆ ˆ ˆ sin cos V t V t n i j b b ′ ′ = + D D At the instant a point on the rim of the wheel reaches its highest point: ˆ ˆ ˆ sin cos V t V t r bn b i j b b ′ ′ ′ = = + D D G Since the coordinate system is moving with the wheel, every point on the rim is fixed in that coordinate system. and 0 r′ = G 0 r′ = G The x y z ′ ′ ′ coordinate system also rotates as the bicycle wheel completes a circle around the track: 2 ˆ ˆ ˆ sin cos V V V t V t n i j b b ω ρ ρ ′ ′ = = + D D D D G The total rotation of the coordinate axes is represented by: 1 2 ˆ ˆ ˆ sin cos V V t V t i j b b ω ω ω ρ V k b ′ ′ ′ = + = + + D D D G G G D 2 ˆ ˆ cos sin V V t V i j b b ω ρ ′ ′ = − D D G t b D 2 2 2 2 ˆ ˆ cos sin V V t V t V r k k b b ω k̂ ρ ρ ′ ′ ′ × = + = D D D G G ′ D 11
- 67. 0 r ω ′ × = GG ˆ ˆ ˆ ˆ sin cos sin cos sin cos V b V t V t V t V t V t V t r k k V j i b b b b b b ω ρ ′ ′ ′ ′ ′ × = − + − D D D D D D D G G D ( ) 2 2 2 2 ˆ ˆ ˆ ˆ sin cos sin cos V V t V t V V t V t r k k i j b b b b ω ω ρ ′ ′ ′ ′ ′ × × = + + − − D D D D D G G G b D ( ) 2 2 ˆ ˆ V V r k n b ω ω ρ ′ ′ × × = − D D G G G Since the origin of the coordinate system is traveling in a circle of radius ρ : 2 ˆ V A k ρ ′ = D D G G G G ( ) 2 r r r r r A ω ω ω ω ′ ′ ′ ′ = + × + × + + × × + D G G G G G G G 2 2 2 ˆ ˆ ˆ ˆ V V V V r k k n k b 2 ρ ρ ρ ′ ′ ′ = + − + D D D G D 2 2 ˆ ˆ 3 V V r k b ρ ′ = − D D G n With appropriate change in coordinate notation, this is the same result as obtained in Example 5.2.2. -------------------------------------------------------------------------------------------------- 12
- 68. CHAPTER 6 GRAVITATIONAL ANDCENTRAL FORCES 6.1 3 4 3 s m V r ρ ρ π = = 1 3 3 4 s m r πρ = ( ) 2 2 4 2 3 3 3 2 4 4 4 3 4 3 2 s Gmm Gm G F m m r πρ πρ = = = 2 1 2 3 3 4 4 3 F F Gm m W mg g πρ = = ( ) 2 11 2 2 3 6 3 1 3 3 2 3 6.672 10 4 11.35 1 10 1 4 9.8 3 10 1 F N m kg g cm kg cm kg W m s g m π − − − − × ⋅ ⋅ × ⋅ = × × ⋅ 3 × × 9 2.23 10 F W − = × 6.2 (a) The derivation of the force is identical to that in Section 6.2 except here r R. This means that in the last integral equation, (6.2.7), the limits on u are R – r to R + r. 2 2 2 2 1 4 R r R r GmM r R F d Rr s + − − = + ∫ s ( ) 2 2 2 2 2 4 GmM R r R r R r R r Rr R r R r − − = + − − + − + − ( ) 2 2 0 4 GmM F r R r R r Rr = + − − + = (b) Again the derivation of the gravitational potential energy is identical to that in Example 6.7.1, except that the limits of integration on s are ( ) ( ) R r R r − → + . ψ θ P Q R r s 2 2 R r R r R G d rR πρ φ + − = − ∫ s ( ) 2 2 R G R r R rR πρ = − + − − r 2 4 R M G G R R π ρ φ = − = − For , r R φ is independent of r. It is constant inside the spherical shell. 6.3 2 ˆr GMm F e r = − 1
- 69. The gravitational forceon the particle is due only to the mass of the earth that is inside the particle’s instantaneous displacement from the center of the earth, r. The net effect of the mass of the earth outside r is zero (See Problem 6.2). 3 r π ρ = 4 3 M 4 ˆ ˆ 3 r r F G mre kre πρ = − = − The force is a linear restoring force and induces simple harmonic motion. r F 2 3 2 2 4 m T k G π π π ω πρ = = = The period depends on the earth’s density but is independent of its size. At the surface of the earth, 3 2 2 4 3 e e e GMm Gm mg R R R π ρ = = ⋅ 4 3 e G g R πρ = 6 2 6.38 10 1 2 2 1. 9.8 3600 e R m hr T h g m s s π π − × = = × ≈ ⋅ 4 r 6.4 2 ˆ g r GMm F e r = − , where 3 4 3 M r π ρ = The component of the gravitational force perpendicular to the tube is balanced by the normal force arising from the side of the tube. The component of force along the tube is cos x g F F θ = The net force on the particle is … 4 ˆ cos 3 F i G mr πρ θ = − cos r x θ = 4 ˆ ˆ 3 F i G mx ik πρ = − = − x As in problem 6.3, the motion is simple harmonic with a period of 1.4 hours. 2
- 70. 6.5 2 2 GMm mv r r =so 2 GM v r = for a circular orbit r, v is constant. 2 r v T π = 2 2 2 2 3 2 4 4 r T r v GM π π = = ∝ 3 r 6.6 (a) 2 r v T π = From Example 6.5.3, the speed of a satellite in circular orbit is … 1 2 2 e gR v r = 3 2 1 2 2 e r T g R π = 1 2 2 3 2 4 e T gR r π = 1 1 2 2 2 2 2 2 3 3 2 2 6 24 3600 9.8 4 4 6.38 10 e e r T g hr s hr m s R R m π π − − × ⋅ × ⋅ = = × 2 6.62 7 e r R = ≈ (b) ( ) 3 3 3 2 2 1 1 2 2 2 60 60 2 2 e e e e R R r T g g R g R π π π = = = 1 3 6 2 2 2 2 2 2 2 2 60 6.38 10 2 9.8 3600 24 m m s s hr hr day π − − × × = ⋅ × ⋅ × ⋅ − 27.27 27 T day day = ≈ 6.7 From Example 6.5.3, the speed of a satellite in a circular orbit just above the earth’s surface is … e R = v g 2 2 e e R R v g π π = = T 3
- 71. This is thesame expression as derived in Problem 6.3 for a particle dropped into a hole drilled through the earth. T 1.4 ≈ hours. 6.8 The Earth’s orbit about the Sun is counter-clockwise as seen from, say, the north star. It’s coordinates on approach at the latus rectum are ( ) ( ) , , x y a ε α = − . The easiest way to solve this problem is to note that 1 60 ε = is small. The orbit is almost circular! 2 2 S GM m mv r r ∴ = and 2 S GM r = v with r a b α = ≈ ≈ when 0 ε ≈ 4 3 10 S GM m v s α ≈ = ⋅ More exactly cos r v v l α β × = = , but 2 ml k α = (equation 6.5.19) Since S k GM m = 2 S l GM α = , hence ( ) 1 2 cos S GM α β α = = l v Or 1 2 1 cos S GM v α β = The angle β can be calculated as follows: 2 2 2 2 1 y x b a + = (see appendix C) 2 2 dy b x dx a y ∴ = − and at ( ) ( ) , , x y a ε α = − so ( ) 2 2 2 2 2 1 dy b a b dx a a ε ε ε α ε = = − = since 2 2 2 b a ε − = 1 here tan dy dx β ε = = or β ε ≈ (small ε) and 1 1 2 2 1 cos S S GM GM v α ε α = ≈ as before. 6.9 ( ) s d F r F F = + 2 s GMm F r = − 2 d d GM m F r = − The net effect of the dust outside the planet’s radius is zero (Problem 6.2). The mass of the dust inside the planet’s radius is: 4
- 72. 3 4 3 d M r π ρ = () 2 4 3 GMm F r mGr r πρ = − − 6.10 1 1 k u e r r θ − = = k du k e d r θ θ − = − 2 2 2 2 k d u k e k d r θ θ − = = u From equation 6.5.10 … ( ) 2 2 1 2 2 1 d u u k u u f u du ml u − + = + = − 2 3 ( ) ( ) 1 2 2 1 f u ml k − = − + u ( ) ( ) 2 2 3 1 ml k f r r + = − The force varies as the inverse cube of r. From equation 6.5.4, 2 r l θ = 2 2 k d l e dt r θ θ − = 2 2 k l e d d r θ θ = t 2 2 1 2 k lt e C k r θ = + 2 1 2 ln 2 klt C k r θ ′ = + θ varies logarithmically with t. 6.11 ( ) 3 3 k f r k r = = u From equation 6.5.10 … 2 3 2 2 2 1 d u ku u ku d ml u θ + = − ⋅ = − 2 ml 2 2 2 1 0 d u k u d ml θ + + = 5
- 73. If 2 1 0 k ml + , 2 2 0 d u cu dθ − = , , for which u a 0 c b e θ = is a solution. If 2 1 0 k ml + = , 2 2 0 d u dθ = 1 du C dθ = 1 2 u c c θ = + 1 2 1 r c c θ = + If 2 1 0 k ml + , 2 2 0 d u cu dθ + = , 0 c ( ) cos u A cθ δ = + 1 2 cos 1 k r A ml θ δ − = + + 6.12 1 1 cos u r r θ = = 2 sin cos du d r θ θ θ = 2 2 2 3 1 1 2sin cos cos d u d r θ θ θ = + θ = 2 2 2 1 2 2cos 1 2 1 1 cos cos cos cos r r θ θ θ θ − θ + = − ( ) 2 2 2 2 3 2 2 1 2 d u u r u r u u dθ = − = − Substituting into equation 6.5.10 … ( ) 2 3 1 2 2 1 2r u u u f u ml u − − + = − ( ) 1 2 2 2 5 f u r ml − = − u ( ) 2 2 5 2r ml f r r = − 6.13 From Chapter 1, the transverse component of the acceleration is … 2 a r r θ θ θ = + If this term is nonzero, then there must be a transverse force given by … ( ) ( 2 ) f m r r θ θ θ = + For r aθ = , and bt θ = ( ) 2 2 0 f mab θ = ≠ Since , the force is not a central field. ( ) 0 f θ ≠ 6
- 74. For r aθ =, and the force to be central, try n bt θ = ( ) ( ) 2 2 2 2 2 2 2 2 1 n n f m ab n t ab n n t θ − − = + − For a central field … ( ) 0 f θ = ( ) 2 1 n n + − = 0 1 3 n = 1 3 bt θ = 6.14 (a) Calculating the potential energy ( ) 2 3 5 4 dv a f r k dr r r − = = − + Thus, 2 2 4 2 4 a V k r r = − + The total energy is … 2 2 2 2 2 1 2 1 1 9 9 0 2 4 2 2 4 k k E T V v k a a a a = + = − + = − = Its angular momentum is … 2 2 2 4 9 2 k l a v constant r 2 θ = = = = Its KE is … ( ) 2 2 2 2 2 2 2 2 2 4 1 1 1 2 2 2 dr dr l T r r r r d d θ θ θ θ = + = + = + r The energy equation of the orbit is … 2 2 2 2 4 2 1 2 0 2 4 dr l a T V r k d r r θ + = = + − + 4 r 2 2 2 4 2 9 2 4 4 dr k a r k d r r θ = + − + 4 r or ( ) 2 2 2 1 9 dr a r dθ = − Letting cos r a φ = then sin d a d d dr φ φ θ θ = − So 2 1 9 d d φ θ = 1 3 φ θ ∴ = 7
- 75. Thus 1 cos 3 r a θ =( ) @ 0 r a θ = = (b) at 3 2 π θ = the origin of the force. To find how long it takes … 0 r → 2 2 2 2 1 1 cos cos 3 3 av v l r a a θ θ θ = = = 2 1 cos 3 a dt d v θ θ = 3 2 2 2 2 0 0 1 3 3 cos cos 3 4 a a T d d v v π π a v π θ θ φ φ = = ∫ ∫ = Since 1 2 2 9 2 k v a = 1 1 2 2 2 2 3 2 2 4 9 4 a T a k k π π = = (c) Since the particle falls into the center of the force v → ∞ (since ) l vr const ⊥ = = 6.15 From Example 6.5.4 … 1 2 1 1 2 c v r v r r = + Letting c v v = V we have 1 2 1 2 1 r r = + V So: 2 2 1 1 1 2 1 2 r r dV dr V r r − = − + − 1 Thus 2 1 1 1 1 1 1 1 1 1 1 1 2 1 1 2 1 dV r r V r r dr r r r r r r r = = + + + 8
- 76. (a) 1 1 1 1 2 dV r V r dr r ≈ (b) ( ) 1 1 1 2 2 60 1% 120%! dr r dV r r V = = = The approximation of a differential has broken down – a correct result can be obtained by calculating finite differences, but the implication is clear – a 1% error in boost causes rocket to miss the moon by a huge factor --- ∼ 2! 6.16 From section 6.5, 0.967 ε = and mi. 6 55 10 r = × From equations 6.5.21ab, 1 1 1 r r ε ε + = − ( ) 6 1 6 1 55 10 2 1 1 0.967 93 10 r mi AU a r r mi ε × = + = = × − − × 1 U 17.92 a A = From equation 6.6.5, 3 2 ca τ = 3 3 2 2 1 17.92 yr AU AU τ − = ⋅ × 3 2 75.9 yr τ = From equation 6.5.21a and 6.5.19 … 2 0 0 1 1 ml r kr α ε = − = − 2 1 mrv k ε = − and k GMm = ( ) 1 2 1 GM v r ε = + From Example 6.5.3 we can translate the factor GM into the more convenient 2 e e GM a v = … with ae the radius of a circular orbit and ve the orbital speed … ( ) ( ) 1 1 2 6 2 2 6 93 10 1 1.967 55 10 e e e a v mi v v r mi ε × = + = × 0 1.824 e v v = Since l is constant … 1 1 rv rv = 1 1 1 1 .967 1.824 0.0306 1 1.967 e e r v v v v r v ε ε − − = = = × = + 6 1 1 2 2 93 10 66,705 1 365 24 e e a mi v m yr day yr hr day π π τ − − × × ≈ = = × ⋅ × ⋅ ph mph ph 5 1.22 10 v = × and 3 1 2.04 10 v m = × 9
- 77. 6.17 From Example6.10.1 … ( ) 1 2 2 2 2 1 si q qd d ε = + − nφ where e v v q = and e r d a = are dimensionless ratios of the comet’s speed and distance from the Sun in terms of the Earth’s orbital speed and radius, respectively (q and d are the same as the factors V and R in Example 6.10.1). φ is the angle between the comet’s orbital velocity and direction vector towards the Sun (see Figure 6.10.1). The orbit is hyperbolic, parabolic, or elliptic as ε is , = , or 1 … i.e., as 2 2 q R − is , = , or 0. 2 2 q R − v v is , = , or 0 as is , = , or 2. 2 q d 6.18 Since l is constant, v occurs at r and v occurs at , i.e. and and form the constancy of l … max min 1 1 v r 1 r max v = min 1 v = v r = 2 min max 1 1 r v v v v v r = = ( ) 2 1 k r v m ε = + (See Example 6.5.4) From equation 6.6.5 … 2 2 k a GM a m π τ = = ( ) 2 min max 1 1 2 a a v v r ε π τ + = ⋅ From equation 6.5.21ab … 1 1 1 r r ε ε + = − . With 2 1 a r r = + : ( ) ( )( ) ( ) 1 1 1 1 1 1 1 1 1 1 1 1 1 ( 1) 2 2 2 1 a r r r r r r ε ε ε ε ε ε + + + − = = + + = + + 1 + = 2 min max 2 a v v π τ = 6.19 As a result of the impulse, the speed of the planet instantaneously changes; its orbital radius does not, so there is no change in its potential energy V. The instantaneous change in its total orbital energy E is due to the change in its kinetic energy, T, only, so 2 2 1 2 2 v v E T mv mv v mv T v v δ δ δ δ δ δ = = = = = 2 E v T v δ δ = 10
- 78. But the totalorbital energy is 2 k E a = − So 2 2 k E a a δ δ = Since planetary orbits are nearly circular ~ k V a − and ~ 2 k T a Thus, a E T a δ δ ≅ and E a T a δ δ = We obtain 2 a v a v δ δ = 6.20 (a) 0 1 τ τ = ∫ V Vdt ( ) k V r r = − From equation 6.5.4, l r2 θ = 2 d dt r l θ = or 2 r d dt l θ = 2 0 0 kr Vdt d l τ π θ = − ∫ ∫ From equation 6.5.18a … ( ) 2 1 1 cos a r ε ε θ − = + ( ) 2 2 0 0 1 1 cos ka d Vdt l τ π ε θ ε θ − = − + ∫ ∫ From equation 6.6.4 … 2 2 2 1 a l π τ ε = − 2 2 0 1 2 1 co k d V a π ε θ s π ε θ − = − + ∫ 2 2 0 2 1 cos 1 d π θ π ε θ ε = + − ∫ , 2 1 ε k V a ∴ = − (b) This problem is an example of the virial theorem which, for a bounded, periodic system, relates the time average of the quantity 0 i i p r τ ⋅ ∑ ∫ to its kinetic energy T. We will derive it for planetary motion as follows: 0 0 0 1 1 1 p rdt mr rdt F rdt τ τ τ τ τ τ ⋅ = ⋅ = ⋅ ∫ ∫ ∫ 11
- 79. Integrate LHS byparts 2 0 0 0 1 1 1 mr r mr dt F rdt τ τ τ τ τ τ ⋅ − = ⋅ ∫ ∫ The first term is zero – since the quantity has the same value at 0 and τ . Thus 2 T F = − ⋅r where denote time average of the quantity within brackets. but dV k r F r r V dr r V − ⋅ = = = = − ⋅∇ hence 2 T V = − but 2 2 V V E T V V = + = − + = hence 2 V = E but 2 k E constant a = − = and 0 1 2 k E Edt E a τ τ = = = − ∫ so 2 k E a = − Thus: k V a = − as before and therefore 1 2 2 k T V a = − = 6.21 The energy of the initial orbit is 2 1 2 2 k k mv E r a − = = − (1) 2 2 1 k v m r a = − Since (1 a r a ) ε = + at apogee, the speed , at apogee is 1 v ( ) ( ) ( ) 2 1 1 2 1 1 1 k k v m a a ma ε ε ε − = − = + + To place satellite in circular orbit, we need to boost its speed to such that c v 2 1 2 2 c a a k mv r r − = − k since the radius of the orbit is ra ( ) 2 1 c a k k v mr ma ε = = + Thus, the boost in speed 1 1 c v v v ∆ = − (2) ( ) ( ) 1 2 1 2 1 1 1 1 k v ma ε ε ∆ = − − + Now we solve for the semi-major axis a and the eccentricity ε of the first orbit. From (1) above, at launch v at , so v = E r R = 12
- 80. 2 2 1 E k v mR a = − and solving for a 2 2 E E R a R mv k = − noting that … (3) E E E E k GM gR mR R = = 3 2 4.49 10 1.426 2 E E E R R a k v gR = = = ⋅ − m The eccentricity ε can be found from the angular momentum per unit mass, l, equation 6.5.19, and the data on ellipses defined in figure 6.5.1 … ( ) ( ) 1 1 2 2 2 2 1 sin E ka k l r v R m m ε α θ θ − = = = = where v , θ are the launch velocity, angle Solving for ε (using (3) above) 2 2 2 2 1 2 sin 0.795 E E v v gR gR ε θ = − − = 0.892 ε ∴ = Inserting these values for a, ε into (2) and using (3) gives (a) ( ) 1 1 2 3 1 2 1 1 1 4.61 10 1 E E R a v gR km s ε ε − ∆ = − − = ⋅ ⋅ + (b) ( ) 3 1 2.09 1 E h a R km ε = + − = ⋅ 0 {altitude above the Earth … at perigee} 6.22 ( ) 2 3 2 2 2 br br br be e e f r k k b r r r − − − − ′ = − − = + r ( ) ( ) 2 f a b f a a ′ = − + From equation 6.14.3, ( ) ( ) ( ) 1 1 2 2 3 3 f a a a f a ψ π π − 2 b − ′ = + = − + 1 ab π ψ = − 13
- 81. 6.23 From Problem6.9, ( ) 2 4 3 GMm f r m r πρ = − − Gr ( ) 3 2 4 3 GMm f r m r πρ ′ = − G ( ) ( ) 3 3 3 2 4 4 2 2 3 3 4 4 1 3 3 a GMma mG f a M f a a GMma mGa a M πρ πρ πρ πρ − − − − + ′ = = − − + From equation 6.14.3, ( ) ( ) 1 2 3 f a a f a ψ π − ′ = + 1 1 3 2 3 2 3 3 4 4 1 4 2 3 3 3 4 4 1 1 3 3 a a M M a a M M πρ πρ ψ π π πρ πρ − − + − + = + = + + 1 2 1 1 4 c c ψ π + = + , 3 4 3 a c M πρ = 6.24 We differentiate equation 6.11.1b to obtain ( ) dU r mr dr = − For a circular orbit at r a , r so = 0 = 0 r a dU dr = = For small displacements x from r a = , and r x a = + r x = From Appendix D … ( ) ( ) ( ) ( ) 2 2 x f x a f a xf a f a ′ ′′ + = + + +… Taking ( ) f r to be dU dr , ( ) 2 2 d U f r dr ′ = Near … r a = 2 2 r a r a dU dU d U x dr dr dr = = = + … + 2 2 r a d U mx x dr = = − This represents a “restoring force,” i.e., stable motion, so long as 2 2 0 d U dr at . r a = 14
- 82. 6.25 ( )3 5 2 4 k f r r r ε ′ = + From equation 6.13.7, the condition for stability is ( ) ( ) 0 3 a f a f a ′ + 2 4 3 5 2 4 0 3 k a k a a a a ε ε − − + + 2 4 0 3 3 k a a ε − + 2 k a ε 1 2 a k ε 6.26 (a) ( ) 2 br e f r k r − = − ( ) 2 3 2 2 2 br br b e f r ke k b r r r r − − ′ = − − − = + From equation 6.13.7, the condition for stability is ( ) ( ) 0 3 a f a f a ′ + 2 2 2 0 3 ba ba e a e k k b a a a − − − + + 2 0 3 3 ba ba e be k k a a − − − + 1 b a 1 a b (b) ( ) 3 k f r r = − ( ) 4 3k f r r ′ = ( ) ( ) 3 4 3 0 3 3 a k a k f a f a a a ′ + = − + = Since ( ) ( ) 3 a f a f ′ + a is not less than zero, the orbit is not stable. 15
