solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 19

COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Chapter 19, Solution 2.
Eq. 19.15: 2
m m n m m nv x a xω ω= =
Given data 2
0.2 m/s 4 m/sm mv a= =
: 0.2 m/sm m n m mv x xω ω= = (1)
2 2 2
: 4 m/sm m n m ma x xω ω= = (2)
Divide Equ. (2) by Equ. (1):
2
4 m/s
20 rad/s
0.2 m/s
nω = =
Eq. (1): ( )0.2 m/s 20 rad/smx=
0.01 mmx = 10 mmmx = !
Frequency
20 rad/s
2 2
n
nf
ω
π π
= = 3.18 Hznf = !

cycle 2 rad
sin , 6
s cycle
m n nx x t
π
ω ω= = ×
sin 12mx x tπ=
12 cos12mx x tπ π=&
2
144 sin12mx x tπ π= −&&
12 4 ft/smxπ =
4
0.1061 ft
12
mx
π
= =
1.273 in.mx = !
( )2 2
Max Acc. 144 0.1061 150.8 ft/sπ= = !

Simple Harmonic Motion
20 lb
0.2222 in.
90 lb/in.
s
W
k
δ = = =
( )( )90 12
41.699 rad/s 2
20
32.2
n
k
f
m
ω π= = = =
 
 
 
(a) Amplitude 0.222 in.s mxδ= = =
0.222 in.mx = !
41.699 rad/s
6.6366
2
f
π
= =
6.64 Hzf = !
(b) ( )( )41.699 rad/s 0.2222 in. 9.2655 in./sm n mv xω= = =
9.27 in./smv = !
( ) ( )22 2
41.699 rad/s 0.2222 in. 386.36 in./sm n ma xω= = =
2
32.197 ft/s=
2
32.2 ft/sma = !

(a) ( )sinm nx x tω φ= +
( )( )2
9000 lb/ft
70 lb 32.2 lb/s
n
k
m
ω = =
64.343 rad/s=
2
0.90765 sn
n
π
τ
ω
= =
0.0977 snτ = !
1
10.240 Hzn
n
f
τ
= =
10.24 Hznf = !
(b) At 0 0 00: 0, 10 ft/st x x v= = = =&
( )( )0 0 sin 0 0m nx x ω φ φ= = + ⇒ =
( )( )0 0 cos 0m n n m nx v x xω ω φ ω= = + =&
Substituting 10 ft/s 64.343 rad/smx=
or 0.1554 ft 1.865 in.mx = =
1.865 in.mx = !
( )( )22
0.15542 ft 64.343 rad/sm m na x ω= =
2
643.4 ft/s=
2
643 ft/sma = !

In Simple Harmonic Motion
(a) 2
m m na x ω=
Substituting ( )2 2
50 m/s 0.058 m nω=
or ( )22
862.07 rad/snω =
29.361 rad/snω =
Now
29.361 rad/s
4.6729 Hz
2 2
n
nf
ω
π π
= = =
Then
( )( )
1 cycle 1
in Hz Hz
1 min 60 s/min 60
f = =
So
( ) 11
6060
Hz 4.6729 Hz
280.37 r/min
Hz
f
= =
and 280 rpm
(b) ( )( )0.058 m 29.361 rad/s 1.7029 m/sm m nv x ω= = =
1.703 m/smv =

(a) ( )sinm ntθ θ ω φ= +
( )
2 2
1.35 s
n
n
π π
ω
τ
= =
4.833 rad/s=
( )cosm n ntθ θ ω ω φ= +&
m m nθ θ ω=&
m m m nv l lθ θ ω= =&
Thus, m
m
n
v
l
θ
ω
= (1)
For a simple pendulum
n
g
l
ω =
Thus,
( )
2
2 2
9.81 m/s
4.833 rad/sn
g
l
ω
= =
0.420 m=
From (1)
( )( )
0.4 m/s
0.42 m 4.833 rad/s
m
m
n
v
l
θ
ω
= =
0.197 rad=
or 11.287°
11.29mθ = °!

(b) Now
ta lθ= &&
Hence, the maximum tangential acceleration occurs when θ&& is
maximum.
( )2
sinm n ntθ θ ω ω φ= − +&&
2
m m nθ θ ω=&&
( ) 2
t m nm
a lθ ω=
or ( ) ( )( )( )2
0.42 m 0.197 rad 4.833 rad/st m
a =
2
1.9326 m/s=
( ) 2
1.933 m/st m
a = !