- 83. 6.27 (See Figure6.10.1) From equation 6.5.18a 2 (1 ) 1 cos a r ε ε θ − = + and the data on ellipses in Figure 6.5.1 (1 ) p a ε = − so (1 ) 1 cos r p ε ε θ + = + For a parabolic orbit, 1 ε = The comet intersects earth’s orbit at r a = . 2 1 cos p a θ = + 2 cos 1 p a θ = − + 6.28 (See Figure 6.10.1) T along the comet’s trajectory inside earth’s orbit d = ∫ t From equation 6.5.4, 2 2 d r r dt l θ θ = = so 2 r d dt l θ = 2 r d T l θ = ∫ From equation 6.5.18a 2 (1 ) 1 cos a r ε ε θ − = + and the data on ellipses in Figure 6.5.1 (1 ) p a ε = − so (1 ) 1 cos r p ε ε θ + = + From equation 6.5.18b, with 1 ε = for a parabolic orbit: 2 1 cos p r θ = + At 2 π θ = the distance to the comet is 2 2 1 cos 2 p r p α π = = = + From equation 6.5.19, 2 ml k α = , where k GMm = , so 2 2 l p GM = As shown in Example 6.5.3, 2 e GM av = For a circular orbit, 2 1 e a yr v π = ( ) ( ) ( ) 1 1 1 3 1 2 2 2 2 2 2 2 e l GMp ap v a p yr π − = = = ( ) ( ) 1 2 2 3 1 2 2 2 4 2 1 cos r d p T a l θ θ θ θ θ π θ θ + + − − − − − = = ⋅ + ∫ ∫ p d yr where 1 2 cos 1 p a θ − = − + from Problem 6.27 16
- 84. ( ) 3 2 3 2 2 2 1cos p d T y a θ θ θ θ π + − = + ∫ r From a table of integrals, ( ) 3 2 1 1 tan tan 2 2 6 1 cos dx x x x = + + ∫ 2 3 2 3 2 1 tan tan 2 3 2 p T y a θ θ π = + r 1 2 1 cos tan 2 1 cos x x x − = + 1 2 1 2 2 2 tan 2 2 p a p a p p a θ − − = = 1 3 3 2 2 2 2 1 3 p a p a p T y a p p π − − = + r 1 3 2 2 2 1 3 p a p a p yr a p p π − − = + 1 2 2 2 1 1 3 p p T y a a π = + − r ) T is a maximum when ( )( 1 2 2p a a p + − is a maximum. ( ) ( ) ( )( )( ) ( ) ( 2 2 2 2 2 2 2 d p a a p p a a p p a dp + − = + − + + − ) 1 ( )( ) 2 3 6 p a a p − = + T is a maximum when 2 a p = . ( ) 1 2 2 1 2 2 77.5 3 2 3 T yr day π π = = = When 0.6 p a = ( ) 2 2.2 .04 0.2088 76.2 3 T yr day π = = = 6.29 ( ) 3 k k V r r r ε = − − 17
- 85. ( ) () 2 2 4 4 3 3 dV k k k f r r dr r r r ε ε = − = + = + ( ) ( ) 2 3 5 5 2 12 2 6 k k k f r r r r r ε ε ′ = − − = − + ( ) ( ) 2 2 2 6 3 f a r f a a r ε ε ′ + = − + From equation 6.14.3, ( ) ( ) 1 2 3 f a a f a ψ π − ′ = + 1 2 2 2 2 2 6 3 3 2 3 3 r r r r ε ε ψ π π ε ε − + + = − = + − For 2 5 R R ε = ∆ , 4000 R mi = , 13 R mi ∆ = ( )( ) 4 2 2 4000 13 2.08 10 5 mi ε = = × For , r R ≈ 2 7 1.6 10 r m = × 2 i 1.0039 180.7 ψ π = = 6.30 2 2 1 2 rel k k V E r m c r = − − + ( ) 2 2 2 2 dV k k k f r E dr r m c r r = − = − + + − 2 ( ) 2 2 1 1 k k f r E r m c r = − + + ( ) 3 2 2 2 2 1 1 1 k k k f r E r m c r r m c r ′ = + + − − 2 k ( ) 3 2 1 3 2 2 k k f r E r m c r ′ = + + ( ) ( ) 2 2 1 3 2 2 1 1 1 k E f a m c a k f a a E m c a + + ′ = − + + ( ) ( ) 2 2 2 3 1 3 2 2 3 1 1 k k E E f a m c a m c a a f a k E m c a + + − − + ′ + = + + 3 18
- 86. 2 2 1 1 1 E m c k E m ca + = + + ( ) ( ) 2 2 3 f a m c E a k f a m c E a ′ + + = + + ( ) ( ) 1 2 3 f a a f a ψ π − ′ = + 1 2 2 1 k a m c E ψ π = + + 6.31 From equation 6.5.18a 2 (1 ) 1 cos a r ε ε θ − = + … (Here θ is the polar angle of conic section trajectories as illustrated by the coordinates in Figure 6.5.1) … and the data on ellipses in Figure 6.5.1 0 (1 ) r a ε = − so 1 1 cos com r r ε ε θ + = + From equation 6.5.18b 1 cos r α ε θ = + and at 0 θ = 0 1 r α ε = + And from equation 6.5.19 2 ml k α = so 2 0 (1 ) ml r k ε = + 1 m m k GMm GM = = From Example 6.5.3, and l r 2 e e GM a v = 2 2 2 2 sin com com v φ = ( ) ( )( ) 2 2 2 2 sin 1 1 1 cos com com com e e r v r a v φ ε ε ε θ + = + + 2 2 1 1 sin 1 cos RV φ ε θ = + ( ) 2 2 1 cos sin 1 RV θ φ ε = − ( ) ( ) 1 1 2 2 2 2 2 2 2 1 sin 1 cos 1 sin 1 RV θ θ φ ε = − = − − 19
- 87. ( ) () 1 2 2 2 2 2 2 2 1 sin sin 2 sin 1 RV RV θ ε φ φ ε = − + − Again from Example 6.5.3 … ( ) 1 2 2 2 2 1 si V RV R ε φ = + − n 2 ( ) 2 2 2 2 1 sin 2 sin RV RV ε φ φ = + − ( ) ( ) 1 2 2 2 2 2 2 1 sin sin sin RV RV θ φ ε = − φ 2 1 sin sin cos RV θ φ φ ε = 2 2 2 cos sin 1 sin sin cos RV RV θ φ θ φ φ − = 2 2 cot tan sin 2 RV θ φ φ = − 1 2 2 cot tan csc2 RV θ φ φ − = − For , 0.5 V = 4 R = , : 30 φ = ( ) 1 2 2 cot tan30 csc60 4 .5 θ − = − ( ) 1 1 1 2 cot cot 3 3 3 2 − − = − = − 30 θ = − 6.32 The picture at left shows the orbital transfer and the position of the two satellites at the moment the transfer is initiated. Satellite B is “ahead” of satellite A by the angle θ0 r2 r1 θ0 1 2 2 r r a + = is the semi-major axis of the elliptical transfer orbit. From Kepler’s 3rd law (Equation 6.6.5) applied to objects in orbit about Earth … 2 2 3 2 4 E a GM π τ = The time to intercept is … 3 3 2 2 1 2 t E E T a GM R g τ π π = = = a since 2 E E g R GM = 20
- 88. Letting and whereh 1 1 E r R h = + 2 E r R h = + 2 1 and h2 are the heights of the 2 satellites above the ground. Inserting these into the above gives … 3 3 3 2 2 1 2 1 2 2 2 2 t E E E E E h h h h T R R R R g R g π π + + = + = + From Example 6.6.2, RE = 6371 km, h1 = 200 mi = 324 km and h2 = r2 - RE = 42,400 km - 36,029 km. Putting in the numbers … Tt = 4.79 hr (b) Thus, 0 0 0 180 1 108 12 t T θ = − = 6.33 The potential for the inverse-cube force law is … ( ) 2 2 k r = V r Letting u r 1 − = , we have (Equation 6.9.3) ( ) 2 2 2 1 1 2 du ml u V u E dθ − + + = ( ) 2 2 2 E V du u d ml θ − = − ( ) ( ) 2 2 2 2 2 2 2 ml d du E V m E V m l u u ml θ = = − − − − du ) r θ b θS θ0 θ0 Now, integrating from up to ( 0 r u = ∞ = ( ) min max u u = = r r … ( ) max 0 2 2 2 0 2 u ml du m E V m l u θ = − − ∫ But 2 0 0 1 , 2 E mv l b = v = , so … max 0 0 0 2 2 2 2 2 2 0 0 1 1 2 2 2 u mbv du m mv ku m b v u θ = − − ∫ max 0 0 0 2 2 2 2 2 2 0 0 1 1 2 2 2 u mbv du m mv ku m b v u θ = − − ∫ 21
- 89. Before evaluating thisintegral, we need to find ( ) 1 max min r − = u , in other words, the distance of closest approach to the scattering center. ( ) ( ) 2 2 min min 0 2 min 1 1 1 2 2 2 k E T r V r mv mv r = + = + = But, the angular momentum per unit mass l is … and substituting for v into the above gives … 0 min l bv r v = = 2 2 0 2 2 min min ml k mv r r + = so … 2 2 0 max 2 2 min 1 mv u r ml k = = + Solving for u … max max 2 2 0 1 u k b mv = + Now we evaluate the integral for θ0 … max max 1 0 2 2 max 0 0 2 2 2 2 2 0 0 0 sin 2 1 u u b b u du u k k k b b b u mv mv mv b π θ − = = = + + − + ∫ Solving for b … ( ) 0 0 2 2 2 0 0 2 4 k b mv θ θ π θ = − But ( 0 1 2 S ) θ π θ = − . Thus, we have … ( ) ( ) 2 0 2 S S S S k b mv π θ θ θ π θ − = − We can now compute the differential cross section … ( ) ( ) ( ) 2 2 2 2 0 sin 2 sin S S S S S S k b db d mv π π θ σ θ θ θ S θ π θ θ − = = − Since 2 sin s s d d π θ θ Ω = we get … ( ) ( ) ( ) 3 2 2 2 2 S S S S S k d bdb d E π θ π σ θ π π θ θ − Ω = = − θ --------------------------------------------------------------------------------------------------- 22
- 90. Chapter 7 Dynamics ofSystems of Particles z 7.1 From eqn. 7.1.1, 1 cm i i i r m m = r ∑ 3 2 1 y x ( ) ( ) 1 2 3 1 1 ˆ ˆ ˆ ˆ ˆ 3 3 cm r r r r i j j k = + + = + + + + k ( ) 1 ˆ ˆ ˆ 2 2 3 cm r i j = + + k ( ) ( ) 1 2 3 1 1 ˆ ˆ ˆ ˆ ˆ 2 3 3 cm cm d v r v v v i j i j dt = = + + = + + + + k ( ) 1 ˆ ˆ ˆ 3 2 3 cm v i j = + + k 3 From eqn 7.1.3, 1 2 i i i p m v v v v = = + ∑ + ˆ ˆ ˆ 3 2 p i j = + + k 7.2 (a) From eqn. 7.2.15, 2 1 2 i i i = ∑ T m v ( ) 2 2 2 2 2 1 2 1 1 1 1 4 2 T = + + + + = (b) From Prob. 7.1, ( ) 1 ˆ ˆ ˆ 3 2 3 cm v i j = + + k ( ) 2 2 2 2 1 1 1 3 3 2 1 2 2 2 9 cm mv = × × + + = 1 3 v (c) From eqn.7.2.8, i i i L r m = × ∑ ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 L i j i j k j k i j k = + × + + × + × + + ˆ 2k omentum i ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 L k i j i i j = − + − + − = − + − 7.3 b g v v v = − Since m s conserved and the bullet and gun were initially at rest: 0 b g mv Mv = + g b v v γ = − , m M γ = ( ) 1 b v v γ = + 1 b v v γ = + 1
- 91. 1 g v v γ γ = − + 7.4 Momentumis conserved: 2 blk v mv m Mv = + 1 2 blk v v γ = m M γ = 2 2 2 1 1 1 2 2 2 2 2 i f v v T T mv m M γ − = − + 2 2 1 1 1 1 2 4 mv M m 4 γ = − − 3 4 4 i f i T T T γ − = − v0 60 o θ v0 / 2 7.5 At the top of the trajectory: ˆ ˆ cos60 2 v v iv i = = Momentum is conserved: 2 ˆ ˆ 2 2 2 2 v m v m im j v = + 2 ˆ ˆ 2 v v iv j = − Direction: 1 2 tan 26.6 v v θ − − = = below the horizontal. Speed: 1 2 2 2 2 1.118 2 v v v = + = v 7.6 When a ball reaches the floor, 2 1 2 mv mgh = . As a result of the bounce, v v ε ′ = . The height of the first bounce: 2 1 2 mgh mv ′ ′ = 2 2 2 2 2 2 v v h h g g ε ε ′ ′ = = = Similarly, the height of the second bounce, h h 2 4 h ε ε ′′ ′ = = 2
- 92. Total distance 24 2 0 2 2 1 2 n n h h h h ε ε ε ∞ = = + + + = − + ∑ … Now 0 1 n n a r ∞ = = − ar ∑ , 1 r . 2 2 2 2 0 2 1 1 2 1 1 1 n n ε ε ε ε ∞ = + − + = − + = − − ∑ total distance 2 2 1 1 h ε ε + = − For the first fall, 2 1 2 gt h = , so 2h g = t For the fall from height h′ : 1 2 2 h h g g ε ′ = = t Accounting for equal rise and fall times: Total time ( ) 2 0 2 2 1 2 2 1 2 n n h h g g ε ε ε ∞ = + + = − + ∑ … = + Total time 2 1 1 h g ε ε + − = -v0/2 4m v0 m 7.7 From eqn. 7.5.5: ( ) ( ) 1 2 1 2 2 1 1 2 m m x m m x x m m ε ε − + + ′ = + 2 ) vt ( ) ( 1 1 1 2 1 2 1 2 m m x m m x x m m ε ε + + − ′ = + 2 vc 1 1 4 4 4 0 5 4 4 2 4 5 c v v m m v m m m v m m m − + + − + − = = + 2 2 v = − 1 1 4 4 4 4 t v m m v m m v m m + + − − = + 2 5 15 4 4 2 5 8 v mv m v m + − = = − Both car and truck are traveling in the initial direction of the truck with speeds 2 v and 8 v , respectively. 3
- 93. 7.8 From eqn.7.2.15, 2 2 2 1 1 2 2 1 1 1 2 2 2 i i i v m v m = = + ∑ T m v Meanwhile: ( ) 2 2 2 2 1 1 2 2 1 2 1 2 1 2 1 1 1 1 2 2 2 2 cm m v m v m m mv v m v v m m m µ + + = + − + ( ) 2 2 2 2 2 2 1 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 m v m v m m v v m m v v v v m = + + ⋅ + + − ⋅ 2 2 1 1 2 2 1 1 2 2 m v m v = + Therefore, 2 2 1 1 2 2 cm v T m v µ = + 7.9 From Prob. 7.8, 2 2 1 1 2 2 cm v T m v µ = + and since v v Q T T′ = − cm cm ′ = : 2 2 1 1 2 2 Q v v µ µ ′ = − From eqn. 7.5.4, v v ε ′ = ( ) 2 2 1 1 2 Q v µ ε = − 7.10 Conservation of momentum: 1 1 1 1 2 2 m v m v m v ′ ′ = + 2 1 1 1 m v v v m ′ ′ = − 2 2 2 2 2 2 1 1 1 2 2 1 1 2m m v v v m m ′ ′ = − + 2 2 ′ v v Conservation of energy: 2 2 1 1 1 1 2 2 1 1 1 2 2 2 m v m v m v ′ ′ = + 2 2 2 2 2 2 1 1 1 1 2 1 2 2 2 2 1 1 1 1 2 2 2 2 m m v m v m v v v m v m 2 ′ ′ ′ = − + + 2 2 2 2 2 1 2 1 1 0 2 m m v m v v m ′ ′ + − = 1 1 2 1 2 2m v v m m ′ = + ( ) 2 2 2 2 1 2 1 1 1 1 1 1 2 2 1 2 1 2 1 1 1 2 2 2 2 m m T T m v m v m v v m m ′ ′ ′ − = − = = + 2 4
- 94. 2 1 1 1 1 2 1 1 1 2 4 1 2 m v TT m T m m v µ µ ′ − = = 7.11 From eqn. 7.2.14, cm cm i i i i L r mv r m v = × + × ∑ 1 1 1 2 2 i i i i r m v r m v r m v × = × + × ∑ 2 From eqn. 7.3.2, 1 1 2 1 1 2 2 1 m m m m1 1 R r r m m r µ + = + = = Since from eqn. 7.3.1, 2 1 2 1 m r r m = − 2 2 m R r µ = − 1 1 2 2 1 2 i i i i r m v R m v R m v m m µ µ × = × + − × ∑ ( ) 1 2 R v v R v µ µ − = × = × L rcm cm mv R v µ = × + × 7.12 Let ms = mass of Sun and ae = semi-major axis of Earth’s orbit then from eqn. 7.3.9c, 1 3 2 2 1 2 s s e m m a yr m m a τ − = + 1 2 3 1 2 3 e s a m m a m m τ = + s ( ) 2 2 1 3 3 3 1 5.6 20 16 365 yr da yr da − = × + 1 0.20 5 e e a a = = a the 4 L 16 ms 20 ms 7.13 (Ignore primes in notation) a) The coordinates of the two primaries, P1 and P2, are shown at left – along with coordinates of and . 5 L 5
- 95. b) ( ) () ( ) ( ) 2 2 1 1 2 2 2 2 2 2 1 , 2 1 x y y x y x y α α α α − V x + = − − − − + + − + (7.4.13) ( )( ) [ ] ( ) [ ] 3 3 2 2 1 1 x x V x x α α α α − − + − ∂ = + ∂ − Now x 0.5 α = − at and 4 L 5 L also, each bracket term in the denominator equals 1 at , 4 L 5 L ( )( ) ( ) ( 1 0.5 0.5 1 0.5 V x α α α α α α α ∂ = − − − + − + − − − ∂ ) 0.5 0.5 0.5 0.5 0 α α α = − + + − + ≡ ( ) [ ] [ ] 3 3 2 2 1 y V y y y α α − ∂ = + ∂ − Again, the denominator in brackets equals 1 @ , 4 L 5 L So, ( ) 3 3 1 2 2 V y α α = − ± + ± − ± ∂ 3 2 ∂ 3 3 3 3 0 2 2 2 2 α α = ± ± ≡ ∓ ∓ Thus ( ) ˆ ˆ , 0 V V V x y i j x y ∂ ∂ + ∂ ∂ ∇ = ≡ n of moment : at L , . 4 5 L 7.14 Conservatio um 4 p p p p m v m v m vα ′ ′ = + 45 o v’ φ v’p cos45 4 cos p v v vα φ ′ ′ = + 4 cos 2 p v v v α φ ′ ′ = − vo 0 sin 45 4 sin p v vα φ ′ ′ = − 4 sin 2 p v vα φ ′ ′ = 2 2 16 2 p p v v v v v α ′ ′ = − + 2 ′ Conservation of energy: 2 2 1 1 1 4 2 2 2 p p p m v m v m v 2 p α ′ ′ = + 16 2 2 4 4 p v v α ′ ′ = − v Subtracting: 2 2 2 5 p p v v v 0 3 v ′ ′ = − − + 6
- 96. ( ) 2 2 22 60 2 62 10 10 p v v v v v ± + ′ = = ± 0, so the positive square root is used. p v′ 0.9288 p v v ′ = 0.657 2 p px py v v v ′ ′ ′ = = = v ( ) ( ) 1 1 2 2 2 2 2 1 1 .9288 2 2 p v v v v α ′ ′ = − = − 0.1853 v v α ′ = .9288 2 tan 2 2 .9288 2 p p p p v v v v v v φ ′ ′ = = = ′ ′ − − − 1 tan 1.9134 62.41 φ − = = cos 0.086 x v v α α v φ ′ ′ = = sin 0.164 y v v α α v φ ′ ′ = − = − 7.15 Conservation of energy: 2 2 2 1 1 1 1 1 4 2 2 2 4 2 p p p p p m v m v m v m v α ′ ′ = + + 2 2 2 2 16 3 4 p v v v α ′ ′ = − From the conservation of momentum eqn of Prob. 7.14: 2 2 2 p p v v v v v α ′ ′ = − + 2 ′ 16 Subtracting: 2 2 2 5 p p v v v 0 2 v ′ ′ = − − + ( ) 2 2 2 2 40 2 42 10 10 p v v v v v ± + ′ = = ± Using the positive square root, since 0 p v′ : 0.7895 p v v ′ = 0.558 2 p px py v v v ′ ′ ′ = = = v ( ) 1 1 2 2 2 2 2 3 1 .75 .7895 16 4 2 p v v v v α ′ ′ = − = − 0.1780 v v α ′ = From the conservation of momentum eqns of Prob. 7.14: .7895 tan 2 2 .7895 p p v v v φ ′ = = ′ − − 7
- 97. 1 tan 1.2638 51.65 φ− = = cos 0.110 x v v α α v φ ′ ′ = = sin 0.140 y v v α α v φ ′ ′ = − = − 7.16 From eqn. 7.6.14, 1 sin tan cos θ φ γ θ = + φ1 and θ are the scattering angles in the Lab and C.M. frames respectively. From eqn. 7.6.16, for Q = 0, 1 2 m m γ = sin n 45 1 1 cos 4 ta θ θ = = + 1 cos sin 4 θ θ + = and squaring … 2 2 1 1 cos cos 1 cos 16 2 θ θ θ + + = − 2 1 15 os cos 0 2 16 θ θ + − 2c = 1 1 15 2 4 2 cos .125 .696 4 θ − ± + = = − ± Since 0 2 π θ , 1 cos .571 55.2 θ − = ≈ 7.17 From eqn. 7.6.14, 1 sin tan cos θ φ γ θ = + From eqn. 7.6.18, 1 2 1 1 2 2 1 1 m Q m m T m γ − = − + 1 2 1 1 1 1 1 0.3015 4 4 4 γ − = − + = sin n 45 .3015 cos ta θ θ = + .3015 cos sin θ θ + = (since sin cos θ θ , ) 45 θ .30152 2 2 sin 2sin cos cos θ θ θ = − + θ Using the identity 2sin cos sin 2 θ θ θ = sin 2 2 1 .3015 0.9091 θ = − = − Since 45 θ , : 2 90 θ 1 2 sin .9091 114.62 θ = = 57.3 θ = 8
- 98. φ P1’ ψ 7.18 Conservation ofmomentum: P1 ( ) 1 1 2 cos cos P P P φ ψ φ ′ ′ = + − ( ) 1 2 0 sin sin P P φ ψ φ ′ ′ = − − From Appendix B for ( ) sin α β + and ( ) cos α β + : P2’ ( ) 1 1 2 cos cos cos sin sin P P P φ ψ φ ψ φ ′ ′ = + + 0 s ( ) 1 2 in sin cos cos sin P P φ ψ φ ψ φ ′ ′ = − − 2 2 2 2 2 2 2 cos cos cos 2cos cos P ( ) 2 2 1 1 sin sin sin sin P P φ ψ φ ψ ′ ′ = + + φ φ ψ ψ + φ ( ) 2 1 2 2 cos cos cos sin sin P P φ ψ φ ψ ′ ′ + + φ ( ) 2 2 2 2 2 2 2 1 2 0 sin sin cos 2sin cos cos sin cos sin P P φ ψ φ ψ φ ψ φ ψ ′ ′ = + − + φ ( ) 2 1 2 2 sin sin cos cos sin P P φ ψ φ ψ φ ′ ′ − − Adding: P P 2 2 2 1 1 2 1 2 2 cos P P P ψ ′ ′ ′ ′ = + + Conservation of energy: 2 2 2 1 1 2 2 2 2 P P P Q m m m ′ ′ = + + ( ) ( ) 2 2 2 1 1 2 1 2 1 1 2 cos 2 2 Q P P P P P m m ψ ′ ′ ′ ′ = − − = 1 2 cos P P Q m ψ ′ ′ = 7.19 2 1 1 2 T m = 1 1 v 2 1 1 1 2 T m v ′ 1 ′ = let 2 1 1 2 1 1 T v r T v ′ ′ = = … ratio of scattered particle to incident particle energy Looking at Figure 7.6.2 … ( ) ( 1 1 1 1 cm cm v v v v v v ′ ′ ′ ′ ⋅ = − ⋅ − ) 2 2 2 1 1 1 2 cos cm cm v v v v v 1 φ ′ ′ ′ = + − hence 2 2 2 1 1 1 2 cm cm v v v v v γ ′ ′ ′ = − + where 1 cos γ φ = 2 2 1 1 2 2 2 1 1 1 2 cm cm v v v v r v v v γ ′ ′ ∴ = − + but 1 1 ′ = v v …the center of mass speeds of the incident and scattered particle are the same. 1 2 1 2 1 1 v m v m m α α ′ = = + + …from equation 7.6.12 where 2 1 m m α = 9
- 99. 1 1 2 1 1 1 cm vm v m m α = = + + Equation 7.6.11 Thus ( ) ( ) ( ) 2 2 2 1 2 1 1 1 r α γ α α α = − + + + + ( ) ( ) ( ) 1 2 1 2 2 2 1 1 2 1 1 1 v r v α γ α α α ′ = − + + + + Simplifying 1 2 2 1 0 1 1 r r γ α α α − − + + + = Let 2 x r = and solving the resulting quadratic for x ( ) 1 2 2 2 1 1 1 1 x γ γ α α α = + − − + + Squaring ( ) ( ) 1 2 2 2 2 2 2 2 1 2 1 2 1 r x γ α γ γ α α = = + − + + − + 1 Now ( ) ( ) 1 2 2 2 2 1 2 2 1 1 1 1 2 1 2 1 1 T r T γ α γ γ α α ∆ = − = − + − + + − + And, after a little algebra, we get the desired solution ( ) 2 2 1 2 1 2 2 1 1 1 T T γ γ γ α α α ∆ = − + + − + + 7.20 From Equation 7.6.15 … ( ) ( ) 1 1 1 1 1 1 2 2 1 2 1 m v m v v m m m m m v γ = = 1 ′ ′ + + Now we solve for 1 1 v v′ … 1 2 1 1 2T v m = and now solving for 1 v′ starting with Equation 7.6.9 … 2 2 1 1 1 1 2 2 v v µ µ ′ = −Q and using 2 1 1 2 m m m ′ = + 1 v v we get … ( ) ( ) 2 1 1 1 2 1 2 1 1 2 1 1 T m v Q Q m m m m = + + = − − ( ) ( ) 2 1 1 1 2 1 2 2 1 1 T v Q m m m m m ′ = − + + . Thus, solving for γ … 10
- 100. ( ) ( )( ) ( ) 1 2 1 1 1 2 2 1 1 2 2 1 1 2 1 2 2 2 1 1 T m m m T m m m Q m m γ = + − + Finally… ( ) 1 1 2 2 1 2 1 1 1 m m Q m m T γ = + − 7.21 The time of flight, τ = constant—so 1 r v τ = ′ but from problem 7.19 above 2 2 1 1 1 1 v r v τ τ γ γ α α ′ = = + + − + As an example, let v1 1 τ = and we have r1 γ = 1 α = pp scattering 2 2 1 3 3 γ γ = + + r 2 α = p – D 2 3 1 15 5 γ γ = + + r 4 α = p – He 2 4 1 143 13 γ γ = + + r 12 α = p – C Below is a polar plot of these four curves. 7.22 From eqn. 7.7.6, u g F F mv vm − = + since v = constant, 0 v = m z v λ λ = = , λ =mass per unit length ( ) g F z λ = g ( ) 2 u v F zg v v g z g λ λ λ = + = + is equal to the weight of a length u F 2 v z g + of chain. 11
- 101. 7.23 3 4 3 m r πρ = 2 4 m r r π ρ = where v z ∝ 2 r z π = k a constant of proportionality r kz = r r kz = + From eqn. 7.7.6, mg mv vm = + ( ) 3 3 2 4 4 4 3 3 r g r z r kz z π ρ π ρ π ρ = + 2 3kz g z r = + 2 3z z g r z k = − + For 0 r = , 2 3z z g z = − A series solution is used for this differential equation: 2 0 n n n z a ∞ = = ∑ z 2 1 ( ) 2 dz dz dz dz d z z z dt dz dt dz dz = = ⋅ = = 2 1 ( ) n n n d z a nz dz − = ∑ 2 1 n n n z a z z − = ∑ 1 1 1 3 2 n n n n n n z a nz g a z − − ∴ = = − ∑ ∑ For 1 n = : 1 1 1 3 2 a g a = − 1 2 7 a g = For 1 n ≠ : 1 3 2 n n na a = − Since n is an integer, for 0 n a = 1 n ≠ 2 2 7 z g = z 3 2 7 7 g z g g z z = − = 12
- 102. 7.24 From eqn.7.7.6, , where m and v refer to the portion of the chain hanging over the edge of the table. mg mv vm = + z m λ = and v where λ is the mass per unit length of chain z = m z λ = and v z = 2 1 ( ) 2 dz dz dz dz d z z z dt dz dt dz dz = = ⋅ = = ( ) 2 1 ( ) 2 d z z g z z z dz λ λ λ = + 2 2 1 ( ) 2 d z z z g dz z = = − Because of the initial condition 0 z b = ≠ , a normal power series solution to this differential equation (…as in Prob. 7.22) does not work. Instead, we use the Method of Frobenius … 2 0 n s n n z a z ∞ + = = ∑ ( ) 2 1 ( ) n s n n d z a n s z dz + − = + ∑ 2 1 n s n n z a z z + − = ∑ ( ) 1 1 1 2 n s n s n n n n z a n s z g a z + − + = + = − ∑ ∑ − Equality can be attained for 0 n a ≠ at n = 0 and n = 3 …otherwise 0 0,3 n a n = ≠ For , 0 n = 1 2 a s a = − 2 s = − ( ) 3 3 1 2 2 n n n n n n z a n z g a z − − = − = − ∑ ∑ For 3 n = , 3 3 1 2 a g a = − 3 2 3 a g = For all 0 n ≠ , 3: ( ) 1 2 2 n n a n a − = − a , . 0 n = 0,3 n ≠ 2 2 2 3 z a z g − = + z 13
- 103. At 0 t =, z , and 0 = z b = 2 2 3 a g b = + 0 b 3 2 3 = − a gb 3 2 2 2 2 3 3 b z g g z = − + z At z a , = ( ) 3 2 3 2 2 2 2 3 3 b g z g a a b a a = − = − 3 ( ) 1 2 3 3 2 2 3 g z a b a = − 7.25 Initially, the upward buoyancy force balances the weight of the balloon and sand. (1) ( ) 0 B F M m g − + = Let m m the mass of sand at time t where 0 ( ) t = − t t ≤ ≤ 1 t m m t = − (2) The velocity of sand relative to the balloon is zero upon release so V in equation 7.7.5 … there is no upward “rocket-thrust.” 0 = As sand is released, the net upward force is the difference between the initial buoyancy force, FB, and the weight of the balloon and remaining sand. Let be the subsequent displacement of the balloon, so equation 7.7.5 reduces to F = ma y ( ) ( ) B dv F M m g M m dt − + = + and using (1) and (2) above we get ( ) ( ) ( ) M m gt dv m gt g dt M m t m t M m t m t + = = − + + − + − whose solution is: ( ) ( ) ln 1 M m gt m t dy t dt m M m t + = = − − − + v g ( ) ln 1 g y C gt kt dt k = − + − ∫ , ( ) m k t M m = + ( ) 2 1 ln 1 2 1 gt tdt C gt kt g k k = − − − − t − ∫ Integrating by parts 14
- 104. ( ) 2 1 1 ln1 1 2 1 gt g C gt kt d k k = − − − − − + − ∫ t kt ( ) ( ) 2 2 1 ln 1 ln 1 2 gt gt g C gt kt k k k k = − − − + + − t = ( ) ( ) 2 2 1 1 ln 1 2 g g C gt kt kt k k − + − − + t but 0 y = at t so 0 = 0 C = ( ) ( ) 2 2 1 1 ln 1 2 gt g y gt kt k k = − + − − kt and at t t = (a) ( ) ( ) 2 2 2 2 ln 2 gt M m m M M m m M = + + + + H M m (b) ( ) ( ) ln M m gt m m M + = + − v M m (c) letting 1 m M ε = we have ( ) ( ) ( 2 2 2 2 1 ln 1 2 gt H ) ε ε ε ε = + − + + ε ( ) 2 2 2 2 2 2 1 2 2 gt ε ε ε ε ε ε ε = + − + − + − … 3 3 2 6 gt H ε ≅ Similarly: ( ) ( ) 1 ln 1 gt v ε ε ε ε = + + − ( ) 2 3 1 2 3 gt ε ε ε ε ε ε − + − − … = + 1 gt 2 ε ≈ (d) H m ; 327 = 1 9.8 v ms− = 7.26 or m m m = −k k t = − Burn-out occurs at time m k T ε = So – the rocket equation (7.7.7) becomes dv dt = − m V (-) since V is oppositely directed to m v 15
- 105. dv Vk dt mk t = + − Thus ( ) 1 ln dt k V m k t C m k t = = − − − ∫ v V where C is a constant. + 1 Now, v so 0@ 0 t = = 1 ln C V m = Hence ( ) ln m kt m − = − v V , 0 m t k ε ≤ ≤ Let y be the displacement at the time t so ( ) 2 ln m kt y V dt C m − = − + ∫ Integrating the above expression by parts ( ) 2 ln m k t t dt y Vt Vk C m m k t − = − − + − ∫ ( ) 2 1 m k t m Vt V dt C m m k t − + − + − ∫ ln = − ( ) ( ) 2 ln m k t Vm Vt Vt m k t C m k − + + − + ln = − since 0 y = at t , 0 = 2 ln Vm m k C − = and we have ( ) ( ) ln m k t V y Vt m k t k m − = + − At burn-out m k t T ε = = so (a) ( ) ( ) ( ) 1 ln 1 mV y D k ε ε ε ε = = + − − (b) ε cannot exceed 1.0 although it can approach 1.0 for small payloads Thus ( ) max lim 1 mV y y k ε ε = = → 7.27 From eqn. 7.7.5, kv mv Vm − = − Since V is opposite in direction from v − = kv mv Vm + V m m k k = − − v v v V m k α = and since m 0 , m k α = − 16
- 106. m m dvm dv dm v m m m dt m dm dt d α α α α α − = = = = dv m v V dm dv m v V α α = − 0 1 m v m dm dv m v V α α = − ∫ ∫ 1 ln ln m v V m V α α α − = − 1 1 m v m V α α = − + 1 1 m m α α = − v V 7.28 From eqn. 7.7.5, mg mv Vm − = − V v since V is opposite in direction from v , − = mg mv Vm + − = mgdt mdv Vdm + dm dt = m so dm dt = m dm mg mdv Vdm m + − = g V dm m m = − + dv 0 e p v m m g V dv dm m m = − + ∫ ∫ mp =payload mass ( ) ln e p p m g m V m m = − + v m f p m m m m = − ≈ f m = fuel mass ln 1 f e f p m g v m V m m = + + ln 1 f e p m v g m m V V + = − m exp 1 f e p m v g m m V V m = − − 17
- 107. For V , e kv = 1 exp1 f p e m g m m k kv m = − − From chap. 2, Section 2.3 … 11 e km v s ≈ For 1 0.01m − = m and s 1 4 k = : ( )( ) 9.8 exp 4 1 1 11,000 .01 4 f p m m = − − − 77 f p m m = 7.29 We can use Equation 7.7.9 to calculate the final velocity attained by the ion rocket during the 100 hour burn. Assuming the rocket starts from rest (even if the ion rocket is turned on while in Earth orbit, the initial rocket speed v 4 0 10 0 c − ≈ ≈ . Thus … 0 ln p m v V m ≈ and 0 2 3 F p p p m m m m m m = + = + = p . The final rocket velocity is a little more than 10% c. ln3 0.1099 v V c ≈ = 4 36.4 0.1099 0.1099 L LY T y c LY yr = = = r 7.30 We again use Equation 7.7.9 … 0 ln ln ln 2 F p ion ion ion p p m m m v V V V m m + ≈ = = for the ion rocket. For the chemical rocket … ln F chem p m m v V m + ≈ p . Setting these two equations equal … ln ln 2 F p chem ion p m m V m + =V . Solving for mF … ( ) ( ) 5 4 0.1 10 10 500 2 2 2 1 ion chem c c V V F p p m m m − + = = = ≈ 0 which demonstrates the virtue of ejecting mass at high velocity! ---------------------------------------------------------------------------------------------------------- 18
- 108. 1 CHAPTER 8 MECHANICS OFRIGID BODIES: PLANAR MOTION 8.1 (a)For each portion of the wire having a mass 3 m and centered at , 2 2 b b − , ( ) 0,0 , and , 2 2 b b … y x b b 1 0 0 2 3 2 3 cm b m b m x m = − + + = 1 0 2 3 2 3 cm b m b m b y m 3 = + + = (b) ( ) 1 2 2 2 ds xdy b y dy = = − ( ) 1 2 2 2 0 1 b cm y y b y m ρ = − ∫ dy ( ) ( ) 1 2 2 2 2 2 0 2 2 1 4 y b y cm b y d b y y b ρ π ρ = = − − − = ∫ 4 y 3 cm b π = From symmetry, 4 x 3 cm b π = (c) The center of mass is on the y-axis. ( ) 1 2 2 2 ds xdy by dy = = ( ) ( ) 3 1 2 2 0 0 1 1 2 2 0 0 2 2 b b cm b b y by dy y dy y by dy y dy ρ ρ = = ∫ ∫ ∫ ∫ 3 y = 5 cm b (d) The center of mass is on the z-axis. ( ) 2 2 2 x y dz bzdz π + = dv r dz π π = = 1
- 109. 2 2 0 0 0 0 bb cm b b z bzdz z dz z bzdz zdz ρ π ρπ = = ∫ ∫ ∫ ∫ 2 3 cm z b = (e) The center of mass is on the z-axis. α is the half-angle of the apex of the cone. is the radius of the base at z = 0 and is radius of a circle at some arbitrary z in a plane parallel to the base. rD r tan r r b b z α = = − D , a constant ( ) 2 2 2 tan dv r dz b z dz π π α = = − 2 3 1 1 tan 3 3 b b 2 m r ρ π πρ = = D α ( ) ( ) 2 2 2 2 3 0 3 0 3 2 tan 3 2 1 tan 3 b b cm z b z dz z b b b ρ π α πρ α − = = − ∫ ∫ z bz z dz + 4 cm b z = 8.2 2 0 0 b cm b cx dx xdx x dx cxdx ρ ρ = = ∫ ∫ ∫ ∫ 2 3 cm b x = 8.3 The center of mass is on the z-axis. Consider the sphere with the cavity to be made of a (i) solid sphere of radius and mass a s M , with its center of mass at , and (ii) a solid sphere the size of the cavity, with mass 0 z = c M − and center of mass at 2 a z = − . The actual sphere with the cavity has a mass s c M m M = − and center of mass . cm z 1 0 s 2 c cm a M m z = − + M 3 4 3 s M a π ρ = , 3 4 3 2 c a M π ρ = 3 3 3 3 1 0 2 2 2 cm a a a a z a = − + − 2
- 110. 3 14 cm a z = 8.4 (a) 22 2 0 3 2 2 z i i i m b b R = = + + ∑ I m 2 6 z mb I = (b) ds rd dr θ = , sin R r θ = 2 z I R ds ρ = ∫ 2 2 4 0 4 sin r b z r I r rdr d π θ π θ ρ θ θ = = = =− = ∫ ∫ 4 2 4 4 sin 4 z b I d π π ρ θ θ − ∫ = 2 sin 2 sin 2 4 d θ θ θ θ = − ∫ 4 1 4 4 2 z b I ρ π = − 2 1 4 m b ρπ = ( ) 2 2 4 z mb I π π = − (c) 2 x ds hdx b dx b = = − Where the parabola intersects the line , y b = ( ) 1 2 x by b = = ± 2 4 2 2 b b y b b x x I x b dx bx dx b b ρ ρ − − = − = − ∫ ∫ 4 4 15 y I b ρ = 2 2 4 3 b b x m b dx b b ρ ρ − = − = ∫ 2 1 5 y I mb = (d) dv 2 RhdR π = h b z = − 3
- 111. 4 ( ) () 1 1 2 2 2 2 R x y bz = + = 1 2 1 2 b dR dz z = ( ) ( ) 1 1 2 2 2 0 1 2 2 b z b I R dv bz bz b z dz z ρ ρ π = = − ∫ ∫ ( ) 2 2 0 1 6 b z 5 I b bz z dz b πρ π = − = ∫ ρ ( ) ( ) 1 1 2 2 0 1 2 2 b b m dv bz b z z ρ ρ π = = − ∫ ∫ dz ( ) 3 0 1 2 b m b b z dz b πρ π = − = ∫ ρ 2 1 3 z I mb = (e) α is the half-angle of the apex of the cone. r is the radius of the base at z = 0 and is radius of a circle at some arbitrary z in a plane parallel to the base. D r , a constant tan R R b b z α = = − D ( ) 2 2 b z dv RhdR π π − = = R zdR b D ( ) Since R = , , and the limits of integration for correspond to z b b z R b − D R dR dz b = − D 0 R R = → D 0 = → ( ) ( ) 2 2 0 2 2 2 z b z R b z R I R dv z ρ ρ π − − = = ∫ ∫ D D R b b b ( ) 4 3 2 2 3 4 4 1 2 3 3 b z R 4 0 10 I b z b z bz z dz πρ = + − + − = ∫ D R b b πρ D dz b − D 2 1 3 m R b ρ π D = 2 3 10 z I mRD = 4
- 112. 5 8.5 Consider thesphere with the cavity to be made of a (i) solid sphere of radius and and mass a s M , with its center of mass at 0 z = , and (ii) a solid sphere the size of the cavity, with mass c M − and center of mass at 2 a z = − . The actual sphere with the cavity has a mass s c M m M = − and center of mass . cm z 3 3 3 4 4 7 4 3 3 2 8 3 s c a m M M a a π ρ π ρ π = − = − = ρ 8 7 s M m = and 1 7 c M m = From eqn. 8.3.2, s c I I I = + 2 2 2 2 5 5 2 s c a M a M I = + 2 2 2 2 8 2 1 31 5 7 5 7 4 70 a I m a m ma = − = 8.6 The moment of inertia about one of the straight edges is 2 z I R dv ρ = ∫ where 2 2 2 R x y = + . From Appendix F … 2 sin dv r dr d d θ θ φ = φ θ r R y x z 2 2 2 2 2 sin R x y r θ = + = Let a = radius of sphere 2 2 2 2 2 0 0 0 sin sin r a z r I r r dr d π π θ φ θ φ d θρ θ θ = = = = = = = ∫ ∫ ∫ φ 4 3 2 0 0 sin 2 r a z r I r drd π θ θ π ρ θ θ = = = = = ∫ ∫ 5 3 2 0 1 sin 10 z I a d π ρπ θ = ∫ θ 3 3 cos sin cos 3 d θ θ θ θ = − ∫ 5 2 30 z I a ρπ = 3 3 1 4 1 8 3 6 m a a π ρ π = = ρ 2 2 5 z I ma = 5
- 113. 6 8.7 For arectangular parallelepiped: dv hdxdy = h is the length of the box in the z-direction z x y 2b 2a ( ) 1 2 2 2 R x y = + ( ) 2 2 2 x a y b z x a y b I R dv x y hdxdy ρ ρ = = =− =− = = + ∫ ∫ ∫ ( ) 3 2 2 2 4 2 3 3 a z a b 2 I h bx dx hab a b ρ ρ − = + = ∫ + ( )( ) 2 2 4 a b h abh ρ ρ = = m ( ) 2 2 3 z m I a b = + For an elliptic cylinder: Again dv hdxdy = , and ( ) 1 2 2 2 R x y = + 2a 2b z y x On the surface, 2 2 2 2 1 x y a b + = 1 2 2 2 1 y b = ± − x a ( ) 1 2 2 2 1 2 2 2 1 2 2 1 y x a y b b z y b y x a b 2 I R dv x y hdxdy ρ = − = =− =− − = = + ∫ ∫ ∫ 3 1 2 2 2 2 3 2 2 2 2 1 2 1 3 b z b y y I h a a y b b ρ − = − + − ∫ dy ( ) ( ) 3 1 2 2 2 2 2 2 2 2 2 2 3 b b b b a a h b y dy b y y b b ρ − − dy + − = − ∫ ∫ From a table of integrals: ( ) ( ) ( ) 3 3 1 2 2 2 2 2 2 2 4 1 2 2 2 3 3 sin 4 8 8 y y b y dy b y b y b y b b − − = − + − + ∫ ( ) ( ) ( ) 1 3 1 2 4 2 2 2 2 2 2 2 1 2 2 2 sin 4 8 8 y b y b b y y dy b y b y b − − = − − + − + y ∫ ( ) 2 4 4 2 2 3 1 2 3 8 8 4 z a a b 2 I h b h ab a b b ρ π π ρ π = + = b + 6
- 114. 7 m h( ) ab ρπ = ( ) 2 2 4 z m I a b = + For an ellipsoid: , dv hdxdy = 2 2 2 R x y = + and on the surface, 2 2 2 2 2 2 1 x y z a b c + + = 1 2 2 2 2 2 1 x y z c a b = ± − − 2b 2a y x 1 2 2 2 2 2 2 2 1 x y h z c a b = = − − In the xy plane 2 2 2 2 1 x y a b + = 1 2 2 2 1 y x a b = ± − z ( ) 1 2 2 2 1 2 2 2 1 2 2 1 2 2 2 2 2 2 1 2 1 y x a y b b z y b y x a b x y I R dv x y c dxdy a b ρ ρ = − = =− =− − = = + − − ∫ ∫ 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 y b z y b c a y a y I a x x dx y a x a b b ρ = =− = − − + − − ∫ ∫ ∫ dx From a table of integrals: ( ) ( ) ( ) 3 1 1 2 4 2 2 2 2 2 2 2 2 1 2 2 sin 4 8 8 x k x k x k x x dx k x k x k − − = − − + − + ∫ ( ) ( ) 1 1 2 2 2 2 2 1 2 2 sin 2 2 x k x k x dx k x k − − = − + ∫ 2 4 2 2 2 2 2 2 2 1 1 8 2 b z b c a y a y I y d a b b ρ π π − = − + − ∫ y 2 2 4 4 2 2 4 2 2 1 4 b z b a y y y I ac y dy b b b ρπ − = − + + − ∫ ( ) 2 2 4 15 z I abc a b ρπ = + For an ellipsoid, 4 3 m abc ρ π = , so ( ) 2 2 5 z m I a b = + 7
- 115. 8 8.8 (See Figure8.4.1) Note that l l ′+ is the distance from 0′ to , defined as 0 d From eqn. 8.4.13, 2 cm k ll′ = ( ) 2 2 2 cm k l ll l l l l ′ ′ + = + = + 2 2 cm k l l + = d 2 d From eqn. 8.4.9b, 2 2 cm k k l = + k l 2 = From eqn. 8.4.6, 2 2 k gl π = D T 2 d g π = D T 8.9 Period of a simple pendulum: 2 a T M π = Period of real pendulum: 2 I T Mgl π = D (eqn. 8.4.5) Where moment of inertia I = distance to CM of physical pendulum l = distance to CM of bob a = radius of bob b = location of CM of physical pendulum: ( ) ( ) ( ) ( ) 2 2 2 2 a b m M m a m a b m l a a m M m M M − + − = = − + = − + − a b + 1 2 2 m m b l a M M a = − − Moment of inertia: ( ) ( ) ( ) 2 2 2 2 2 5 5 bob 2 I M m a M m b M m a b = − + − = − + 2 2 2 2 1 1 5 m b Ma M a = − + ( ) 2 2 2 1 1 1 3 3 rod m b I m a b Ma M a = − = − 2 2 2 2 2 1 1 1 1 5 3 bob rod m b m b I I I Ma M a M a ∴ = + = − + + − letting m M α = and b a β = 8
- 116. 9 ( ) () 1 2 2 2 2 2 1 1 1 1 5 3 2 1 2 2 Ma Mga α β α β π α αβ − + + − = − − D T (a) ( ) ( ) 2 2 2 1 1 1 1 1 5 3 1 12 1 2 2 T T α β α β α α αβ − + + − = ≈ − − D − to 1st order inα (b) 10 m g = 1 M kg = 1.27 a m = 5 b cm = 0.01 α = 0.0394 β = 1 1 0.9992 12 T α ≈ − = D T (actually 0.9994 using complete expression) 8.10 The period of the “seconds” pendulum is 2 2 2 I Mgl π = = T s The period of the modified pendulum is 2 2 20 I n M gl n π ′ ′ = = ′ ′ − T where I′, M ′ , l refer to parameters of pendulum with m attached and ′ is the number of seconds in a day. n ( 24 60 60 = × × ) 2 m I I ml ′ = + m Ml ml l M + ′ = where l is the distance of the attached mass from the pivot point. m m So ( ) 2 2 2 2 20 1 m m 2 I ml I n M gl Ml ml π π π − ′ + − = = ′ ′ + g ( ) 2 2 m m Mgl ml Ml ml g π + + = Thus ( ) ( ) 2 2 2 20 1 m m Ml ml g n Mgl ml π + − = + Or 2 2 1 40 1 m m l l n l gl α π α + − ≈ + 1 m M α = Solving for α gives the approximate result 9
- 117. 10 2 40 1 m m l n l m M l g α π =≅ − Letting l m; l ; we obtain 1.3 m = 1.0 = m 3 1.15 10 α − ≅ ⋅ 2 cm I ma ( l mass ⊥ = 8.11 (a) al in rim) 2 2 2 2 rim I ma ma ma ⊥ = + = 2 2 2 rim I a T mga g π π ⊥ ∴ = = 2 z x y cm 2 I I I ma = + = I = ( ) cm I⊥ = (b) 2 2 cm ma I ∴ = hence 2 2 2 3 2 2 rim ma I ma ma = + = 3 2 2 2 rim I a T mga g π π ∴ = = 8.12 cm mx mg T = − T G mg G a cm I aT ω = cm x aω = 2 2 5 cm I ma = 2 1 2 2 5 5 cm cm cm cm I x mx mg mg ma mg mx a a a ω = − = − = − 5 7 cm x g = x 8.13 When two men hold the plank, each supports 2 mg . When one man lets go: cm mg R mx − = and 2 cm l R I ω = From table 8.3.1, 2 12 cm ml I = 2 12 6 2 Rl R ml ml ω = ⋅ = 10
- 118. 11 3 2 cm l R x m ω = = 3 3 R mg R m R m − = = 4 mg R = 6 6 4 end R mg x l l ml m ω = = = 3 2 end x g = 8.14 For a solid sphere: 3 4 3 s M a π ρ = and 2 2 5 s s I M a = ( ) 2 2 2 5 cm s k a = For subscript c representing a solid sphere the size of the cavity, from eqn. 8.3.2: s c I I I = + 3 2 3 2 5 2 4 2 4 2 4 31 5 3 5 3 2 2 5 3 32 a a I a a a π ρ π ρ = ⋅ ⋅ π ρ = − 3 3 3 4 4 4 7 3 3 2 3 8 a a m a π ρ π ρ π = − = ⋅ ρ 2 2 2 31 8 31 5 32 7 70 cm I m = = ⋅ = ⋅ 2 a k a From eqn. 8.6.11, for a sphere rolling down a rough inclined plane: 2 2 sin 1 cm cm g k a x θ = + ( ) 2 2 2 2 1 1 cm s cm s k a k x a + + x = 2 7 1 5 5 31 101 1 70 70 + = = + 98 x = 101 s x 11
- 119. 12 8.15 Energy isconserved: a x m2 m1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 x m x m x I m gx m gx E + + + − = 1 2 2 1 2 0 I m xx m xx xx m gx m gx a + + + − = ( ) 1 2 1 2 2 m m g x I m m a − = + + 8.16 While the cylinders are in contact: 2 cos cm r mv f mg R r θ = = − R G mg G b a r a b = + , so 2 cos cm mg R a b θ mv = − + vcm From conservation of energy: ( ) ( ) 2 2 1 1 cos 2 2 cm a b mv I mg a b mg ω θ + = + + + From table 8.3.1, 2 1 2 I ma = cm v a ω = ( )( 2 2 2 2 1 1 1 1 cos 2 2 2 cm cm v mv ma mg a b a ) θ + = + − ( ) 2 4 1 cos 3 cm mv mg a b θ = − + When the rolling cylinder leaves, 0 R = : ( ) 4 cos 1 cos 3 mg mg θ θ = − 7 4 cos 3 3 θ = 1 4 cos 7 θ − = 8.17 2 mx N = 1 my N mg = − 12
- 120. 13 2 2 1 1 22 cm mv I mgy mgy ω + + = D cos 2 l x θ = , sin 2 l x θ θ = − , ( ) 2 cos sin 2 l x θ θ θ θ = + sin 2 l y θ = , cos 2 l y θ θ = , ( ) 2 sin cos 2 l y θ θ θ θ = − + 2 2 2 2 2 2 2 sin cos 2 2 cm l l v x y 4 l θ θ θ θ θ = + = − + = 2 12 ml I = , and ω θ = 2 2 2 2 1 1 sin sin 2 4 2 12 2 2 l ml l l m mg m θ g θ θ θ + + = D ( ) 2 sin sin 3 l g θ θ θ = − D ( ) 1 2 3 sin sin g l θ θ = − D θ ( ) ( ) 1 2 1 3 3 3 sin sin cos cos 2 2 g g l l g l θ θ θ θ θ − = − − = − D θ ( ) 2 3 3 cos sin sin sin cos 2 2 ml g g N mx l l θ θ θ θ = = − − + − D θ Separation occurs when 2 0 N = : 1 sin sin sin 0 2 θ θ θ − − = D , 1 2 sin sin 3 θ θ − = D 8.18 x R mx = y R mg my − = sin 2 l x θ = , cos 2 l x θ θ = , ( ) 2 sin cos 2 l x θ θ θ θ = − + cos 2 l y θ = , sin 2 l y θ θ = − , ( ) 2 cos sin 2 l y θ θ θ θ = − + 2 2 1 1 cos 2 2 2 cm l l mv I mg mg θ θ + + = 2 2 cm l v θ = , 2 12 ml I = 13
- 121. 14 ( ) 2 22 2 1 1 cos 2 4 2 12 2 m l ml l mg θ θ θ + = − ( ) 2 1 cos 3 l g θ θ = − ( ) 1 2 3 1 cos g l θ θ = − ( ) 1 2 1 3 3 3 1 cos sin sin 2 2 g g l l g l θ θ θθ − = − = θ ( ) ( ) 3 3 sin 1 cos cos sin 2 2 x ml g g R l l θ θ θ θ = − − + ( ) 3 sin 3cos 2 4 x mg R θ θ = − ( ) 3 3 cos 1 cos sin sin 2 2 y ml g g R mg l l θ θ θ θ = − − + 2 2 3 s cos cos 2 2 y mg R mg in θ θ θ = − − + ( ) 2 3cos 1 4 y mg R θ = − The reaction force constrains the tail of the rocket from sliding backward for : 0 x R 3cos 2 0 θ − 1 2 cos 3 θ − The rocket is constrained from sliding forward for 0 x R : 1 2 cos 3 θ − 8.19 sin cos mx mg mg θ µ θ = − − ( ) sin cos x g θ µ θ = − + x Since acceleration is constant, 2 1 2 x x t xt = + D : mg G f G θ ( ) 2 sin cos 2 gt x v t θ µ θ = − + D Meanwhile ( ) 2 2 cos 5 mg a I ma µ θ ω = = ω 5 cos 2 g a µ θ ω = 5 cos 2 g t a µ θ ω = 14
- 122. 15 The ball beginspure rolling when v aω = … ( ) 5 cos sin cos 2 g v v xt v g t a t a µ θ θ µ θ = + = − + = D D ( ) 2 2sin 7 cos v g t θ µ θ = + D At that time: ( ) ( ) ( ) 2 2 2 2 4 sin cos 2 2sin 7 cos 2 2sin 7 cos v v g x g g θ µ θ θ µ θ θ µ θ + = − + + D D ( ) ( ) 2 2 sin 6 cos 2 2sin 7 cos v x g θ µ θ θ µ θ + = + D 8.20 mx mg µ = x g µ = x gt µ = , and 2 1 2 x gt µ = 2 2 5 I ma mga ω ω µ = = − 5 2 g a µ ω = − 5 2 g t a µ ω ω = − D Slipping ceases to occur when v aω = … 5 2 gt a gt µ ω µ = − D 2 7 a t g ω = D 2 1 2 2 7 a x g g ω µ µ = D 2 2 2 49 a x g ω µ = D 8.21 Let the moments of inertia of and A B be 2 1 2 a a I M a = and 2 1 2 b b I M b = . The angular velocity of is A α while that of B is β α φ − + (remember that in two dimensions, angular velocity is the rate of change of an angle between an line or direction fixed to the body and one fixed in space). For rolling contact, lengths traveled along the perimeters of the disks and A B must be equal to the arc length traveled along the track C. ( )( ) 2 a b a b φ β α = = + −φ 15
- 123. 16 so that ( ) () 2 2 a b a b α φ = and + + ( ) ( ) 2 2 a a b b a b α β + = + After some algebra … the angular velocity of B is found to be … 2 B a b α ω β α φ = − + = For , we take moments about O and for A B we take moments about its center. Call T and T the components of the reaction forces tangent to and A B A B (the “upward-going” T acts on disk A B . The “downward-going” acts on disk A) A T Thus K TA A a I α − = (Torque on Disk A) ( ) 2 A B B B T b T b I I a b α β α φ − − = − − + = − (Torque on Disk B) ( )( ) 1 2 A B B B T T M a b M a α φ − = + − = α (Force on Disk B) Eliminate T and A B T ( ) 2 2 2 2 4 4 A B B K b I M a b a b 2 I α = + + Integrating this equation gives : ( ) 2 2 2 2 4 4 A B B Kt b I M a b a I b 2 α = + + Putting A ω α = at t gives t = D ( ) 2 2 2 2 4 4 A A B B b Kt b I M a b a I ω = + + D 2 Putting in values for A I and B I gives 2 4 3 2 2 A A B Kt a M M ω = + D Since the angular velocity of B is 2 B a b α ω β α φ = − + = , we have 2 3 2 2 2 B A A B a Kt b ab M M ω ω = = + D 16
- 124. 17 8.22 From section8.7 (see Figure 8.7.1), the instantaneous center of rotation is the point . If O x is the distance from the center of mass to O and 2 l is the distance from the center of mass to the center of percussion O′, then from eqn. 8.7.10 … 2 2 1 2 12 cm I l Ml x M M = = = 12 l 6 l x = 8.23 In order that no reaction occurs between the table surface and the ball, the ball must approach and recede from its collision with the cushion by rolling without slipping. Using a prime to denote velocity and rotational velocity after the collision: ˆ cm cm P v v m ′ = − ( ) ˆ cm cm P h a I ω ω ′ = − − The conditions for no reaction are vcm cm aω = and vcm cm aω ′ ′ = . ( ) ˆ ˆ cm cm P P a a h m I ω ω − = − − a ( ) 1 1 a h a m I = − 2 2 5 I ma = 2 2 7 5 5 ma h a ma = + = a 2 d a = 7 10 h d = 8.24 During the collision, angular momentum about the point 0 is conserved: m v l Iθ ′ ′ = D m v l I θ ′ ′ = D After the collision, energy is conserved: 2 1 cos cos 2 2 2 l l I mg m gl mg m gl θ θ θ ′ ′ ′ ′ − − = − − D D ( ) 2 2 2 1 1 cos 2 2 m v l ml g m I θ ′ ′ l ′ ′ = − + D D 17
- 125. 18 18 2 2 3 ml I m l ′′ = + ( ) 1 2 2 2 1 2 1 cos 2 3 ml ml v g m l m l m l θ ′ ′ ′ ′ = − + + ′ ′ D D 8.25 The effect of rod BC acting on rod AB is impulse + P̂′. The effect of AB on BC is P̂′ − . 1 ˆ ˆ mv P P′ = + 2 ˆ mv P′ = − ( 1 ˆ ˆ 2 l ) I P P ω ′ = − 2 ˆ 2 l I P ω ′ = − 2 12 ml I = ( 1 6 ˆ ˆ P P ml ω ′ = − ) 2 6 P̂ ml ω ′ = − 1 1 2 B l v v ω = − 2 2 2 B l v v ω = + ( ) ˆ ˆ 6 ˆ ˆ 2 B P P l v P m ml ′ + P′ = − − ˆ 6 ˆ 2 B P l v P m ml ′ ′ = − + − ( ) ˆ ˆ ˆ ˆ ˆ ˆ 3 3 P P P P P P ′ ′ ′ + − − = − − ′ 8 2 ˆ ˆ P P ′ = ˆ ˆ 4 ′ P P = 1 ˆ 5 4 P m = v 2 ˆ 4 P v m = − 1 ˆ 9 2 P ml ω = 2 ˆ 3 2 P ml ω = − ˆ B P v m = − -----------------------------------------------------------------------------------------------------------
- 126. CHAPTER 9 MOTION OFRIGID BODIES IN THREE DIMENSIONS 9.1 (a) ( ) 2 2 xx I y z dm = + ∫ dm dxdy ρ = and m a2 2 ρ = z ( ) 2 2 0 0 0 y a x a xx y x I y dx ρ = = = = = + ∫ ∫ dy 2a a x = 2 0 2 a a y dy ρ ∫ 2 4 2 3 3 xx ma I a ρ = = y ( ) 2 2 2 2 0 0 y a x a yy y x I x z dm x dxd ρ = = = = = + = ∫ ∫ ∫ y 3 0 8 3 a a dy ρ = ∫ 4 2 8 4 3 3 yy a m I ρ = = a From the perpendicular axis theorem: 2 5 3 zz xx yy ma I I I = + + 2 0 0 y a x a xy yx y x I I xydm xy dx ρ = = = = = = − = − ∫ ∫ ∫ dy 2 0 4 2 a a ydy ρ = −∫ 2 4 2 xy yx ma I I a ρ = = − = − 0 xz zx yz zy I I xzdm I = = − = = = ∫ I (b) 2 5 α = cos , 1 cos 5 β = , cos 0 γ = From equation 9.1.10 … 2 2 2 4 4 1 2 1 2 3 5 3 5 2 5 5 ma ma ma I = + + − 2 2 15 ma = (c) ( ) ( ) ˆ ˆ ˆ cos cos 2 5 i j i ĵ ω ω ω α β = + = G + From equation 9.1.29 … 1
- 127. 2 2 2 22 ˆ ˆ 3 2 2 5 5 5 5 ma ma ma ma L i j ω ω ω ω = ⋅ + − + − + ⋅ G 2 4 3 ( ) 2 ˆ ˆ 2 6 5 ma L i ω = + G j (d) From equation 9.1.32: ( ) 2 2 2 1 1 2 2 2 1 2 5 6 5 ma T L ma ω ω 5 ω ω = ⋅ = ⋅ + = G G 9.2 (a) ( ) ˆ ˆ ˆ 3 i j k ω ω = + + G ( ) 2 2 2 2 2 2 12 3 xx rod yy zz m a I I ma I I = = = = = a -a a -a -a a 0 xy I xydm = − = ∫ since, for each rod, either x or or both are 0. The same is true for the other products of inertia. y From equation 9.1.29: ( ) 2 2 ˆ ˆ ˆ 3 3 L ma i j k ω = ⋅ + + G From equation 9.1.32, ( ) 2 2 2 2 2 1 2 1 1 1 2 3 3 3 3 ma ma T 2 ω ω ω = + + = (b) From equation 9.1.10, with the moments of inertia equal to 2 2 3 ma and the products of inertia equal to 0: ( ) 2 2 2 2 2 2 cos cos cos 3 3 ma 2 I ma α β γ = + + = (c) For the x-axis being any axis through the center of the lamina and in the plane of the lamina, and the y-axis also in the plane of the lamina … xx yy I I = due to the similar geometry of the mass distributions with respect to the x- and y-axes. From the perpendicular axis theorem: zz xx yy I I I = + 2 zz xx I I = From Table 8.3.1, 2 6 zz ma I = 2 12 xx ma I = 2
- 128. 9.3 (a)From equation9.2.13, 2 tan 2 xy xx yy I I I θ = − From Prob. 9.1, 2 3 xx ma I = , 2 4 3 yy ma I = , 2 2 xy ma I = − tan 2 1 θ = 2 45 22.5 θ θ = = D D The 1-axis makes an angle of with the x-axis. 22.5D (b) From symmetry, the principal axes are parallel to the sides of the lamina and perpendicular to the lamina, respectively. 9.4 (a)From symmetry, the coordinate axes are principal axes. From Table 8.3.1: ω G z x ( ) ( ) 2 2 2 1 13 2 3 12 12 m I a a m = + = a ( ) 2 2 2 2 10 3 12 12 m I a a m = + = a ( ) 2 2 2 3 5 2 12 12 m I a a m = + = a y ( ) 1 2 3 ˆ ˆ ˆ 2 3 14 e e e ω ω = + + G From equation 9.2.5, 2 2 2 2 2 1 13 10 4 5 9 2 12 14 12 14 12 14 T ma ma ma 2 ω ω ω = ⋅ + ⋅ + ⋅ 2 2 7 24 ma ω = (b)From equation 9.2.4, 2 2 1 2 3 13 10 2 5 3 ˆ ˆ ˆ 12 12 12 14 14 14 L e ma e ma e ma2 ω ω ω = ⋅ + ⋅ + ⋅ G ( ) 2 1 2 ˆ ˆ ˆ 13 20 15 12 14 ma L e e ω = + + G 3 e ( )( ) ( )( ) ( )( ) ( ) ( ) 1 1 2 2 2 2 2 2 2 2 1 13 2 20 3 15 cos 1 2 3 13 20 15 L L ω θ ω + + ⋅ = = + + ⋅ + + G G ( ) 1 2 98 0.9295 11,116 = = 21.6 θ = D 9.5 (a)Select coordinate axes such that the axis of the rod is the 3-axis, its center is at the 3
- 129. origin, and ω G liesin the 1, 3 plane. ω G α 3 2 1 From Table 8.3.1, 2 1 2 12 ml = = I I , I3 0 = ( ) 1 3 ˆ ˆ sin cos e e ω ω α = + G α From equation 9.2.4, ( ) 2 2 1 1 ˆ ˆ sin 0 0 sin 12 12 ml ml L e e ω ω α α = + + = G L G is perpendicular to the rod, and 2 sin 12 ml ω L α = G (b) Since ω G is constant, from equations. 9.3.5 … 1 0 0 N = + ( )( ) 2 2 0 cos sin 12 ml N ω α ω α = + 3 0 0 N = + 2 2 2 ˆ sin cos 12 ml N e ω α α = G G N is perpendicular to the rod ( direction) and to 3 ê L G (e direction), and 1 ˆ 2 2 sin cos 12 ml ω N α α = G 9.6 From Problem 9.4 … ( ) 1 2 3 ˆ ˆ ˆ 2 3 14 e e e ω ω = + + G 2 1 13 12 I ma = , 2 2 10 12 I ma = , and 2 3 5 12 I ma = From eqns. 9.3.5: ( )( ) ( ) ( ) 2 2 2 1 0 2 3 5 10 14 12 28 ma ma N ω ω = + − = − 2 5 ( )( ) ( ) ( ) 2 2 2 2 0 3 1 13 5 14 12 28 ma ma N ω ω = + − = 2 4 ( )( ) ( ) 2 2 3 0 1 2 10 13 14 12 28 ma ma N 2 2 ω ω = + − = − ( ) 1 2 2 2 2 2 2 2 2 5 4 1 42 28 28 ma ma N ω ω = + + = G 9.7 Multiplying equations 9.3.5 by 1 ω , 2 ω , and 3 ω ,respectively … ( ) 1 1 1 1 2 3 3 2 0 I I I ω ω ω ω ω = + − ( ) 2 2 2 1 2 3 1 3 0 I I I ω ω ω ω ω = + − 4
- 130. ( ) 3 33 1 2 3 2 1 0 I I I ω ω ω ω ω = + − Adding, 0 1 1 1 2 2 2 3 3 3 I I I ω ω ω ω ω = + + ω ( ) 2 2 2 1 1 2 2 3 3 1 0 2 d I I I dt ω ω ω = + + From equation 9.2.5, [ ] rot d T dt = 0 rot T constant = Multiplying equations 9.3.5 by 1 1 I ω , 2 2 I ω , and 3 3 I ω , respectively: ( ) 2 1 1 1 1 1 2 3 3 2 0 I I I ω ω ω ω ω = + − I ( ) 2 2 2 2 2 1 2 3 1 3 0 I I I ω ω ω ω ω = + − I ( ) 2 3 3 3 3 1 2 3 2 1 0 I I I ω ω ω ω ω = + − I Adding, 0 2 2 2 1 1 1 2 2 2 3 3 3 I I I ω ω ω ω ω = + + ω ( ) 2 2 2 2 2 2 1 1 2 2 3 3 1 0 2 d I I I dt ω ω ω = + + From equation 9.2.4, ( ) 2 d L dt = 0 2 L constant = 9.8 From equations 9.3.5 for zero torque … ( ) 1 1 2 3 3 2 0 I I I ω ω ω = + − ( ) 2 2 1 3 1 3 0 I I I ω ω ω = + − From the perpendicular axis theorem, 3 1 2 I I I = + 1 1 2 3 1 0 I I ω ω ω = + 2 2 1 3 2 0 I I ω ω ω = − Multiplying by 1 1 I ω and 2 2 I ω , respectively: 1 1 1 2 3 0 ω ω ω ω ω = + 2 2 1 2 3 0 ω ω ω ω ω = − Adding, ( ) 2 2 1 1 2 2 1 2 1 2 d dt 0 ω ω ω ω ω ω = + = + 2 2 1 2 constant ω ω + = If 1 2 I I = , from the third of Euler’s equations … ( ) 3 3 0 I ω = 3 constant ω = 9.9 (a)From symmetry … 3 s I I = and 1 2 I I I = = From the perpendicular axis theorem, 3 1 2 I I I = + or 2 s I I = From eqn 9.5.8, ( ) 2 1 cos ω α − Ω = 5
- 131. For , 45 α =D 2 ω Ω = For 2 1s π ω = , 1 2 2 1.414 T s s π = = = Ω ω G 45o 3 2 1 1 T is the period of precession of ω G about 3 ê . , From equation 9.6.12 1 1 2 2 2 2 2 2 2 2 2 1 1 1 cos 1 1 1 2 s I I φ ω α ω = + − = + − 5 2 φ ω = 2 2 0.4 0.632 T s s π φ = = = G is the period of wobble of about 2 T 3 ê L . (b) 2 2 2 1 2 17 12 16 16 12 m a m I I I a = = = + = ⋅ a ( ) 2 2 2 3 2 12 12 s m m I I a a = = + = ⋅ a From equation 9.5.8, 2 1 15 1 17 17 2 2 16 ω ω − = Ω = 1 2 17 2 1.603s 15 T s π = = = Ω From equation 9.6.12, 1 2 2 2 2 2 2 16 1 1 1 17 2 φ ω ⋅ = + − 1.5072 φ ω = 2 2 1 0.663s 1.5072 T s π φ = = = 9.10 From equation 9.6.10, tan tan s I I θ α = . Since s I I , θ α and the angle between the axis of rotation ( ) ω G and the G invariable line ( is ) L α θ − . From your favorite table of trigonometric identities … ( ) tan tan tan 1 tan tan α θ α θ α θ − − = + 6
- 132. ( ) 2 1 tan tan 1tan s s I I I I α α θ α − − = + ( ) 1 2 tan tan tan s s I I I I α α θ α − − − = + 9.11 α θ − is a maximum for s I I a maximum. For 2 s I I = , ( ) 2 tan tan 2 tan α α θ α − = + ( ) ( ) ( ) 2 2 2 2 2 2 2 tan 1 2tan 2 tan tan 2 tan 2 tan 2 tan d d α θ α α α α α α − − = − = + + + At the maximum, ( ) 2 tan 0 2 tan tan d d α θ α α − = = − 1 tan 2 54.7 α − = = D ( ) ( ) 2 1 1 2 2 2 tan 1 4 1 2 2 α θ − − ≤ = + 1 2 tan 19.5 4 α θ − − ≤ = D 9.12 (a) From Problem 9.10, for s I I , the angle between ω G and L G is … ( ) 1 2 tan tan tan s s I I I I α α θ α − − − = + From Problem 9.9(a), 2 s I I = and tan tan 45 1 α = = D ( ) 1 1 1 1 tan tan 18.4 2 1 3 α θ − − − = = = + D (b) From Prob. 9.9(b), 2 2 12 s ma I = ⋅ and 2 17 16 12 ma I = ⋅ 1 1 17 2 15 16 tan tan 17.0 17 49 2 16 α θ − − − − = = = + D 7
- 133. 9.13 From Example9.6.2, 0.2 α ′′ = and 1.00327 s I I = It follows from equation 9.6.10 that, for s I I , the angle between ω G and is: L G 1 tan tan s I I α θ α α − − = − For α so small, tanα α ≈ , and 1 tan tan tan s s s I I I I I I α α α − ≈ ≈ s s s I I I I I α α θ α α − − ≈ − = .00327 0.2 0.00065 1.00327 α θ − ≈ = arcsec 9.14 From Table 8.3.1, 2 2 ma = s I and 2 4 ma I = 3 ê ω ω = D G and 3 ˆ ˆ 4 ma P e ω = G 3 P G 0 ω G ω G Selecting the 2-axis through the point of impact, during the collision … 2 2 3 ˆ ˆ ˆ 4 4 ma ma Ndt r p ae e e1 ˆ ω ω = × = × = ∫ G G G G The only non-zero component of is . During the collision N 1 N 2 0 ω = , so integrating the first of Euler’s equations 9.3.5 … 2 1 1 N dt I 1 ω = ∫ Immediately after the collision, 2 1 2 4 4 ma ma ω ω ω = = 1 ( 3 ω still is equal to ω , and 2 0 ω = ) G After the collision, . From Prob. 9.8, with 0 N = 1 2 I I = , for 0 N = G , 3 constant ω = and . 2 2 1 2 constant ω ω + = ( ) 1 2 2 2 1 2 3 tan constant ω ω α ω + = = Using the values of i ω immediately after the collision … ( ) 1 2 2 0 tan 1 ω α ω + = = 45 α = D 8
- 134. From equation 9.5.8,with ( ) 1 1 1 3 ˆ ˆ e e e ω ω ω ω = + = + D ˆ G G : ( )( )( ) 2 1 2 cos45 ω ω Ω = − = D From equation 9.6.12 … ( )( ) 1 2 2 2 2 1 2 1 cos 45 5 φ ω ω = + − = D 9.15 ( ) 1 1 2 2 3 3 ˆ ˆ ˆ N c c e e e ω ω ω ω = − = − + + G G Say the 3 axis is the symmetry axis, 3 s I I = , 1 2 I I I = = For the third component of equations 9.3.5 … 3 3 0 s c I ω ω − = + ω G N G α 3 1 3 3 s c I ω ω = − ( ) 3 3 ln s c t I ω ω = − D ( ) 3 3 s ct I e ω ω − = D 2 For the first two components of equations 9.3.5 … ( ) 1 1 2 3 s c I I ω ω ω ω I − = + − , and ( ) 2 2 1 3 s c I I I ω ω ω ω − = + − Rearranging terms and multiplying by 1 ω and 2 ω respectively … ( ) 2 1 1 1 1 2 3 0 s c I I I ω ω ω ω ω ω + + − = ( ) 2 2 2 2 1 2 3 0 s c I I I ω ω ω ω ω ω + − − = Adding, ( ) 2 2 1 1 2 2 1 2 0 c I ω ω ω ω ω ω + + + = ( ) ( ) 2 2 1 2 2 2 1 2 1 2 d c dt I ω ω ω ω + = − + ( ) 2 2 1 2 2 2 1 2 2 ln c t I ω ω ω ω + = − + D ( ) ( ) 2 2 2 2 2 1 2 1 2 ct I e ω ω ω ω − + = + D ( ) ( ) 1 1 2 2 2 2 2 2 1 2 1 2 ct I e ω ω ω ω + = + D ( ) ( ) 1 1 1 2 2 2 1 2 3 tan tan s ct I I e ω ω α α ω − − + = = D 9
- 135. For s I I , 1 1 0 s I I − and α decreases with time. 9.16 (a) From table 8.3.1… 2 2 s ma I = about the symmetry (spin) axis. 2 4 cm disk a I m = about an axis through the center of mass of the disk and ⊥ to the spin axis. From the parallel axis theorem, equation 8.3.21, … S G a / 2 a 45o z y x 2 2 4 2 disk ma a I m = + about the pivot point. 2 2 3 rod m a I = moment of inertia of the rod about the pivot point. Thus … 2 2 2 2 2 4 2 2 3 3 disc rod ma a m a I I I m m = + = + + = a From equation 9.7.10, ( ) 1 2 2 2 4 cos 2 cos s s I S I S MglI I θ φ θ + − = 1 2 2 2 1 4 2cos s s I I Mgl S S I I I θ φ θ = ± − cos 2 2 3 2 2 4 3 s ma I ma I = = , 1 cos cos45 2 θ = = D 1 2 9 9 2 2 2 8 80 3 m a m Ml cm ma I a − + = = = ( ) ( ) 1 2 2 2 2 1 2 1 1 3 3 9 1 60 900 900 4 980 4 4 80 2 2 2 cm rpm rpm cm s s φ π − − = ± − or 15.1rpm φ = 939rpm 10
- 136. (b) From equation9.7.12, 2 2 4 s I S Mg ≥ lI 4 3 s I I = , 1 2 3 3 3 2 2 2 20 2 s m a Ml cm ma I a − = = = ( ) 2 2 1 2 1 4 3 4 60 4 980 20 3 2 s s Mgl I rpm S cm cm s I I s π − − − ≥ ⋅ = ⋅ 267 S r ≥ pm 9.17 (Neglecting the pencil head) From Table 8.3.1, 2 2 2 2 8 s b m mb I = = The moment of inertia of the pencil about an axis thru its center of mass and ⊥ to its symmetry axis from equation 8.3.26 ... 2 2 2 2 4 12 16 12 c b a b I m m = + = + 2 a From the parallel axis theorem, 2 2 2 2 16 c a b I I m m = + = + 3 a a b From equation 9.7.12, 2 2 4 s MglI S I ≥ 2 2 2 2 4 4 2 16 3 64 a b a mg m S m b + ≥ ( ) ( )( ) 1 1 2 2 2 2 2 2 1 2 2 980 20 16 16 1 20 2 16 3 2 16 3 1 ga b a S r b − ≥ + = + ad s ⋅ 1 18,294 2910 S rad s rps − ≥ ⋅ = 9.18 s rim spokes hub I I I I = + + 2 2 2 rim rim ma I m a = = 3 2 3 12 spokes spokes a ma I m = = , assuming the spokes to be thin rods 0, assuming its radius is small hub I = 11
- 137. From the perpendicularaxis theorem, equation 8.3.14, 2 s I I = From equation 9.10.14, ( ) 2 2 s s Imga S I I ma + 1 2 1 2 2 2 1 6 2 1 7 12 mga g S a ma ma = + 9 For rolling without slipping, v aS = 1 2 1 2 1 1 30 6 32 6 2 3.55 19 19 12 ga v ft s ft s − − × × = ⋅ = × ⋅ If the spokes and hub are neglected, 2 2 s ma I = 1 2 1 2 2 2 1 2 3 2 mga g S a ma ma = + 1 2 1 2 1 1 30 32 2 3.65 3 3 12 ga v ft s ft s − − × = ⋅ = ⋅ × 9.19 From equations 9.3.5 with 0 N = G … ( ) 1 1 3 2 2 3 0 I I I ω ω ω + − = ( ) 2 2 1 3 1 3 0 I I I ω ω ω + − = ( ) 3 3 2 1 1 2 0 I I I ω ω ω + − = Differentiating the first equation with respect to t: ( )( ) 1 1 3 2 2 3 2 3 0 I I I ω ω ω ω ω + − + = From the second and third equations: ( ) 3 1 2 1 2 I I I 3 ω ω ω − = , and ( ) 1 2 3 1 3 I I I 2 ω ω ω − = 12
- 138. ( ) ( )( ) 1 2 3 1 2 2 1 1 3 2 1 2 1 3 3 2 0 I I I I I I I I I ω ω ω − − + − + = ω ω 1 1 1 0 K ω ω + = , ( )( ) ( )( ) 3 2 2 1 3 2 3 1 2 2 1 2 3 1 3 1 2 I I I I I I I I K I I I I ω ω − − − − = − + (a) For 3 ω large and 2 0 ω = , so 1 0 K 1 1 1 0 K ω ω + = is the harmonic oscillator equation. 1 ω oscillates, but remains small. Motion is stable for initial rotation about the 3 axis if the 3 axis is the principal axis having the largest or smallest moment of inertia. (b) For 3 0 ω = and 2 ω large, 1 0 K so 1 1 1 0 K ω ω + = is the differential equation for exponential growth of 1 ω with time: 1 1 k t k t Ae Be ω − = + 1 = . Motion is unstable for the initial rotation mostly about the principal axis having the median moment of inertia. 9.20 0 since either xy i i i i I m x y = ∑ i x or is zero for all six particles. Similarly, all the other products of inertia are zero. Therefore the coordinate axes are principle axes. i y (0,0,c) (a,0,0) ( ) ( ) ( ) 2 2 2 2 2 2 0 0 xx i i i i I m y z m b b c c = + = + + + − + + − ∑ ( ) 2 2 2 xx I m b c = + (0,b,0) ( ) 2 2 2 yy I m a c = + ( ) 2 2 2 zz I m a b = + 2 2 2 2 2 2 0 0 2 0 0 0 0 b c I m a c a b + = + + I ( ) ( ) ( ) 1 2 3 1 1 2 2 2 2 2 2 2 2 ˆ ˆ ˆ a ae be ce b a b c a b c c ω ω ω = + + = + + + + G G I G From equation 9.1.28, L Iω = ( ) 2 2 2 2 1 2 2 2 2 2 2 0 0 2 0 0 0 0 b c a m L a c a b c a b c ω + b = + + + + G 13
- 139. ( ) ( ) () ( ) 2 2 2 2 1 2 2 2 2 2 2 2 a b c m L b a b c c a b ω + = + + + + G a c From equation 9.1.32, 1 2 ω = ⋅ T L G G ( ) [ ] ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 1 2 2 a b c m T a b c b a b c c a b ω a c + = + + + + ( ) ( ) ( 2 2 2 2 2 2 2 2 2 2 2 2 2 m T a b c b a c c a b a b c ω = + + + + + + ) + ( ) 2 2 2 2 2 2 2 2 2 2 2m T a b a c a b c ω = + + + b c + 9.21 For Problem 9.1: 2 1 1 0 3 2 1 4 0 2 3 5 0 0 3 I ma − = − I , 2 1 5 0 ω ω = G 2 1 1 0 3 2 2 1 4 0 1 2 3 5 0 5 0 0 3 ma L I ω ω − = = − G I G 2 2 2 1 1 3 2 6 1 4 1 1 2 3 3 5 5 6 0 0 0 ma ma ma ω ω − = − + = = 2 5 ω ( ) 2 1 1 1 2 1 0 2 2 2 5 6 5 0 ma T L ω ω ω = ⋅ = G G ( ) 2 2 2 2 2 2 0 60 15 ma ma ω ω = + + = 14
- 140. For Problem 9.4: 2 130 0 0 10 0 12 0 0 5 ma I = I , 1 2 14 3 ω ω = G 2 13 0 0 1 0 10 0 2 12 14 0 0 5 3 ma L I ω ω = = G I G 2 13 20 12 14 15 ma ω = ( ) 2 13 1 1 1 2 3 20 2 2 14 12 14 15 ma T L ω ω ω = ⋅ = G G ( ) ( ) 2 2 2 2 7 13 40 45 24 14 24 ma ma ω ω = + + = 9.22 Since the coordinate axes are axes of symmetry, they are principal axes and all products of inertia are zero. ω G 2c 2b 2a z x From Table 8.3.1, ( ) ( ) ( ) 2 2 2 2 2 2 12 3 xx m m I b c b c = + = + ( ) 2 2 3 yy m I a c = + , ( ) 2 2 3 zz m I a b = + y 2 2 2 2 2 2 0 0 0 0 3 0 0 b c m I a c a b + = + + I ( ) ( ) ( ) 1 2 3 1 2 2 2 2 2 2 2 ˆ ˆ ˆ a ae be ce b a b c a b c c ω ω ω = + + = + + + + G G I G From equation 9.1.28, L Iω = ( ) 2 2 2 2 1 2 2 2 2 2 2 0 0 0 0 3 0 0 b c a m L a c a b c a b c ω + b = + + + + G ( ) ( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 2 2 3 a b c m L b a b c c a b ω + = + + + + G a c From equation 9.1.32, 1 2 T ω = ⋅ T L G G 15
- 141. ( ) [ ] () ( ) ( ) 2 2 2 2 2 2 2 2 2 2 1 2 3 a b c m T a b c b a b c c a b ω a c + = + + + + ( ) ( ) ( ) ( 2 2 2 2 2 2 2 2 2 2 2 2 2 6 m T a b c b a c c a b a b c ω = + + + + + + ) + ( ) ( ) 2 2 2 2 2 2 2 2 2 2 3 m T a b a c a b c ω = + + + b c + With the origin at one corner, from the parallel axis theorem: ( ) ( ) ( ) 2 2 2 2 2 2 4 3 3 xx m b c m I m b c b c + = + + = + ( ) 2 2 4 3 yy m I a c = + , ( ) 2 2 4 3 zz m I a b = + xy I xydm xy dV ρ = − = − ∫ ∫ 2 2 2 2 2 0 0 0 8 x a y b z c xy x y z I xydxdydz a b c ρ ρ = = = = = = = − = − ∫ ∫ ∫ , so ( )( )( ) 2 2 2 8 m a b c abc ρ ρ = = xy I mab = − xy I mac = − , yz I mbc = − ( ) ( ) ( ) 2 2 2 2 2 2 4 3 4 3 4 3 b c ab ac I m ab a c bc ac bc a b + − − = − + − − − + I 9.23 (See Figure 9.7.1) ( ) ( ) ( ) z x y z z z z L L L L ′ ′ = + + ′ ( ) ( ) sin cos sin sin cos z y z s L L L I I S θ θ φ θ θ ′ ′ = + = + θ 9.24 (See Figure 9.7.1) If the top precesses without nutation, it must do so at θ θ = D whereV ( ) θD is a minimum of ( ) V θ … ( ) 0 dV d θ θ θ θ = = D ( ) ( ) 2 2 cos cos 2 sin z z L L V I θ mgl θ θ θ ′ − = + (See equation 9.8.7) 16
- 142. ( ) () 2 2 3 cos cos sin cos sin 0 sin z z z z z L L L L L dV mgl d I θ θ θ θ θ θ θ θ θ ′ ′ ′ = − − + − = − D D D D D D D = let cos z z L L γ θ ′ = − D Then and solving for 2 2 cos sin sin 0 z L mglI γ θ γ θ θ ′ − + D D 4 = D γ 1 2 2 2 sin 4 cos 1 1 2cos z z L mglI L θ θ γ θ ′ ′ = ± − D D D now is large since z L ′ ψ is large and the precession rate is small, so we can expand the term in square root above and use the (-) solution since γ must be positive … 2 2 sin 2 cos 1 1 2cos z z L mglI L θ θ γ θ ′ ′ ≈ − + D D D 2 sin cos z z z mglI L L L θ γ θ ′ ′ = − ≈ D D From equations 9.7.2, 9.7.5 and 9.7.7 … ( ) 2 2 0 0 sin cos cos z s L I I 0 φ θ φ θ ψ = + + θ 0 and … 0 0 cos ( cos )cos z s L I θ φ θ ψ θ ′ = + so … ( ) 2 2 2 sin cos cos cos cos s s s s I I I I I γ θ θ φ ψ θ φ θ ψ = + + − − D D D D θD ( ) 2 2 2 sin sin sin cos z s mglI mglI I L I θ θ φ θ ψ φ θ ′ = ≈ = + D D D D and since 0 ψ , we can ignore the φ term in the denominator and we have … s mgl I φ ψ ≈ Hence, if ψ large, 1 0 θ θ θ = = and 1 s mgl I θ θ φ ψ = = the top will precess without nutation at 1 θ θ = D the place where ( ) min θ = V . 9.25 1 2 3 2 a b b + = 1 2 1 sin30 2 b b = = D 1 2 3 a b = 2 3 a b = 2 2 2 a H b2 = + 2 2 2 2 2 2 2 2 3 3 a a H a b a = − = − = 17
- 143. 2 3 H a = Thus, thecoordinates of the 4 atoms are: Oxygen: ( ) 0,0, H Hydrogen: 1 1 , ,0 ; , ,0 2 2 a a b b − − − Carbon: ( ) 2 ,0,0 b (a) The axes 1, 2, 3 are principal axes if the products of inertia are zero. 1 1 0 0 2 2 xy i i i a a I m x y b b − − = = + − + − + ≡ ∑ 0 0 0 0 0 0 yz I − = + + + ≡ The 1,2,3 axes are principal axes. 0 0 0 0 0 xz I − = + + + ≡ (b) Find principal moments ( ) 2 2 2 2 2 1 67 16 1 1 2 2 6 xx i i i a a 2 I I m y z H = = + = + + = ∑ a ( ) ( ) ( ) 2 2 2 2 2 2 2 1 89 16 1 1 12 6 yy i i i 2 1 2 I I m x z H b b b = = + = + + + = ∑ a ( ) 2 2 2 2 2 2 2 3 1 1 14 0 2 2 zz i i i a a 2 2 3 I I m x y b b b = = + = + + + + + = ∑ a 2 67 0 0 6 89 0 0 6 14 0 0 3 I a = (c) 3 1 2 I I I therefore rotation about the1-axis is unstable (see discussion Sec. 9.4) ---------------------------------------------------------------------------------------------------------- 18
- 144. CHAPTER 10 LAGRANGIAN MECHANICS 10.1Solution … ( ) ( ) ( ) 0, x t x t t αη = + ( ) ( ) ( ) 0, x t x t t αη = + where ( ) 0, sin x t t ω = and ( ) 0, cos x t t ω ω = 2 1 2 T m = x 2 2 1 1 2 2 V kx mω = = 2 x so: ( ) ( ) 2 1 2 2 2 2 t t m J x α ω = − ∫ x dt ( ) ( 2 1 2 2 2 cos sin 2 t t m t t ω ω αη ω ω αη = + − + ∫ ) dt ( ) ( ) ( ) ( ) 2 2 1 1 2 2 2 2 2 2 2 cos sin cos sin 2 2 t t t t t m m J t t dt m t t dt dt α α ω ω ω αω η ω ωη ω η ω η = − + − + − ∫ ∫ ∫ 2 1 t Examine the term linear in α : ( ) ( ) 2 2 2 2 1 1 1 1 cos sin cos sin sin 0 t t t t t t t t dt t t tdt tdt η ω ωη ω η ω ωη ω ωη ω − = + − ∫ ∫ t t ≡ ∫ (1st term vanishes at both endpoints: ( ) ( ) 2 1 0 t t η η = = ) so ( ) ( ) 2 2 1 1 2 2 2 1 1 cos2 2 2 t t t t J m tdt m α ω ω α η ω η = + − ∫ ∫ 2 2 dt [ ] ( ) 2 1 2 2 2 2 2 1 1 1 sin 2 sin 2 4 2 t t m t t m ω ω ω α η ω η = − + − ∫ dt which is a minimum at 0 α = 10.2 V mgz = ( ) 2 2 2 1 2 T m x y z = + + ( ) 2 2 2 1 2 L T V m x y z mgz = − = + + − L mx x ∂ = ∂ , L my y ∂ = ∂ , L mz z ∂ = ∂ d L mx dt x ∂ = ∂ , d L my dt y ∂ = ∂ , d L mz dt z ∂ = ∂ 1
- 145. 0 L L x y ∂∂ = = ∂ ∂ , L mg z ∂ = − ∂ From equations 10.4.5 … 0 i i L d L q dt q ∂ ∂ − = ∂ ∂ , 0 mx = mx const = , 0 my = my const = mz mg = − 10.3 Choosing generalized coordinate x as linear displacement down the inclined plane (See Figure 8.6.1), for rolling without slipping … x a ω = 2 2 2 2 2 1 1 1 1 2 7 2 2 2 2 5 10 x T mx I mx ma mx a ω = + = + = 2 For V at the initial position of the sphere, 0 = V m sin gx θ = − 2 7 sin 10 L T V mx mgx θ = − = + 7 5 L mx x ∂ = ∂ , 7 5 d L mx dt x ∂ = ∂ sin L mg x θ ∂ = ∂ 7 sin 5 mx mg θ = 5 sin 7 x g θ = From equations 8.6.11 - 8.6.13 … 5 sin 7 cm x g θ = 10.4 (a) For x the distance of the hanging block below the edge of the table: 2 2 1 1 2 2 x mx m = + = 2 x x T m and V mg = − x L T 2 V mx mgx = − = + 2 L mx x ∂ = ∂ , 2 d L mx dt x ∂ = ∂ L mg x ∂ = ∂ 2mx mg = 2 g x = 2
- 146. (b) 2 2 1 22 m x m x m x2 T m ′ ′ = + = + and 2 2 2 x x m gx m g mgx x l l ′ ′ = − − = − − g V m 2 2 2 2 m m L T V m x mgx x l ′ ′ = − = + + + g ( ) 2 L m m x x ∂ ′ = + ∂ , ( ) 2 d L m m x dt x ∂ ′ = + ∂ L m mg x g x l ′ ∂ = + ∂ ( ) 2 m g m m x mg x l ′ ′ + = + 2 g ml m x x l m m ′ + = ′ + 10.5 The four masses have positions: m1 : 1 2 x x + m2 : 1 1 l x x2 − + m3 : 2 2 l x x3 − + m4 : 2 2 3 l x l x3 − + − ( ) ( ) 2 1 1 2 V g m l x x = − − + 1 1 2 m x x + + ( ) ( ) 3 2 2 3 4 2 2 3 3 m l x x m l x l x + − + + − + − ( ) ( 2 2 1 1 2 2 1 2 1 2 T m x x m x x = + + − + ) ) ( ) ( 2 2 3 2 3 4 2 3 m x x m x x + − + + − − ( ) ( ) ( ) 2 2 1 1 2 2 1 2 3 2 3 1 1 1 2 2 2 L T V m x x m x x m x x = − = + + − + + − + 2 ( ) ( ) ( ) ( ) 2 4 2 3 1 1 2 2 1 2 3 4 3 3 4 1 . 2 m x x gx m m gx m m m m gx m m cons + − − + − + + − − + − + t ( ) ( 1 1 2 2 1 2 1 L m x x m x x x ∂ = + − − + ∂ ) ) 2 = + ( ) ( 1 2 1 1 2 m m x m m x + − ( ) ( 1 2 1 1 2 1 d L m m x m m x dt x ∂ = + + − ∂ ) 2 ( ) 1 2 1 L g m m x ∂ = − ∂ ( ) ( ) ( ) 1 2 1 1 2 2 1 2 m m x m m x g m m + + − = − 3
- 147. ( ) () ( ) ( 1 1 2 2 1 2 3 2 3 4 2 3 2 L m x x m x x m x x m x x x ∂ = + + − + − − + − − − ∂ ) ( ) 1 2 3 4 2 L g m m m m x ∂ = + − − ∂ ( ) ( ) ( ) ( ) 1 2 1 1 2 3 4 2 4 3 3 1 2 3 4 m m x m m m m x m m x g m m m m − + + + + + − = + − − ( ) ( 3 2 3 4 2 3 3 L m x x m x x x ∂ = − + − − − ∂ ) ( ) 3 4 3 L g m m x ∂ = − ∂ ( ) ( ) ( ) 4 3 2 3 4 3 3 4 m m x m m x g m m − + + = − For , m m , , and 1 m m = 2 4 = 3 2 m = m 4 m m = : , 1 2 5 3 3 mx mx mg − = − ( ) 1 2 3 5 x x g = − 1 2 3 3 8 2 mx mx mx mg − + − = , 2 3 3 mx mx mg − + = ( ) 3 2 1 3 x x g = + Substituting into the second equation: 2 2 2 9 9 1 1 8 2 5 5 3 3 x g x x g − + + − − = g 2 88 8 15 15 x g = , 2 11 g x = 1 3 10 6 5 11 11 x g g = − = − 3 1 12 4 3 11 11 x g g = = Accelerations: : 1 m 1 2 5 11 x x g + = − : 2 m 1 2 7 11 x x g − + = : 3 m 2 3 3 11 x x g − + = : 4 m 2 3 5 11 x x g − − = − 10.6 See figure 10.5.3, replacing the block with a ball. The square of the speed of the ball is calculated in the same way as for the block in Example 10.5.6. 2 2 2 2 cos v x x xx θ ′ ′ = + + The ball also rotates with angular velocity ω so … 4
- 148. 2 2 1 11 2 2 2 T mv I M ω = + + 2 x For rolling without slipping, x a ω ′ = . 2 2 5 I ma = ( ) 2 2 2 1 1 2 cos 2 5 T m x x xx mx M θ ′ ′ ′ = + + + + 2 1 2 x sin V mgx θ ′ = − , for V at the initial position of the ball. 0 = 2 2 1 7 2 cos 2 5 L T V m x x xx θ ′ ′ = − = + + 2 1 sin 2 Mx mgx θ ′ + + 7 cos 5 L mx mx x θ ∂ ′ = + ′ ∂ , 7 cos 5 d L mx mx dt x θ ∂ ′ = + ′ ∂ sin L mg x θ ∂ = ′ ∂ 7 cos sin 5 mx mx mg θ θ ′+ = ( ) 5 sin cos 7 x g x θ θ ′ = − cos L mx mx Mx x θ ∂ ′ = + + ∂ , ( ) cos d L m M x mx dt x θ ∂ ′ = + + ∂ 0 L x ∂ = ∂ ( ) cos 0 m M x mx θ ′ + + = ( ) 2 5 5 sin cos cos 0 7 7 m M x mg mx θ θ θ + + − = ( ) 2 5 sin cos 5 cos 7 mg x m m M θ θ θ = − + 10.7 Let x be the slant height of the particle … 2 2 2 v x x 2 ω = + t θ ω = x ( ) 2 2 2 1 1 2 2 T mv m x x 2 ω = = + sin sin V mgx mgx t θ ω = = ( ) 2 2 1 sin 2 L T V m x x mgx t ω ω = − = + − L mx x ∂ = ∂ , d L mx dt x ∂ = ∂ 2 sin L mx mg t x ω ω ∂ = − ∂ 2 sin mx mx mg t ω ω = − 2 sin x x g t ω ω − = − 5
- 149. The solution tothe homogeneous equation , is 2 0 x x ω − = t t x Ae Be ω ω − = + Assuming a particular solution to have the form sin p x C t ω = , 2 2 sin sin sin C t C t g t ω ω ω ω ω − − = − 2 2 g C ω = 2 sin 2 t t g x Ae Be t ω ω ω ω − = + + At time t , 0 = x x = D and 0 x = x A B = + D 0 2 g A B ω ω ω = − + 2 1 2 2 g A x ω = − D 2 1 2 2 g B x ω = + D ( ) ( ) 2 2 1 1 2 2 2 2 t t t t g g x x e e e e t ω ω ω ω sin ω ω ω − − = + − − + D From Appendix B, we use the identities for hyperbolic sine and cosine to obtain 2 2 cosh sinh sin 2 2 g g x x t t t ω ω ω ω ω = − + D 10.8 In order that a particle continues to move in a plane in a rotating coordinate system, it is necessary that the axis of rotation be perpendicular to the plane of motion. For motion in the xy plane, k̂ ω ω = G . ( ) ( ) ˆ ˆ ˆ ˆ ˆ v v r ix jy k ix jy ω ω ′ ′ = + × = + + × + G G G G G ( ) ( ˆ ˆ v i x y j y x) ω ω = − + + ( ) 2 2 2 2 1 2 2 2 2 m T mv v x x y y y y x x ω ω ω ω = ⋅ = − + + + + G G 2 2 L T V = − ( ) L m x y x ω ∂ = − ∂ , ( ) d L m x y dt x ω ∂ = − ∂ , ( ) 2 L V m y x x x ω ω ∂ ∂ = + − ∂ ∂ ( ) ( ) 2 V m x y m y x x ω ω ω ∂ − = + − ∂ ( ) 2 2 x F m x y x ω ω = − − ( ) L m y x y ω ∂ = + ∂ , ( ) d L m y x dt y ω ∂ = + ∂ , 6
- 150. ( ) 2 L V mx y y y ω ω ∂ ∂ = − + − ∂ ∂ ( ) ( ) 2 V m y x m x y y ω ω ω ∂ + = − + − ∂ ( ) 2 2 y F m y x y ω ω = + − For comparison, from equation 5.3.2 ( 0 0 A = G and 0 ω = G ) G G ( ) 2 F ma m v m r ω ω ω ′ ′ = + × + × × G G G G G G ′ ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 F m ix jy m k ix jy m k k ix jy ω ω ω = + + × + + × × + ( ) 2 2 x F m x y x ω ω = − − ( ) 2 2 y F m y x y ω ω = + − 10.9 Choosing the axis of rotation as the z axis … ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ v v r ix jy kz k ix jy kz ω ω ′ ′ = + × = + + + × + + G G G G G ˆ ( ) ( ) ˆ ˆ ˆ v i x y j y x kz ω ω = − + + + ( ) 2 2 2 2 2 2 1 2 2 2 2 m T mv v x x y y y y x x z ω ω ω ω = ⋅ = − + + + + + G G 2 L T V = − ( ) L m x y x ω ∂ = − ∂ ( ) d L m x y dt x ω ∂ = − ∂ , ( ) 2 L V m y x x x ω ω ∂ ∂ = + − ∂ ∂ ( ) ( ) 2 V m x y m y x x ω ω ω ∂ − = + − ∂ ( ) 2 2 x F m x y x ω ω = − − ( ) L m y x y ω ∂ = + ∂ , ( ) d L m y x dt y ω ∂ = + ∂ ( ) 2 L V m x y y y ω ω ∂ ∂ = − + − ∂ ∂ ( ) ( ) 2 V m y x m x y y ω ω ω ∂ + = − + − ∂ ( ) 2 2 y F m y x y ω ω = + − L mz z ∂ = ∂ , d L mz dt z ∂ = ∂ , L V z z ∂ ∂ = − ∂ ∂ z V mz F z ∂ = − = ∂ From equation 5.3.2 ( and 0 0 A = G 0 ω = G ) … 7
- 151. ( ) 2 F mam v m r ω ω ω ′ ′ = + × + × × G G G G G G ′ G G ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 F m ix jy kz m k ix jy kz m k k ix jy kz ω ω ω ˆ = + + + × + + + × × + + ( ) 2 2 x F m x y x ω ω = − − ( ) 2 2 y F m y x y ω ω = − − z F m = z 10.10 ( ) 2 2 2 1 2 T m r r θ = + θ r ( ) 2 1 cos 2 V k r l mgr θ = − − D ( ) ( ) 2 2 2 2 cos 2 2 m k L T V r r r l mgr θ θ = − = + − − + D L mr r ∂ = ∂ , d L mr dt r ∂ = ∂ ( ) 2 cos L mr k r l mg r θ θ ∂ = − − + ∂ D ( ) 2 cos mr mr k r l mg θ = − − + D 2 L mr θ θ ∂ = ∂ sin L mgr θ θ ∂ = − ∂ ( ) 2 sin d mr mgr dt θ θ = − 10.11 (See Example 4.6.2) ( ) 2 sin 2 4 a x θ θ = + ( ) 1 cos2 4 a y θ = − at 0 θ = 0 x y = ( ) 2 2 1 2 T m x y = + V mgy = ( ) ( ) cos2 s2 x 1 co 2 2 a aθ θ θ θ = + θ = + sin 2 2 a y θ θ = ( ) ( ) 2 2 2 2 1 cos2 sin 2 1 cos2 8 4 ma mga L T V θ θ θ θ = − = + + − − [ ] ( ) 2 2 1 2cos2 1 1 cos2 8 4 ma mga θ θ θ + − − = + 2 2 2 2 cos sin 2 2 ma mga θ θ θ = − where we used the trigonometric identies … 2 2cos 1 cos2 θ θ = + and 2 2sin 1 cos2 θ θ = − 8
- 152. Let sin s aθ = so cos s aθ θ = 2 2 2 1 1 2 2 2 2 ms mg L s ms a = − = − 2 ks where mg k a = The equation of motion is thus 0 k s s m + = or 0 g s s a + = -a simple harmonic oscillator 10.12 Coordinates: cos sin x a t b ω θ = + sin cos t b y a ω θ = − sin cos x a t b ω ω θ = − + θ cos sin y a t b ω ω θ = + θ ( ) 2 2 1 2 L T V m x y mgy = − = + − ( ) ( 2 2 2 2 2 sin sin cos 2 m a b b a t mg a t b ) ω θ θ ω θ ω ω + − − − θ = + ( ) ( ) 2 cos d L mb mba t dt θ ω θ ω θ ω θ ∂ = + − − ∂ ( ) cos sin L mb a t mgb θ ω θ ω θ ∂ = − − ∂ θ The equation of motion 0 L d L dt θ θ ∂ ∂ − = ∂ ∂ is ( ) cos sin 0 a g t b b θ ω θ ω θ − − + = Note – the equation reduces to equation of simple pendulum if 0 ω → . 10.13 Coordinates: ( ) cos cos x l t l t ω θ ω = + + ( ) sin sin y l t l t ω θ ω = + + ( ) ( ) sin sin x l t l t ω ω θ ω θ ω = − − + + ( ) ( ) cos cos y l t l t ω ω θ ω θ ω = + + + ( ) ( ) 2 2 2 2 2 1 1 ... 2 2 L T m x y ml ω θ ω = = + = + + + ( ) ( ) ( ) cos cos cos sin sin t t t 2 ... sin sin cos sin cos ml t t t ω θ ω ω θ ω ω θ + + ω θ ω ω θ + − 9
- 153. ( ) () 2 2 2 2 1 cos 2 ml ml ω θ ω ω θ ω θ = + + + + 0 L d L dt θ θ ∂ ∂ − = ∂ ∂ ( ) ( ) 2 2 sin sin 0 ml ml θ ωθ θ ω θ ω θ − + + = (a) 2 sin 0 θ ω θ + = (b) The bead executes simple harmonic motion ( ) 0 θ about a point diametrically opposite the point of attachment. (c) The effective length is 2 g l ω = 10.14 v j ( ) ˆ ˆ ˆ cos sin at l i j θ θ θ = + + G ( ) ˆ ˆ cos sin il j at l θ θ θ + θ = + ( ) 2 2 2 2 2 2 2 2 1 cos 2 sin sin 2 2 m T mv v l a t atl l θ θ θ θ θ = ⋅ = + + + G G θ 2 1 cos 2 V mg at l θ = − ( ) 2 2 2 2 2 2 sin cos 2 2 m a L T V l a t atl mg l t θ θ θ θ = − = + + − − 2 sin L ml matl θ θ θ ∂ = + ∂ , 2 sin cos d L ml mal matl dt θ θ θ θ ∂ = + + ∂ θ cos sin L matl mgl θ θ θ θ ∂ = − ∂ 2 sin cos cos sin ml mal matl matl mgl θ θ θ θ θ θ + + = − θ sin 0 a g l θ θ + + = For small oscillations, sinθ θ ≈ 0 a g l θ θ + + = 2 2 l T a g π π ω = = + D 10.15 (a) 2 1 1 2 2 T mx I 2 θ = + and 2 2 5 I ma = 2 2 1 1 2 2 2 5 T mx ma 2 θ = + V mg = − x 10
- 154. t θ θ = x T 22 2 1 1 2 5 V mx ma mgx θ = − = + + L T The equation of constraint is … ( , ) 0 f x x a θ θ = − = The 2 Lagrange equations with multipliers are … 0 L d L f x dt x x λ ∂ ∂ ∂ − + = ∂ ∂ ∂ 0 L d L f dt λ θ θ θ ∂ ∂ ∂ − + = ∂ ∂ ∂ L mx x ∂ = ∂ d L mx dt x ∂ = ∂ L mg x ∂ = ∂ f x λ λ ∂ = ∂ 0 mg mx λ − + = and from the θ-equation … mg 2 2 0 5 ma a θ λ − − = Differentiating the equation of constraint … x a θ = and substituting into the above … 5 7 x g = and 2 7 mg λ = − (b) The generalized force that is equivalent to the tension T is … 2 7 x f Q m x λ λ ∂ = = = − ∂ g 10.16 For the velocity of a differential mass element, dm, of the spring at a distance v′ G x′ below the support … x v v x ′ ′ = G G , m dm dx x ′ ′ = ( ) 2 2 2 2 0 0 1 1 1 1 2 2 2 2 m x x m T mv v dm mx x d x x ′ x ′ ′ ′ ′ = + = + ∫ ∫ 2 2 1 1 2 2 3 m T mx ′ = + x ( ) 2 0 1 2 m V k x l mgx gx d ′ m ′ ′ = − − − ∫ ( ) 2 0 1 2 x m k x l mgx gx dx x ′ ′ ′ − − ∫ = − ( ) 2 1 2 2 m V k x l mgx gx ′ = − − − ( ) 2 2 1 1 2 3 2 2 m m L T V m x k x l m gx ′ ′ = − = + − − + + 3 L m m x x ′ ∂ = + ∂ , 3 d L m m x dt x ′ ∂ = + ∂ , 11
- 155. ( ) 2 L m kx l m g x ′ ∂ = − − + + ∂ ( ) 3 2 m m m x k x l m ′ ′ + = − − + + g For 2 g m y x l m k ′ = − − + , 0 3 m m y ky ′ + + = The block oscillates about the point 2 g m x l m k ′ = + + … with a period … 2 3 2 m m k π π ω T ′ + = = D 10.17 Note: 4 objects move – their coordinates are labeled i x : The coordinate of the movable, massless pulley is labeled p x . Two equations of constraint: ( ) ( ) 1 1 1 , 0 p p f x x x l x = − − = ( ) ( ) ( ) 2 2 3 2 3 , , 2 0 p p f x x x x x x l′ = + − + = 2 2 2 1 1 1 1 2 2 2 3 3 3 1 1 1 2 2 2 L T V m x m gx mx m gx m x m gx = − = − + − + − 3 0 j j j i i i f L d L q dt q q λ ∂ ∂ ∂ − + = ∂ ∂ ∂ ∑ Thus: (1) 1 1 1 1 0 m g m x λ − − + = (2) 2 2 2 2 0 m g m x λ − − + = (3) 3 3 3 2 0 m g m x λ − − + = Now – apply Lagrange’s equations to the movable pulley – note 0 p m → So: (4) 1 2 1 2 0 p p f f x x λ λ ∂ ∂ + = ∂ ∂ or 1 2 2 0 λ λ − = Now - p x can be eliminated between 1 f and 2 f ( ) 2 1 2 3 1 2 2 2 f f x x x l l′ = = + + − + = 0 (5) Thus x x 2 3 1 2 0 x + + = With a little algebra – we can solve (1) –(5) for the 5 unknowns 1 x , 2 x , 3 x , 1 λ , and 2 λ 12
- 156. 1 1 2 3 2 11 1 1 4 g m m m λ = + + 2 1 2 3 1 1 1 1 4 g m m m λ = + + As the check, let m m and 2 3 m = = 1 2 m m = . Thus, there is no acceleration and 1 2mg λ = 10.18 (See Example 5.3.3) (a) ˆr r re = ˆ ˆ r r re r eθ θ = + Constraint: ( ) 0 f t θ θ ω = − = so … θ ω = and 0 θ = ( ) 2 2 2 1 2 T m r r θ = + L = 0 L d L r dt r ∂ ∂ − ∂ ∂ = and 0 L d L f dt λ θ θ θ ∂ ∂ ∂ − + = ∂ ∂ ∂ 2 r r ω = 2 2 0 mrr mr θ θ λ − − + = t t r Ae Be ω ω − = + ( ) 0 r 0 = apply constraint t t r Ae Be ω ω ω ω − = − ( ) 0 r l ω = so … 0 A B + = 2mrr λ ω = A B l ω ω ω − = 2A l = 2 l A = 2 l B = − thus .. ( ) 2 t t l e ω ω − = − r e sinh cosh r l t r l t ω ω ω = = 2 2 2 sinh cosh m l t t λ ω ω = ω now at so r l = t T = 1 1 0 sinh 1 T .88 ω ω − = = (b) There are 2 ways to calculate F … (i) See Example 5.3.3 … 2 F m x ω ′ = ( 2 t l x e e ω ω − ′ = − ) t and ( ) 2 t t l e ω ω x e ω − ′ = + 2 2 cosh F m l t ω ω = (ii) f λ θ ∂ ∂ is the generalized force, in this case – a torque T, acting on the bead f T rF λ θ ∂ = = ∂ 13
- 157. 2 2 2 sinhcosh sinh f m l t t F r l t λ ω ω θ ω ∂ = = ∂ ω 2 2 cosh m l t ω ω = 10.19 (See Example 4.6.1) The equation of constraint is ( ) , 0 f r r a θ = − = ( 2 2 2 2 m T r r ) θ = + cos V mgr θ = r a θ ( ) 2 2 2 cos 2 m L r r mgr θ θ = + − 0 L d L f r dt r r λ ∂ ∂ ∂ − + = ∂ ∂ ∂ 0 L d L f dt λ θ θ θ ∂ ∂ ∂ − + ∂ ∂ ∂ = 1 f r ∂ = ∂ 0 f θ ∂ = ∂ Thus 2 cos 0 mr mg mr θ θ − − + λ = λ = 2 sin 2 0 mgr mr mrr θ θ θ − − = Now , r r so r a = 0 = = 2 cos 0 ma mg θ θ − + 2 sin 0 mga ma θ θ − = sin g a θ θ = and d d θ θ θ θ = so sin g d d a θ θ θ = ∫ ∫ θ or 2 cos 2 g g a a θ θ = − + hence, ( ) 3cos 2 mg λ θ = − and when 0 λ → particle falls off hemisphere at 1 2 cos 3 θ − = D 10.20 Let 2 x mark the location of the center of curvature of the movable surface relative to a fixed origin. This point defines the position of m2. 1 r marks the position of the particle of mass m1. r is the position of the particle relative to the movable center of curvature of m2. Kinetic energy 2 2 2 2 1 1 1 1 2 2 x m r = + T m ( ) ( ) 2 2 1 1 2 2 2 2 2 x r x r x r x r r r ⋅ = + ⋅ + = + + ⋅ but ˆ ˆ r e r e r r θ θ = + ( ) 2 2 2 2 2 2 2 1 2 2 1 1 ˆ ˆ 2 2 2 r T m x m x r r x re r eθ θ θ ∴ = + + + + ⋅ + 2 2 2 2 2 2 2 1 2 2 2 1 1 2 sin 2 cos 2 2 m x m x r r x r x r θ θ θ θ = + + + + − 14
- 158. Potential energy sin Vmgy mgr θ = − = − L T V = − Equation of constraint ( ) , 0 f r r a θ = − = 1 f r ∂ ∴ = ∂ 0 f θ ∂ = ∂ 2 1 f x ∂ = ∂ Lagrange’s equations 0 i i i L d L f x dt x x λ ∂ ∂ ∂ − + ∂ ∂ ∂ = x2) 2 2 1 1 1 sin cos 0 d m x m x m r r dt θ θ θ − − − − = 2 2 2 1 1 1 sin sin cos cos sin 0 d m x m x m r r r r r dt θ θ θ θ θ θ θ θ θ − − + + − + = using constraint ; r a constant = = 0 r r = = ( 2 1 2 1 2 sin cos m x a m m ) θ θ θ θ = − + + (1) θ) 2 2 2 2 2 2 sin sin cos sin cos cos 0 r rr x r x r x r xr x r gr θ θ θ θ θ θ θ θ θ θ − − − − − + + + = using constraint ; r a constant = = 0 r r = = ) 2 sin cos 0 x g a a θ θ θ − − + = (2) r) 2 2 2 2 1 cos sin sin sin 0 r x x r x g m λ θ θ θ θ θ θ θ − + − + + + + = Apply constraint… 2 2 1 cos sin 0 x a g m λ θ θ θ + + + = (3) Now, plug solution for 2 x (1) into equation (2): ( ) 2 1 1 2 sin cos sin cos 0 m g m m a θ θ θ θ θ θ − + − + θ = ( ) 1 1 2 2 1 sin sin cos cos m m g f f a θ θ θ θ θ − − − 0 θ = where 1 1 1 2 m m f m m = + Now 2 1 2 d d d d d dt d dt d d θ θ θ θ θ θ θ θ θ θ = = = = So – let 2 x θ = 1 1 2 2 1 sin 2 sin cos cos 0 m m dx g f xf d a θ θ θ θ − − − − θ = Let 1 2 1 sin m y f θ = − ( ) sin 2 d dx dy g y x d d a d θ θ θ θ + = 15
- 159. Hence: ( )( 2 sin g x d a ) d y θ = and ( ) ( ) ( ) ( ) sin 0 0 2 sin yx yx g d yx d a θ θ θ θ θ = = = ∫ ∫ D but at so 2 0 x θ = = 0 θ = D ( ) 2 sin g yx a θ θ = ( ) 1 2 2 2 sin 1 sin m g x a f θ θ θ θ = = − Now we can solve for θ and plug 2 θ , θ into (3) to obtain ( ) λ θ 1 2 2 1 si 2 1 s m d g d d a d f θ θ θ n in θ θ θ = = − After some algebra – yields ( ) ( ) 1 1 2 2 2 1 sin cos 1 sin m m f g a f θ θ θ θ + = − The solution for ( ) λ θ is thus determined from equation (3) 1 2 2 1 sin cos cos sin m g f a m a λ θ θ θ θ θ θ θ − + + = − collecting terms: ( ) 1 1 2 2 1 1 cos sin cos sin m m g f f a m a λ θ θ θ θ θ θ − − + = − Plug θ , θ into the above --- plus --- a lot of algebra yields … ( ) 2 1 1 1 2 2 2 2 1 2 sin sin cos 1 sin sin 1 sin m m m m f f f m g f θ θ θ θ λ θ θ θ − − − = + − where 2 2 1 2 m m f m m = + As a check, let so it is immoveable … then and 2 m → ∞ 2 1 m f → 1 0 m f → and we have ( ) 1 3sin m g λ θ θ − → … which checks 10.21 (a) ( ) 2 2 2 2 1 1 2 2 T mv m R R z φ = = + + 2 ( ) 2 2 2 2 2 m L T V R R z V φ = − = + + − L mR R ∂ = ∂ , d L mR dt R ∂ = ∂ , 2 L V mR R R φ ∂ ∂ = − ∂ ∂ 16
- 160. 2 V mR mR R φ ∂ =− ∂ 2 R mR mR Q φ − = 2 L mR φ φ ∂ = ∂ , 2 2 d L mR mRR dt φ φ φ ∂ = + ∂ , L V φ φ ∂ ∂ = − ∂ ∂ 2 2 V mR mRR Qφ φ φ φ ∂ + = − = ∂ L mz z ∂ = ∂ , d L mz dt z ∂ = ∂ , L V z z ∂ ∂ = − ∂ ∂ z V mz Q z ∂ = − = ∂ G For , using the components of F ma = G a G from equation 1.12.3: ( ) 2 R F m R Rφ = − , ( ) 2 F m R R φ φ φ = + z , z F m = From Section 10.2, since R and are distances, and are forces. However, since z R Q z Q φ is an angle, Qφ is a torque. Since Fφ is coplanar with and perpendicular to R , Q RF φ φ = … and all equations agree. (b) 2 2 2 2 2 2 2 sin v r r r θ φ θ = + + ( ) 2 2 2 2 2 2 sin 2 m L T V r r r V θ φ θ = − = + + − L mr r ∂ = ∂ , d L mr dt r ∂ = ∂ , 2 2 2 sin L V mr mr r r θ φ θ ∂ ∂ = + − ∂ ∂ 2 2 2 sin r V mr mr mr Q r θ φ θ ∂ − − = − = ∂ 2 L mr θ θ ∂ = ∂ , 2 2 mr mrr dt d L θ θ θ ∂ = + ∂ , 2 2 sin cos L V mr φ θ θ θ θ ∂ ∂ = − ∂ ∂ 2 2 2 2 sin cos V mr mrr mr Qθ θ θ φ θ θ θ ∂ + − = − = ∂ 2 2 sin L mr φ θ φ ∂ = ∂ , 2 2 2 2 sin 2 sin 2 sin cos d L mr mrr mr dt φ θ φ θ θφ θ φ ∂ = + + ∂ θ , L V φ φ ∂ ∂ = − ∂ ∂ 2 2 2 2 sin 2 sin 2 sin cos V mr mrr mr Qφ φ θ φ θ θφ θ θ φ ∂ + + = − ∂ = is a force . r Q r F Qθ and Qφ are torques. Since φ is in the xy plane, the moment arm for φ is sin r θ ; i.e. Q r sin F φ φ θ = . Q rF θ θ = . From equation 1.12.14 … 17
- 161. ( ) 2 22 sin r m r r r F φ θ θ − − = ( ) 2 2 sin cos m r r r Fθ θ θ φ θ θ + − = ( ) sin 2 sin 2 cos m r r r Fφ φ θ φ θ θφ θ + + = The equations agree. 10.22 For a central field, ( ) r = V V , so 0 V V θ φ ∂ ∂ = = ∂ ∂ and V F r ∂ = − ∂ . For spherical coordinates, using equation 1.12.12: 2 2 2 2 2 2 sin v r r r 2 φ θ θ = + + ( ) ( ) 2 2 2 2 2 2 sin 2 m L T V r r r V r φ θ θ = − = + + − L mr r ∂ = ∂ , d L mr dt r ∂ = ∂ , 2 2 2 sin L V mr mr r r φ θ θ ∂ ∂ = + − ∂ ∂ 2 2 2 sin r mr mr mr F φ θ θ − − = 2 L mr θ θ ∂ = ∂ , 2 2 d L mr mrr dt θ θ θ ∂ = + ∂ , 2 2 sin cos L mr φ θ θ θ ∂ = ∂ 2 2 2 2 sin co mr mrr mr θ θ φ θ θ + − s 0 = 2 2 sin L mr φ θ φ ∂ = ∂ , 2 2 2 2 sin 2 sin 2 sin cos d L mr mrr mr dt φ θ φ θ φθ θ φ ∂ = + + ∂ θ 0 L φ ∂ = ∂ 2 2 2 2 sin 2 sin 2 sin cos 0 mr mrr mr φ θ φ θ φθ θ θ + + = 10.23 Since θ α = = constant, there are two degrees of freedom, and r θ . , v r r v r = sin φ φ α = ( ) 2 2 2 2 1 1 sin 2 2 T mv m r r φ α = = + cos V mgr α = ( ) 2 2 2 2 sin cos 2 m L T V r r mgr φ α α = − = + − L mr r ∂ = ∂ , d L mr dt r ∂ = ∂ , 2 2 sin cos L mr mg r φ α α ∂ = − ∂ 2 2 sin cos mr mr mg φ α α = − 2 2 sin L mr φ α φ ∂ = ∂ , 0 L φ ∂ = ∂ 18
- 162. ( ) 2 sin 0 d mr dt φα = Say constant 2 sin mr φ α = = A 2 3 cos mr mg mr α = − A 2 1 2 d dr d r r r dt dr dr = = = r ( ) 2 2 3 cos 2 m dr d r mg dr m r α = − A 2 2 2 cos 2 2 mr mgr C mr α = − − + A The constant of integration C is the total energy of the particle: kinetic energy due to the component of motion in the radial direction, kinetic energy due to the component of motion in the angular direction, and the potential energy. ( ) 2 2 cos 2 U r mgr mr α = + A For , A , and turning points occur at 0 φ ≠ 0 ≠ 0 r = Then 2 2 0 co 2 mgr C mr α = − − + A s ( ) 2 3 2 cos 0 2 mg r Cr m α − + = A The above equation is quadratic in r ( 2 r4 ∝ A ) and has two roots. 10.24 Note that the relation obtained in Problem 10.23, ( ) 2 3 2 cos 0 2 l mg r Cr m α − + = , defines the turning points. For the particle to remain on a single horizontal circle, there must be two roots with r . Thus ( divided into the above expression leaves a term that is linear in r. r = D ) 2 r r − D ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 r r c r rr r r Cr m r r r rr r C r r r r C r r r C r r r C r r C r m r r C + − − + − + − + − − − − − + − − − + − − D D D D D D D D D D D D D D D D D A A For the remainder to vanish, both terms must equal zero. 2 2 4 2 r rC r − − = D D D 0 3 2 C r = D 19
- 163. 2 3 2 2 0 2 rCr m − + = D D A 2 2 2 2 r mr = D D A And with 2 sin mr φ α = D D A 1 2 2 1 sin mr φ α = D D From Prob. 10.23, 2 3 cos mr mg mr α = − A For small oscillations about r , r r D ε = + D . r ε = 3 3 3 3 1 1 1 3 1 1 r r r r r ε ε − = + ≈ − D D D D 2 3 3 1 c m m mr r ε os g ε α = − − D D A 2 2 4 3 3 cos m m mr mr g ε ε α + = − D D A A 2 4 2 2 2 2 3 sin m r T π π π ω φ = = = D D A 1 3 α 10.25 ( ) ( ) 2 2 2 2 1 2 2 x y m L mv qv A x y z q xA yA zA = + ⋅ = + + + + + G G z x L mx qA x ∂ = + ∂ , x dA d L mx q dt x dt ∂ = + ∂ Using the hint, x x x dA A A A x y z dt x y z ∂ ∂ ∂ = + + ∂ ∂ ∂ x y x z A A L A q x y z x x x x ∂ ∂ ∂ ∂ = + + ∂ ∂ ∂ ∂ y x x x x z A A A A A A mx q x y z q x y z x y z x x x ∂ ∂ ∂ ∂ ∂ ∂ + + + = + + ∂ ∂ ∂ ∂ ∂ ∂ y x x A A A A mx q y z z x y z x ∂ ∂ ∂ ∂ = − − − ∂ ∂ ∂ ∂ G G ( ) ( ) z y q y A z A × − ∇× G G = ∇ G G ( ) x q v A ∇× G = × 20
- 164. ( )x mx qv B = × G G Due to the cyclic nature of the Cartesian coordinates, i.e., i j ˆ ˆ ˆ k × = … ( )y my q v B = × G G G G ( )z mz q v B = × G G Altogether, ( ) mr q v B = × G 10.26 V mgz = ( ) 2 2 2 1 2 T m x y z = + + x T p mx x ∂ = = ∂ , x p x m = similarly, y p y = and m z p z m = ( ) 2 2 2 1 2 x y z H T V p p p mgz m = + = + + + x x p H x p m ∂ = = ∂ 0 x H p x ∂ = = − ∂ , x p constant = ( ) 0 x d p mx mx dt = = = similarly, y p constant = , or 0 my = y y p H y p m ∂ = = ∂ z z H p z p m ∂ = = ∂ z H mg p z ∂ = = − ∂ , ( ) z d p mz mz mg dt = = = − These agree with the differential equations for projectile motion in Section 4.3. 10.27 (a) Simple pendulum … cos V mgl θ = − 2 2 1 2 T ml θ = 2 T p ml θ θ θ ∂ = = ∂ , 2 p ml θ θ = 2 2 cos 2 p H T V mgl ml θ θ = + = − 2 p H p ml θ θ θ ∂ = = ∂ sin H mgl pθ θ θ ∂ = = ∂ − 21
- 165. (b) Atwood’s machine… ( ) 1 2 V m gx m g l x = − − − 2 2 2 1 2 2 1 1 1 2 2 2 x T m x m x I a = + + [Includes the pulley] 1 2 2 T I p m m x x a ∂ = = + + ∂ , 1 2 2 p x I m m a = + + ( ) 2 1 2 2 1 2 2 2 p H T V m m gx m g I m m a = + = − − − + + l 1 2 2 H p x I p m m a ∂ = = ∂ + + ( ) 1 2 H m m g x ∂ = − − = − ∂ p , ( ) 1 2 p m m g = − (c) Particle sliding down a smooth inclined plane … sin V mgx θ = − 2 1 2 T m = x T p mx x ∂ = = ∂ , p x m = 2 sin 2 p H T V mgx m θ = + = − H p x p m ∂ = = ∂ sin H mg p x θ ∂ = − = − ∂ , sin p mg θ = 10.28 (a) r F ( , ) ( , ) i i i L T q q T q t = − …note, potential energy is time dependent. V F r ∂ = − ∂ so t k V Fdr e r β − = = − ∫ ( ) 2 2 2 1 1 2 2 T mv v m r r θ = ⋅ = + G G 22
- 166. r L p mr r ∂ = = ∂ r p r m = 2 L pmr θ θ θ ∂ = = ∂ 2 p mr θ θ = 2 2 2 2 2 t r i i r p p k H p q L p r p e m mr r β θ θθ − = − = + − − − ∑ Substituting for and r θ … 2 2 2 2 2 t r p p k H e m mr r β θ − = + − (b) H = T + V … which is time-dependent (c) E is not conserved 10.29 Locate center of coordinate system at C.M. The potential is independent of the center of mass coordinates. Therefore, they are ignorable. (r1,θ1) L T V = − ( ) ( ) ( ) 2 2 2 2 2 2 2 1 1 1 1 2 2 2 2 1 2 1 1 1 2 2 2 m r r m r r k r r l θ θ = + + + − + − where l is the length of the relaxed spring and k is the spring constant. 1 1 1 r L p mr r ∂ = = ∂ 1 1 1 r p r m = 2 1 1 1 L 1 p mr θ θ θ ∂ = = ∂ 1 1 2 1 1 p m r θ θ = (r2,θ2) … and similarly for m2 i i H p q = − ∑ L 2 2 2 2 2 1 2 1 2 1 2 2 2 1 1 1 2 2 2 1 ( ) 2 2 2 2 2 r r p p p p H k m m r m m r θ θ = + + + − + − r r l Equations of motion … First, θ1 , θ2 are ignorable coordinates, so 1 2 , p p θ θ are each conserved. ( ) 1 1 2 1 r H 1 1 p k r r l m r r ∂ = − = + − = − ∂ ( ) 2 1 2 2 r H 2 2 p k r r l m r r ∂ = − = + − = − ∂ 1 2 r r p p = 0 r p ∆ = r p ∆ = constant. The radial momenta are equal and opposite. 23
- 167. 10.30 2 2 11 2 2 L mx k = − x 2 2 2 1 1 1 2 2 1 1 0 2 2 t t t t t t Ldt Ldt mx kx dt δ δ δ = = = − ∫ ∫ ∫ ( ) 2 1 0 t t mx x kx x dt δ δ = − ∫ d x x dt δ δ = ( ) ( ) 2 2 2 1 1 1 t t t t t t d mx xdt mx x dt mxd x dt δ δ δ = = ∫ ∫ ∫ Integrating by parts: ( ) 2 2 2 1 1 1 t t t t t t mx xdt mx x x d mx δ δ δ = − ∫ ∫ 0 x δ = ( ) at and 1 t 2 t ( ) mx t mxdt = = d dt d mx d 2 2 1 1 t t t t mx xdx x mx dt δ δ = − ∫ ∫ ( ) 2 1 0 t t mx x kx x dt δ δ = − − ∫ 0 mx kx + = 10.31 (a) 2 2 0 2 1 v L m c c = − − −V Let 2 2 1 1 v c − γ = 0 x L m x p x γ ∂ = ≡ ∂ This is the generalized momentum for part (b). Thus, Lagrange’s equations for the x-component are … 0 x d L L d V p dt x x dt x ∂ ∂ ∂ − = − = ∂ ∂ ∂ … and so on for the y and z components. (b) H vi i i p = − ∑ L but … 0 i i p v m γ = So … 2 2 2 0 2 0 i i p c m c H V m c γ γ = + ∑ + 2 2 2 0 2 0 m c p c H V m c γ γ = + + 24
- 168. ( ) 2 22 4 0 2 0 1 H p c m c m c γ = + V + (c) Now, if T then we have … 2 0 m c γ = 2 2 2 2 2 4 2 4 0 0 2 4 0 1 p c p c m c m c m c + = + ( ) 2 2 2 2 2 4 2 4 2 2 2 0 0 0 2 4 0 1 1 m v c m c m c v c m c γ γ = + = + 2 2 2 4 2 4 2 2 4 0 0 2 2 2 2 1 1 1 1 v c m c m c m c v c v c γ = + = = − − 0 V Thus … 2 0 H m c V T γ = + = + (d) 2 2 2 0 0 2 2 2 1 1 ... 1 2 m c v T m c v c c = ≈ + − + 2 2 0 0 1 2 T m v m ≈ + c ------------------------------------------------------------------------------------------------------------ 25
- 169. 1 Chapter 11 11.1 (a) 2 2 ( ) 2 kk V x x x = + 2 2 k V kx x ′ = − At equilibrium, 0 V′ = 1 3 x k = 2 3 2k V k x ′′ = + 1 3 2 3 x k V k k k = ′′ = + = 0 Stable (b) ( ) bx V x kxe− = bx bx V ke bkxe − − ′ = − At equilibrium 0 bx bx ke bkxe − − − = 1 x b = 2 bx bx bx V bke bkxe b kxe − − ′′ = − − + − 1 1 1 1 2 0 x b V bke bke bke − − − = ′′ = − + = − Unstable (c) ( ) 4 2 2 ( ) V x k x b x = − ( ) 3 2 4 2 V k x b x ′′ = − At equilibrium ( ) 3 2 4 2 k x b x 0 − = 0, 2 x b = ± ( ) 2 2 12 2 V k x b ′′ = − 2 0 2 x V kb = ′′ = − 0 Unstable ( ) 2 2 2 2 6 2 4 x b V k b b kb =± ′′ = − = 0 Stable (d) for case (a) 2 3k m ω = 2 2 2 3 3 m T s k π π π ω = = = for case (c) at 2 x b = ± 2 2 4kb m ω = 2 2 2 4 m T s kb π π π ω = = = 11.2 ( ) 2 2 ( , ) 2 4 V x y k x y bx by = + − −
- 170. 2 2 ( )2 ( 2 ) V V k x b k y b x y ∂ ∂ = − = − ∂ ∂ at equilibrium 2 x b and y b = = 2 2 2 2 2 2 0 V V V k k x x y y ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ 2 = k 11 12 21 22 2 0 0 2 k k k k k = = = = 11 12 2 21 22 4 0 k k k k k = − 0 The equilibrium is stabe. 11.3 2 1 ( ) 2 V x kx = − ( ) ( ) dV x dx F x kx mx mx dx dx = − = = = kx dx mx dx = 0 0 x v x kxdx mx dx = ∫ ∫ ( ) 2 2 2 0 2 2 k v x x m − = ( ) 1 2 2 2 0 dx v k m x x dt = = − ( ) 0 1 2 2 2 0 0 ; x t x dx dt where k m x x α α = = − ∫ ∫ ( ) 2 2 0 0 ln ln x x x x t α + − − = 2 2 0 0 t x x x x eα + − = 2 2 2 2 0 0 0 2 t t 2 x x x e xx e x α α − = − + 0 0 cosh 2 t t e e x x x α α t α − + = = 11.4 Let the length of the unstretched, elastic cord be d. Then 2l y 2 2 2 d l y = + ( ) 2 1 2 2 V k d l mg = − − y m
- 171. 3 ( ) 2 22 2 2 4 4 8 4 2 k V l y l l y l m = + − + + − gy ( ) 2 2 2 2 2 2 2 V k l y l l y mg = + − + − y The first term, , is an additive constant to the potential energy, so with appropriate adjustment of the zero reference point … 2 4kl ( ) 2 2 2 ( ) 2 2 V y k y l l y mgy = − + − ( ) 1 2 2 2 ( ) 2 2 2 dV y k y yl l y mg dy − = − + − At equilibrium, the above expression is zero, so … 2 2 4 4 0 kly ky mg l y − − + = 2 2 4 4 kly ky mg l y − = + 2 2 2 2 2 2 2 2 2 16 16 8 k l y k y kmgy m g l y − + = + ( ) 2 4 3 2 2 2 2 2 2 2 2 2 2 2 16 8 16 16 8 0 k y kmgy k l m g k l y kl mgy l m g − + + − − + = 4 2 2 2 3 2 4 4 2 4 2 2 2 0 2 16 2 16 y mg m g mg m g y y y l kl k l kl k l − + − + 2 = letting y u l = and 4 mg a kl = 4 3 2 2 2 2 2 u au a u au a − + − + = 0 11.5 ( ) 1 2 V mg h h = + d φ θ θ b a h1 h2 1 cos h b θ = ( ) 2 sin h d θ ϕ = + ( ) 2 sin cos cos sin h d θ ϕ θ = + ϕ cos sin b a d d θ ϕ ϕ = = 2 sin cos h b a θ θ ϕ = + ( )cos sin V mg a b b θ θ θ = + + ( )( ) sin sin cos V mg a b b b θ θ θ θ ′= + − + +
- 172. 4 [ ] sin cos Vmg a b θ θ θ ′= − + [ ] cos sin cos V mg b b a θ θ θ θ ′′= − − ( ) 0 V mg b a ′′ = − Equilibrium … Stable Unstable a b a b 11.6 2 4 cos 1 ... 2! 4! θ θ θ = − + − 3 5 sin ... 3! 5! θ θ θ θ = − + − ( ) 2 4 3 5 1 ... ... 2! 4! 3! 5! V mg a b b θ θ θ θ θ θ = + − + − + − + − 2 4 3 ... 2 24 a b a b V mg a b θ θ − − = + − + − For a = b, 4 2 ... 12 a g a θ = − V m + 3 ... 3 mga V θ ′=− + terms in higher order of θ 2 ... V mgaθ ′′=− + 2 ... V mgaθ ′′′=− + 2 0 V mga ′′′′=− ∴ Equilibrium is unstable 11.7 The center of mass (CM) of the hemisphere is 3 8 a from the flat side (see Equation 8.1.8). The height of CM) above the point of contact between the two hemispheres is designated by h2 in the figure. h1 is the height of the point of contact above the ground. a φ θ h1 θ b h2 ( ) ( 1 2 3 cos cos cos 8 V mg h h mg b a a ) θ θ θ = + = + − + ϕ and a b ϕ θ =
- 173. 5 ( ) () 1 2 3 cos cos 8 b V mg h h mg a b a a θ θ θ = + = + − + ( ) 3 sin sin 8 a a b a b V mg a b a a θ θ + + ′= − + + Equilibrium occurs at 0o θ = ( ) 2 3 cos cos 8 a a b a b V mg a b a a θ θ + + ′′= − + + ( ) ( )( 2 0 3 3 5 8 8 a a b mg V mg a b a b b a a a + ′′= − + + = + − ) 5a 0 0 3 V for b ′′ Therefore, the equilibrium is stable for 3 5 b a 11.8 From Problem 11.4, we have ( ) 1 2 2 2 2 2 2 2 2 ( ) 2 2 2 2 1 y V y k y l l y mgy k y l mgy l = − + − = − + − Expanding the square root for small y l … 1 2 2 2 2 2 1 1 1 1 ... 2 8 y y y l l l 4 4 + = + − + 4 2 2 2 2 ( ) 2 2 4 y V y k y l y mgy l ≈ − − + − 3 2 2k V y l ′ ≈ −mg at equilibrium, 1 2 3 0 2 mgl k ′= ⇒ = V y 2 2 6k V y l ′′= ( ) 1 2 3 2 2 1 2 3 3 3 2 2 6 6 2 2 y mgl k k mgl mg V k l k l = ′′ = = 1 2 1 3 3 3 6 6 2 2 V k g g m m l l m ω ′′ = = = 1 6 k
- 174. 6 11.9 From Problem 11.5,V m ( ) 0 g b a ′′ = − ( ) 0 V g b a m ω ′′ = = − ( ) 0 2 2 T g b a π π ω = = − 11.10 From Problem 11.7, ( )( 0 3 5 8 mg b b a ′′= + − ) a V a ( )( ) 0 0 2 2 4 3 5 m a T g a b b a V π π = = + − ′′ 11.11 2 2 1 1 2 2 cm T mv Iϕ = + a b φ θ ( ) cm v b a θ = − 2 2 5 I ma = The relationship between the angles, and θ ϕ (see Figure), can be determined from the condition that there is no slipping as the ball rolls in the hemisphere, so the length of roll measured along the ball, ( a ) θ ϕ + , must equal the length of roll measured along the hemisphere, bθ … so we have … ( ) ( ) b a and b a θ θ ϕ θ θ ϕ = + ∴ = + and … b a a ϕ θ − = ( ) ( ) ( ) 2 2 2 2 2 2 2 1 1 2 1 1 2 2 5 2 b a T m b a ma m b a a θ 2 2 5 θ θ − = − + = − + ( )cos V mg b a θ =− − ( )sin @ 0o V mg b a equilibrium θ θ ′= − =
- 175. 7 ( ) () 0 cos V mg b a V mg b a θ ′′ ′′= − ∴ = − ( ) ( ) ( ) 0 2 5 7 7 5 mg b a V g M b a m b a ω ′′ − = = = − − ( ) 0 7 2 2 5 b a T g π π ω − = = 11.12 The potential energy of the satellite shaped like a “thin rod” is (See Example 11.2.2, Figure 11.2.1) … e M dm V G r = − ∫ but 2 m dm dx a = where 2a is the length of the rod. 2 2 a a e e a a GM GM m m d V dx r a a r − − = − =− ∫ ∫ x ( ) 1 2 2 2 0 0 2 cos r r x r x φ = + + ( ) ( ) 1 1 2 2 2 2 0 1 cos r r x ε φ = + + where 0 2 2 0 2xr r x ε = + ( ) ( ) 1 2 1 2 2 2 0 1 cos 2 a e a dx GM m V a r x ε φ − − + =− + ∫ For ( ) 1 2 2 2 2 0 0 0 0 0 2 , , 1 c x r x r x r and r r r os ε ε φ + ≈ ≈ ≈ + Thus, for small ε , the expression for the potential energy, V, can be approximated … 2 2 0 0 0 1 2 3 2 1 cos cos 2 2 8 a e a GM m x x V d ar r r φ φ − ≈− − + ∫ x 2 3 2 0 0 3cos 2 2 0 2 2 e GM a V a ar r φ ≈− + + 3 2 2 2 0 0 cos 1 2 e GM a V r r φ ≈− + 2 2 2 3 0 0 0 2cos sin sin 2 2 2 e e GM GM a a V r r r φ φ φ ′≈ =
- 176. 8 Equilibrium @ 0 φ= 2 3 0 cos2 e GM a V r φ ′′≈ and 2 0 3 0 e GM a r ′ ≈ V 2 1 3 M I ma = = 0 3 0 3 e V G M r ω ′′ = = M 3 0 0 2 2 3 e r T GM π π ω = = 11.13 The amplitude of the symmetric component is A1 and the amplitude of the anti-symmetric component is A2 (See Equations 11.3.19a through 11.3.20b). [ ] 2 2 1 1 2 1 (0) (0) 4 A x x = + [ ] 0 1 1 2 1 (0) (0) 2 2 A A x x = + = [ ] 2 2 2 2 1 1 (0) (0) 4 A x x = − [ ] 0 2 2 1 1 (0) (0) 2 2 A A x x = − = 1 2 A A ∴ = From Equation 11.3.18 the solution for x1 is … ( ) 0 1 1 ( ) cos cos 2 A 2 x t t t ω ω = + (The phase 2 δ is 180o , which insures that 1 0 (0) x A = ) ( ) ( ) 0 1 2 1 2 1( ) 2cos cos 2 2 2 A t t x t ω ω ω ω + − = Letting … 1 2 1 2 2 and 2 ω ω ω ω + − = ∆ = ω ( ) 1 0 ( ) cos cos x t A t t ω = ∆ From Equation 11.3.18, the solution for x2 is … ( ) 0 2 1 ( ) cos cos 2 A 2 x t t t ω ω = − ( ) ( ) 0 1 2 1 2 2 ( ) 2sin sin 2 2 2 A t t x t ω ω ω ω + − = ( ) 2 0 ( ) sin sin x t A t t ω = ∆
- 177. 9 11.14 At time t= 0 and short times thereafter … cos 1 t ∆ ≈ and sin 0 t ∆ ≈ . Thus, … 1 0 2 cos 0 x A t and x ω ≈ ≈ . This situation occurs again when 2 t π ∆ = . 1 1 2 2 2 1 1 2 2 2 k k k m m ω ω ′ − + ∆ = = − 1 1 2 2 1 2 1 1 2 k k m k ′ ∆ = + − for k k , ′ 1 2 2 1 2 1 1 1 ... 1 2 k k k k k k ′ ′ ′ + − ≈ + + − = 1 2 1 2 k k m k ′ ∆ = 1 2 1 2 2 2 2 m k k T k k k π π π ω = = = ′ ′ ∆ 2 1 2k T T k = ′ 11.15 ( ) 2 2 2 1 2 1 2 T ml θ θ = + ( ) ( ) 1 2 1 cos 1 cos k V mgl r θ θ = − + − − 1 0 2 sin sin r r l l θ θ = + − ( ) 1 1 2 1 0 2 cos sin sin sin kl V mgl r l l 1 θ θ θ θ θ ∂ = − ∂ + − ( ) ( ) 2 2 2 1 1 1 2 3 2 1 0 2 1 0 2 sin 2 cos cos sin sin sin sin kl kl V mgl r l l r l l θ θ θ θ 1 θ θ θ ∂ = − − ∂ + − + − θ 1 θ 2 θ l r 2 2 11 2 3 1 0 0 1 2 2 V k k mg r θ θ θ = = ∂ = = ∂ l l − ( ) 2 2 1 2 3 2 1 0 2 2 cos cos sin sin kl V r l l θ θ θ θ 1 θ θ ∂ = ∂ ∂ + −
- 178. 10 2 2 12 3 12 2 1 0 0 0 1 2 1 2 2 V V k r θ θ θ θ θ θ θ θ = = = = ∂ ∂ = = ∂ ∂ ∂ ∂ 2 kl = ( ) 2 2 2 2 0 2 cos sin sin sin kl V mgl r l l 1 θ θ θ θ θ ∂ = + ∂ + − ( ) ( ) 2 2 2 2 2 2 2 3 2 2 0 2 1 0 2 sin 2 cos cos sin sin sin sin kl kl V mgl r l l r l l θ θ θ θ 1 θ θ θ ∂ = − − ∂ + − + − θ 2 2 22 2 3 2 0 0 1 2 2 V k k m r θ θ θ = = ∂ = = ∂ l gl − Thus, from Equation 11.3.37a or b, we have 2 2 11 1 12 1 2 22 2 1 2 2 V k k k θ θ θ θ = + + But from Equation 11.3.9, for the coupled oscillator, we have ( ) 2 2 2 1 1 1 2 2 1 1 1 2 2 2 2 V kx k x x x x kx ′ = + − + + 2 1 The forms of the potential energy function are similar with … 11 22 12 k k k k and k k ′ ′ = + = = − In the case here … 2 3 0 2kl k mgl and k r ′ = = − -----------------------------------------------------------------------------------------
- 179. 11 Chapter 11 (continued) 11.16 22 1 1 2 2 1 1 2 2 T m x m x = + ( ) 2 2 2 1 1 2 1 2 2 1 1 1 2 2 2 V k x k x x k x ′ = + − + L T V = − 1 1 1 d L m x dt x ∂ = ∂ ( ) 1 1 2 1 1 L k x k x x x ∂ ′ = − + − ∂ ( ) 1 1 1 1 2 1 0 m x k x k x x ′ + − − = 2 2 2 d L m x dt x ∂ = ∂ ( ) 2 2 2 1 2 L k x k x x x ∂ ′ = − − − ∂ ( ) 2 2 2 2 2 1 m x k x k x x ′ + + − 2 1 1 2 2 2 0 m k k k k m k k ω ω ′ ′ − + + − = ′ ′ − − + + ( ) ( ) ( )( ) 4 2 1 2 2 1 1 2 1 2 0 m m m k k m k k k k k k k ω ω ′ ′ ′ ′ − + + + + + + − 2 ′ = ( ) ( ) ( ) 4 2 1 2 2 1 1 2 1 2 1 2 0 m m m k k m k k k k k k k ω ω ′ ′ − + + + + + ′ = ( ) ( ) ( ) ( ) ( ) 2 2 1 1 2 2 1 1 2 1 2 1 2 1 2 2 1 2 4 0 2 m k k m k k m k k m k k m m k k k k k m m ω ′ ′ ′ ′ + + + ± + + + − + + = = ′ 2 For 1 2 1 2 , 2 , , 2 , m m m m k k k k k k ′ = = = = = ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 2 2 2 3 4 6 4 4 2 2 6 2 2 m k m k mk mk m k k m ω + ± + − + = 2 10 6 4 4 k k m m ω = ± 0 0 0 2 k and where m ω ω ω ω = =
- 180. 12 11.17 k k 2m m x2 x1 Note: As discussedin Section 3.2, the effect of any constant external force on a harmonic oscillator is to shift the equilibrium position. x1 and x2 are the positions of the harmonic oscillator masses away from their respective “shifted” equilibrium positions. ( ) 2 2 1 2 1 1 2 2 2 T mx m x = + ( ) 2 2 1 2 1 1 2 2 V kx k x x = + − 1 L T V = − ( ) 1 1 2 1 1 , d L L mx kx k x x dt x x ∂ ∂ = =− + ∂ ∂ 1 − 1 1 2 2 0 mx kx kx + − = ( ) 2 2 2 2 2 , d L L mx k x x dt x x ∂ ∂ = =− ∂ ∂ 1 − 2 2 1 2 0 mx kx kx + − = The secular equation (11.4.12) is thus 2 2 2 0 2 m k k k m k ω ω − + − = − − + 2 4 2 2 2 2 5 2 m mk k k ω ω − + + 0 = The eigenfrequencies are thus … 2 5 17 4 k m ω ± = The homogeneous equations (Equations 11.4.10) for the two components of the jth eigenvector are … 2 1 2 2 2 0 2 j j a m k k a k m k ω ω − + − = − − + For the first eigenvector (the anti-symmetric mode, j = 1) … Inserting 2 1 5 17 4 k m ω + = into the first of the two homogeneous equations yields 11 21 5 17 2 4 k k a ka + − + = 21 11 3 17 4 a a − = Letting a11 = 1, then a21 = -0.281 (Thus, in the anti-symmetric normal mode, the amplitude of the vibration of the second mass is 0.281 that of the first mass and 180o out of phase with it.)
- 181. 13 For the secondeigenvector (the symmetric mode, j = 2) … Inserting 2 2 5 17 4 k m ω − = into the first of the two homogeneous equations yields 12 22 5 17 2 4 k k a ka − − + = 22 12 3 17 4 a a + = Letting a12 = 1, then a22 = 1.781 (Thus, in the symmetric normal mode, the amplitude of the vibration of the second mass is 1.781 that of the first mass and in phase with it.) The two eigenvectors (Equation 11.4.13 and see accompanying Table) are … ( ) ( 11 1 1 1 1 21 1 cos cos 0.281 a Q t a ) 1 t ω δ ω = − = − δ − ) 2 t ( ) ( 12 2 2 2 2 22 1 cos cos 1.781 a Q t a ω δ ω = − = δ − 11.18 ( ) 2 2 2 1 1 1 1 2 2 T ml m l l2 θ θ φ = + + ( ) 1 1 2 cos cos cos V mgl mg l l θ θ φ =− − + For small angular displacements … ( ) 2 2 2 2 2 1 1 2 1 2 2 1 1 2 2 2 m m L T V l l l mgl mgl 2 θ φ θ θ φ = − ≈ + + + − + − ( ) 2 1 1 1 2 , 2 d L L ml ml l l mgl dt 1 θ θ φ θ θ ∂ ∂ = + + = − ∂ ∂ θ 0 1 2 2 2 l l g θ φ θ + + = ( ) 2 1 2 2 , d L L ml l l mgl dt θ φ φ φ φ ∂ ∂ = + = − ∂ ∂ 1 2 0 l l g θ φ φ + + = The secular equation (Equation 11.4.12) is … 2 2 1 2 2 2 1 2 2 2 0 l g l l l g ω ω ω ω − + − = − − + ( ) 4 2 2 1 2 1 2 1 2 2 2 2 l l g l l g l l ω ω − + + − = 4 0 ω Solving for the eigenfrequencies 2 ω … ( ) ( ) ( ) 2 2 2 1 2 1 2 1 2 2 2 2 1 2 1 2 1 2 1 2 2 4 8 2 g l l g l l l l g g l l l l l l l l ω + ± + − = = + + +
- 182. 14 The homogeneous equations(Equations 11.4.10) for the two components of the jth eigenvector are … 2 2 1 1 2 2 2 2 1 2 2 2 0 j j a l g l a l l g ω ω ω ω − + − = − − + Inserting the larger eigenfrequency (the (+) solution for 2 ω above) into the upper homogeneous equation yields the solution for the components of the 1st eigenvector … ( ) 2 2 1 1 11 2 1 21 2 2 l g a l a ω ω − + − = 0 2 2 2 1 2 1 2 1 2 1 2 11 21 2 1 2 1 l l l l l l l l ga ga l l + + + + + + − = 2 2 2 2 1 1 2 21 11 2 l l l l a a l − − + = for the higher frequency, anti-symmetric mode. Inserting the smaller eigenfrequency (the (-) solution for 2 ω ) into the upper homogeneous equation yields the solution for the components of the 2nd eigenvector … ( ) 2 2 1 2 12 2 2 22 2 2 l g a l a ω ω − + − = 0 2 2 2 1 1 2 22 12 2 l l l l a a l − + + = for the lower frequency, symmetric mode. Again, we let a11 = 1 and a21 = 1, since only ratios of the components of a given eigen vector can be determined. The two eigenvectors are thus (Equation 11.4.12 and accompanying table) ( 2 2 1 1 2 1 1 2 2 1 cos Q t l l l l l ) 1 ω δ = − − − + … anti-symmetric ( 2 2 2 2 2 1 1 2 2 1 cos Q t l l l l l ) 2 ω δ = − − + + … symmetric As a check, set and compare with the solution for Example 11.3.1. 1 2 l l = = l 11.19 ( ) 2 2 2 1 2 3 1 2 T m x x x = + + x3 x2 x1 ( ) ( ) 2 2 2 2 1 2 1 3 2 3 1 2 V k x x x x x x = + − + − + L T V = − 0 i i d L L dt x x ∂ ∂ − = ∂ ∂
- 183. 15 ( ) 1 12 1 0 mx kx k x x + − − = 1 1 2 2 0 mx kx kx + − = ( ) ( ) 2 2 1 3 2 0 mx k x x k x x + − − − = 1 2 2 3 2 0 kx mx kx kx − + + − = ( ) 3 3 2 3 0 mx k x x kx + − + = 2 3 3 2 0 kx mx kx − + + = The secular equation (Equation 11.4.12) is … 2 2 2 2 0 2 0 0 2 m k k k m k k k m k ω ω ω − + − − − + − − − + = 0 2 0 ( ) ( ) 3 2 2 2 2 2 2 m k k m k ω ω − + − − + = ( ) 2 2 2 2 0, 2 2 m k or m k k ω ω − + = − + − = 2 2 0 2 2 k m ω ω = = 2 2 2 m k ω − + = ± k ( ) ( ) 2 2 0 2 2 2 2 k m ω ω = ± = ± From Equation 11.5.17 (N = 1 and n = 3) 1 0 2 sin 8 π ω ω = Because 1 cos 2 2 sin θ θ − = 1 0 0 1 cos 4 2 2 2 π ω ω ω − = = 1 2 2 − 1 0 2 2 ω ω = − 2 0 0 2 2 2 sin 2 2 8 2 π 0 ω ω ω = = = ω 3 0 0 3 1 cos3 2 sin 2 8 2 4 π π ω ω ω − = = 3 0 0 2 1 2 2 2 ω ω ω = + = + 2
- 184. 16 11.20 Generalized coordinates: X,s a ( ) θ = θ y x (x,y) X M m a and sin (1 cos ) x X a y a θ θ = + = − cos sin x X a y a θ θ θ = + = θ ( ) 2 2 1 1 2 2 T MX m x y = + + 2 ( ) ( 2 2 2 1 1 cos sin 2 2 MX m X a a θ θ θ θ ) = + + + ( ) 1 cos V mgy mga θ = = − For small oscillations, in terms of the generalized coordinates X and s … ( ) 2 2 2 1 1 1 2 2 2 T MX m X s V mg a ≈ + + ≈ s ( ) 2 2 2 1 1 1 2 2 2 L T V MX m X s mgs a = − ≈ + + − Lagrange’s equations of motion yield … ( ) 0 0 g X s s M m X ms a + + = + + = Assuming … i t i t X Ae s Be ω ω = = we obtain the matrix equation … ( ) 2 2 2 2 0 A g a B M m m ω ω ω ω − = + Setting the determinant of the above matrix equal to zero yields… 2 1 0 ω = , ( ) 2 2 g m M M ω + = The mode corresponding to 1 ω is … 0 0 0 0 g a A B − = implies that 0 A B = = Thus, mode 1 exhibits no oscillation! It is a pure translation with … 0 θ = and 1 2 X A t A = + The mode corresponding to 2 ω is … ( ) 2 2 2 2 0 A g a B ω ω + − =
- 185. 17 or… 2 2 2 2 B m M Ag a m ω ω + = = − − Setting , we have for the 2 1 A = nd mode … 2 i t X e ω = and 2 i t m M e m s a ω θ + = = − This mode corresponds to an oscillation about the CM where … ( ) ( m M X ms ma ) θ + = − = − The normal mode vectors are … 1 2 1 0 A t A Q + = and 2 2 1 i t m M m Q e ω = + − 11.21 (a) We can solve for the normal modes using Equation 11.4.9 … ( ) 2 0 ω − = K M a K and M are the potential energy and kinetic energy matrices respectively. a is a two- component vector whose elements are the amplitudes of the coordinates q. The kinetic energy in Example 11.3.2, assuming small displacements from equilibrium, are … ( ) ( ) 2 2 2 1 1 2 2 2 T mX m X r Xr θ θ ≈ + + + ( ) ( ) 2 2 1 1 1 2 2 2 2 2 T mX m Xr m r θ θ ≈ + + or, in matrix form θ X M m r 1 2 T = qMq where X rθ = q Thus, 2 1 1 1 m = M The potential energy is … ( ) 2 2 1 2 2 g V kX m r r θ ≈ + ( ) ( ) 2 2 2 2 0 1 2 1 2 2 2 2 2 k mg V mX m r m X r m kr θ ω 2 θ ≈ + = + where 2 0 2 k k M m m ω = = +
- 186. 18 Thus The abovematrix equation is thus … 2 0 2 0 0 1 mω = K = 1 2 2 0 2 2 0 2 1 0 0 1 1 1 a m m a ω ω − = ( ) 2 2 2 0 1 2 2 2 2 0 2 0 a a ω ω ω ω ω ω − − = − − From the top equation in the matrix equation above, we get … ( ) 2 2 2 0 1 2 2 0 a a ω ω ω − − ( ) 2 2 0 2 2 1 2 a a ω ω ω − = The amplitudes for the two normal modes can be found by setting 2 2 1 2 or 2 ω ω ω = . In each case we set a , which we are free to do since only ratios of the amplitudes can be determined. 1 1 = For ( ) 2 2 1 2 2 2 0 ω ω = = − ω we obtain … ( ) ( ) ( ) 2 2 0 1 2 2 1 2 1 2 2 2 2 2 2 a ω ω ω − − − = = − = Thus, for mode 1, the lower frequency, symmetric mode … 1 1 1 2 i t e ω = q For ( ) 2 2 2 2 2 2 0 ω ω = = + ω we obtain … ( ) ( ) ( ) 2 2 0 2 2 2 1 2 1 2 2 2 2 2 2 a ω ω ω − + − = = + = − Thus, for mode 2, the higher frequency, anti-symmetric mode … 2 2 1 2 i t e ω = − q (b) In this case, 1 m or M m m M + . We also assume that the spring is “slack”, i.e., k g M m r + , an assumption not stated in the problem (which needs to be rectified in the next edition, I suppose) Also, , so we let 2mg kr ≠ 2 2 0 0 k k and g M m M r ω = ≈ Ω + = The kinetic energy is
- 187. 19 ( ) ( ) 2 22 1 1 2 2 2 T MX m X r Xr θ θ ≈ + + + ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 1 1 1 1 2 2 2 2 2 2 2 T M m X m r Xr MX m Xr m r θ θ θ ≈ + + + ≈ + + θ and the M-matrix is M m m m ≈ M The potential energy is ( ) 2 2 2 0 0 1 1 1 1 2 2 2 2 g V kX m r M m r θ ω ≈ + = + 2 Ω The K-matrix is thus 2 0 2 0 0 0 M m ω ≈ Ω K To solve for 2 ω , we set 2 0 ω − = K M ( ) ( ) 2 2 2 0 2 2 2 0 0 M m m m ω ω ω ω ω − − = − Ω − which yields … 2 2 4 2 2 2 2 4 2 0 0 0 0 0 mM m ω ω ω ω ω ω Ω + − Ω − − = ( ) ( ) 4 2 2 2 2 2 0 0 0 0 0 M m M M ω ω ω ω − − + Ω + Ω = Neglecting m with respect to M and simplifying yields … ( ) 4 2 2 2 2 2 0 0 0 0 0 ω ω ω ω − + Ω + Ω = Solving for 2 ω … ( ) ( ) 2 2 2 2 2 2 0 0 0 0 0 0 2 4 2 2 ω ω ω ω + Ω − Ω + Ω = ± 2 ( ) ( ) 2 2 2 0 0 0 0 2 2 2 ω ω ω + Ω − Ω = ± 2 2 0 Ω Thus, we get … 2 2 2 1 0 2 and ω ω ω = = Now, we solve for the amplitudes a of the normal mode vectors … ( ) 2 0 ω − = K M a ( ) ( ) 2 2 2 0 1 2 2 2 2 0 0 M m a a m m ω ω ω ω ω − − = − Ω − Using the first of the above matrix equations for 2 2 1 2 0 ω ω ω = = gives … 2 1 0 1 a and a = = Using the second equation for 2 2 2 ω ω 2 0 = = Ω gives …
- 188. 20 1 2 0 1 aand a = = Thus, the normal modes are approximately … 0 0 1 2 1 0 0 1 i t i t e and e ω Ω = = Q Q Note that we could have guessed this almost immediately. The above assumption is tantamount to omitting the cross elements . This completely eliminates the small coupling between the two oscillators, which reduces the matrix to the purely diagonal terms … 2 m ω − 2 ω − K M ( ) ( ) 2 2 0 2 2 0 0 0 M m ω ω ω − Ω − , which leads directly to the above solution. 22. m m 2m k K k θ3 θ2 θ1 We “scale” the force constants and masses to 1 unit, namely, 1 1 m and k = = . Let k 1 K and l ′ = = such that 3l = circumference So: ( ) 2 2 2 1 2 3 1 2 2 T θ θ θ = + + ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 2 2 1 1 3 3 1 2 3 3 2 1 1 1 1 1 1 2 2 2 2 2 2 V K 2 K θ θ θ θ θ θ θ θ θ θ θ = − + − + − + − + − + −θ Collecting terms … ( ) ( ) 2 2 2 1 2 3 1 2 1 3 1 4 2 1 2 1 4 4 4 2 V K K 2 3 K θ θ θ θ θ θ θ θ θ = + + + + − − − K - Matrix M - Matrix 2 0 0 0 1 0 0 0 1 = M 4 -2 -2 = -2 2+2K -2K -2 -2K 2+2K K The ttansformation matrix A that diagonalizes these matrices is made up of the three eigenvectors Qi whose amplitudes are ai … we guess that … 1. Uniform rotation 1 2 3 1 1 1 1 or θ θ θ = = = a
- 189. 21 2. Anti-symmetric oscillationof 23, while 1 remains fixed 1 3 2 1 0 0; 1 1 or θ θ θ = = − = − a 3. Anti-symmetric oscillation of 12 together with respect to 3 1 3 2 1 1 1; 1 1 or θ θ θ = = = − − a Thus, and A A = 1 0 1 1 1 1 = 1 1 -1 0 1 -1 1 -1 -1 1 -1 -1 4 We can now diagonalize Kand M … diag 4 -2 -2 -2 2+2K -2K -2 -2K 2+2K = AKA K 1 1 1 1 0 1 = 0 1 -1 1 1 -1 1 -1 -1 1 -1 -1 0 0 0 4 0 0 0 4 8 0 2 0 0 16 0 0 diag diag K Likewise = + = K M The eigenfrequencies are 2 i diag diag i ω = K M 2 2 2 2 2 3 0; 2 4 ; 4 K ω ω ω = = + = The general solution can be generated from the following table … 1 2 1 1 2 1 2 3 1 2 0 Q Q Q a a a a a a a θ θ θ − − − 3 3 3 3 a where ( ) 1 1 1 1 1 1 cos 1 a t ω δ = − Q , ( ) 2 2 2 2 0 1 cos 1 a t ω δ = − − Q , ( ) 3 3 3 3 1 1 cos 1 a t ω δ = − − − Q Thus, ( ) ( ) 1 1 2 2 3 3 3 cos cos a t a t θ ω δ ω δ = − + − ( ) ( ) ( ) 1 1 2 2 2 2 2 3 3 3 cos cos cos a t a t a t θ ω δ ω δ ω δ = − + − − − ( ) ( ) ( ) 3 1 2 2 2 2 2 3 3 3 cos cos cos a t a t a t θ ω δ ω δ ω δ = − − − − −
- 190. 22 Initial conditions are:1 0 2 3 1 2 3 10 ; 0; 0 o θ θ θ θ θ θ θ = = = = = = = The conditions generate 6 equations with 6 unknowns and solving gives … 0 0 1 1 cos cos 2 2 t t 3 θ θ θ ω ω = + 0 0 2 1 cos cos 2 2 t t 3 θ θ θ ω ω = − 0 0 2 1 cos cos 2 2 t t 3 θ θ θ ω ω = − 11.23 See Ex. 11.4.1, page 497-498 The amplitudes of the eigenvectors are … 1 2 3 1 1 1 1 0 2 1 1 1 m M = = = − − a a a K - Matrix M - Matrix 0 0 0 0 0 0 m M m = M K -K 0 = -K 2K -K 0 -K K K 1 1 1 0 1 1 1 1 0 1 2 1 0 2 1 2 1 0 1 1 1 diag K K K K K m M m M K K − = = − − − − − − − K AKA 2 2 0 0 0 0 2 0 0 2 8 8 diag K K K m M K m M = + K 1 1 1 0 0 1 1 1 1 0 1 0 0 1 0 2 1 2 1 0 0 1 1 1 diag m M m M m M m = = − − − − M AMA 2 2 0 0 0 2 0 0 0 2 4 diag m M m m m M + = + M So, we have … 2 1 0 ω = , 2 2 2 2 K K m m ω = = , and … ( ) ( ) 2 2 2 2 2 3 2 1 4 4 2 8 8 2 4 1 2 K m M m M K K m M K m M m m M m m M ω + + + + = = + +
- 191. 23 ( ) ( ) () 2 2 3 1 2 1 2 1 2 m M K K m M m m M m ω + = = + + 11.24 y2 a a α α M x2 m y1 x1 y3 x3 m Select coordinates (xi , yi) as shown, then … ( ) ( ) 2 2 2 2 2 2 1 1 3 3 2 2 1 1 2 2 T m x y x y M x y = + + + + + The potential energy depends only on the compression (or stretching) of the two springs connecting each m to M (hydrogen to sulfur). Let 1 a δ and 2 a δ be incremental changes in the distances a or and or . We have … 1 (1 2) HS → 2 a (3 HS 2) → ( ) ( ) ( ) 1 1 2 2 1 1 2 2 1 sin cos 2 a x x y y x x y y δ α α = − + − = − + − 1 ( ) ( ) ( ) 2 2 3 2 3 2 3 2 1 sin cos 2 a x x y y x x y y δ α α = − + − = − + − 3 ( ) ( ) 2 2 1 2 1 2 V k a a δ δ = + We can reduce the degrees of freedom from 6 to 3 by ignoring the two translational modes and the rotational mode. Thus we consider only vibrational modes. The coordinates must obey the following constraints … No center of mass motion: ( ) ( ) 1 3 2 1 3 2 0 0 m y y My and m x x Mx + + = + + = No angular momentum about any point. We choose that point to be the sulfur atom (M)… 3 1 3 1 sin sin sin sin 0 my a my a mx a mx a α α α − − − α =
- 192. 24 3 1 31 0 y y x x − − − = We introduce three generalized coordinates Q, q1 and q2 that should be close to what we guess would be normal modes … 1 3 1 1 3 2 1 3 ; ; Q x x q x x q y y = + = − = + Solve for xi , yi in terms of Q, q1 and q2 using the above 3 equations of constraint … ( ) ( ) 1 1 2 1 3 3 1 1 ; ; ; 2 2 m 1 x Q q x Q x x x Q q M = + =− = − = − ( ) ( ) 1 2 2 2 3 1 1 ; ; 2 2 m y q Q y q y Q q M = − =− = + 2 Thus, the kinetic energy, in terms of these generalized coordinates, is … 2 2 1 2 1 1 1 1 2 4 4 m T m Q mq m q M M µ = + + + 2 where 2 M m µ = + is the mass of the H2S molecule The potential energy is … 2 2 2 2 2 1 2 2 1 1 1 1 1 2 8 8 4 m V k Q kq k q k q M M µ µ = + + + − 1 2 q M Note, that in the Lagrangian , the only cross term is one involving q q . therefore, Q is a normal mode with eigenfrequency given by the ratio … L T V = − 1 2 2 1 Q Q Q k m K M m M ω = = + Constructing the residual 2x2 K and M matrices involving only q1 and q2 terms, which we will call Kq and Mq gives … 2 2 1 1 1 1 0 4 2 M and 0 M M M µ µ µ µ − = = − Kq Mq We have omitted the factors k and m, which we’ll replace in the final solution, remembering that the eigenfrequencies that we find as a solution to 2 -ω Kq Mq = 0 will be multiples of k / m. ( ) 2 2 1 2 1 0 4 2 2 M - M M M ω µ ω µ µ µ ω − − = = − − Kq Mq which reduces to … ( ) 2 2 2 2 1 M ω ω µ − + = 0 which has the non-trivial solution … ( ) 2 1,2 1 1 2 k M m ω µ = + in which, we have put back the factor k / m. These two modes are “degenerate.”
- 193. 25 Plugging 2 ω backinto the matrix equation ( ) 2 - = ω Kq Mq q 0 3 y 3 gives … 1 2 1 3 1 q q or x x y = − − + = + which can be satisfied in a variety of ways, for example, with 1 3 1 x x and y y = = − , etc Pictorially, the three normal modes are … Q-mode: ant-symmetric about y-axis: 1 3 1 3 x x and y y = = − 2 1 Q k m m M ω = + Breathing mode: symmetric about y-axis: 1 3 1 3 x x and y y = − = ( ) 2 1 1 2 k M m ω µ = + Stretching mode: symmetric about y-axis: 1 3 1 3 x x and y y = − = ( ) 2 2 1 2 k M m ω µ = + -------------------------------------------------------------------------------------------------- 11.25 (a) Plug each of these functions into the wave equation and it is satisfied! (b) ( ) ( ) 1 2 i t i k i t ikx Q q q e e e e ω ω ω +∆ − +∆ − = + = + k x ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 i t k x i t k x i t i k k x e e e e ω ω ω ω − ∆ − ∆ + ∆ − ∆ +∆ − +∆ = + The real part of the above is … ( ) ( ) 2cos cos 2 2 t k x k Q t ω ω ω ∆ − ∆ 2 k x ∆ ∆ = + − + ( ) ( ) ( ) 2cos cos 2 t k x t kx ω ω ∆ − ∆ ≈ − (c) The group speed is … g dx u dt = (the phase of the amplitude remains the same)
- 194. 26 g u k ω ∆ = ∆ 11.26 From Equation 11.5.17... 3 2 4 1 1 1 1 2 3 4 sin sin sin sin 10 10 10 10 , 1.28, 2.62, 3.08 sin sin sin sin 10 10 10 10 N Nπ π π ω ω ω ω π π π π ω ω ω ω = = = = = = π =
- 195. 27 11.27 F k = ∆lis the tension in the cord 1 l d n + ∆ = + l is its stretched length From the equation following Equation 11.5.7… F K d = From Equation 11.6.4c … ( ) 2 k v l l m = + ∆ l ∆ ( ) 1 1 2 2 trans k v l l m = + ∆ l ∆ for transverse waves For longitudinal vibrations, we use Y, the tension in the cord per unit stretched length ( ) ( ) k l Y k l l l ∆ = = ∆ + ∆ l l + ∆ Y K d = (Equation 11.6.8) ( ) 2 2 k l l nd kd v m n m + ∆ = = (Equation 11.6.9a) ( ) 1 nd n d l l ≈ + = + ∆ ( ) 1 2 long k v l m = + l ∆ 11.28 From Equation 11.6.7b … 1 2 F v µ = As in problem 11.27, F k = ∆l 1 2 trans k l v µ ∆ = From Equation 11.6.9b … 1 2 Y v µ = and from problem 11.27 … ( ) l l Y k = + ∆ so … ( ) 1 2 long k l l µ + ∆ = v
- 196. 28 11.29 The general solutionto the wave equation (Equation 11.6.10) that yields a standing wave of any arbitrary shape can be obtained as a linear combination of standing sine waves of a form given by Equation 11.6.14, i.e., ( ) 1 2 ( , ) sin cos sin n n n n n n x y x t A t B t π ω ω λ ∞ = = + ∑ where 2 n n T π ω = since the speed of a wave is … 1 2 0 2 n n n n F v T λ ω λ π µ = = = we have … 1 2 0 2 n n F π ω µ λ = But the wavelength of the standing wave is constrained by the fixed endpoints of the string, i.e. … 2 n l n λ = , so … 1 2 0 n F n l π ω µ = Now, at t = 0, the wave starts from rest in the configuration specified, so… ( ,0) sin n n x y x B l π = ∑ From the discussion of Fourier analysis in Appendix G or in Section 3.9, the Fourier coefficients are given by … 0 2 ( ,0)sin l n n x B y x dx l l π = ∫ Since the string starts from rest, we have … 0 ( , ) sin 0 n n t y x t n x A t l π ω = ∂ = ≡ ∂ ∑ Therefore, 0 n A = The initial configuration is shown in the Figure P11.29. Thus … 2 ( ,0) 0 2 a l y x x x l =
- 197. 29 ( ) 2 ( ,0) 2 al y x l x x l l = − We can now determine Bn … ( ) 2 0 2 2 2 2 sin sin l l n l a n x a n x B x dx l x l l l l l π π = + − ∫ ∫ dx Using integration tables, we obtain the result … ( ) 1 2 2 2 8 1 sin 1,3,5,... 2 n n a n B n n π π − = − = 0 2,4,6,... n B n = = Thus, the general solution is … 2 8 1 3 3 1 5 ( , ) cos sin cos sin cos sin ... 9 25 a vt x vt x vt x y x t l l l l l l π π π π π π π = − + 5 − None of the harmonics which have a node at the midpoint have been stimulated … only the odd harmonics have been excited. 11.30 By analogy with the generation of the traveling sine wave of Equation 11.6.14 from Equation 11.6.13, we get … ( ) ( ) ( ) ( ) 2 3 3 4 1 ( , ) sin sin sin sin 9 x vt x vt x vt x vt a y x t l l l l π π π π π + − + = + − + − ( ) ( ) 5 5 1 sin sin ... 25 x vt x vt l l π π + − + + ----------------------------------------------------------------------------------------------------