Simple Harmonic Motion:
2
60 lb/ft
6.2161 rad/s
50 lb
32.2 ft/s
n
k
m
ω = = =
6.2161
Hz
2 2
n
nf
ω
π π
= =
(a) 2.4 in. 0.2 ftmx = = 0.2 ftmx = !
0.989 Hznf = !
( )( )22
0.2 ft 6.2161 rad/sm m na x ω= =
2
7.728 ft/s=
(b) f m sF ma mgµ= =
or
2
2
7.728 ft/s
0.240
32.2 ft/s
m
s
a
g
µ = = = !

6 lb/in. 72 lb/ft, 55 in./s, 4 lb.mk v W= = = =
: 0
k
F ma kx mx x x
m
= − = + =&& &&
Thus: 2 72
579.6 24.025 rad/s
4
32.2
k
m
ω ω= = = =
 
 
 
Eq. (19.15): m mv x ω=
( )55 in./s 24.025 rad/smx=
2.2845 in.mx = 2.28 in.mx = !
( )( )2 2 2
2.2845 in. 579.6 rad /sm ma x ω= =
2
1324.1 in./sma = 2
110.3 ft/sma = !

( )60cos 10 45sin 10
3
x t t
π
π π
 
= + − 
 
( )60cos 10 45 sin10 cos cos10 sin
3 3
t t t
π π
π π π
 
= + − 
 
22.5sin10 21.02886cos10t tπ π= + (1)
Now
sin(10 ) sin10 cos cos10 sinm m mx t x t x tπ φ π φ π φ+ = + (2)
Comparing (1) and (2) gives
22.5 cos , 21.02866 sinm mx xφ φ= =
(a)
2 2
0.2 s
10
n
n
π π
τ
ω π
= = = !
(b) 2 2 2
(22.5) (21.02866)mx = +
30.8 mmmx = !
(c)
21.02866
tan
22.5
φ = 0.7516 rad 43.1φ = = °!

At both 600 rpm and 1200 rpm, the maximum acceleration is just equal to g.
(a) 600 rpm 62.832 rad/sω = =
Eq. (19.15):
( )
2
2
62.832
m m m
g
a x xω= =
SI:
( )
3
2
9.81
2.4849 10 m
62.832
mx −
= = × 2.48 mmmx =
US:
( )2
32.2
0.008156 ft
62.832
mx = = 0.0979 in.mx =
(b) 1200 rpm 125.664 rad/sω = =
Eq. (19.15):
( )
2
2
125.664
m m m
g
a x xω= =
SI:
( )
6
2
9.81
621.2 10 m
125.664
mx −
= = × 0.621 mmmx =
US:
( )2
32.2
0.002039 ft
125.664
mx = = 0.0245 in.mx =

Simple Harmonic Motion, thus
( )sinm nx x tω φ= +
400 N/m
16.903 rad/s
1.4 kg
n
k
m
ω = = =
Now (0) 0 sin(0 ) 0mx x φ φ= = + ⇒ =
Then
(0) cos(0 0)m nx x ω= +&
or 2.5 m/s ( )m(16.903 rad/s) 0.14790 mm mx x= ⇒ =
Then ( ) ( )0.14790 m sin 16.903 rad/sx t =  
(a)
At ( )0.06 m: 0.06 m (0.14790 m)sin 16.903 rad/sx t = =  
or
1 0.06 m
sin
0.14790 m
0.02471 s
16.903 rad/s
−  
 
 = =t
0.0247 st = !
(b) Now
( )cosm n nx x tω ω=&
( )2
sinm n nx x tω ω= −&&
Then, for 0.024713 st =
( )( ) ( )( )0.1479 m 16.903 rad/s cos 16.903 rad/s 0.024713 sx  =  &
2.285 m/s= 2.29 m/sx =& !
And
( )( ) ( )( )2
0.1479 m 16.903 rad/s sin 16.903 rad/s 0.024713 sx  = −  &&
2
17.143 m/s= − 2
17.14 m/sx =&& !

Referring to the figure of Problem 19.12
( )cosm n nx x tω ω φ= +&
( )2
sinm n nx x tω ω φ= − +&&
Using the data from Problem 19.13: 0, 0.1479 m, 16.903 rad/sm nxφ ω= = =
, , arex x x& &&
And ( ) ( )0.14790 m sin 16.903 rad/sx t =  
So, at 0.9 s,t =
( ) ( )( )0.1479 m sin 16.903 rad/s 0.9 sx  =  
0.0703 m= 70.3 mmx = !
( )( ) ( )( )0.1479 m 16.903 rad/s cos 16.903 rad/s 0.9 sx  =  &
2.19957 m/s= − 2.20 m/sx =& !
( )( ) ( )( )2
0.1479 m 16.903 rad/s sin 16.903 rad/s 0.9 sx  =  &&
2
20.083 m/s= − 2
20.1 m/sx =&& !

(a)
sin( )m nx x tω φ= +
( )
( )2
9000 lb/ft
70 lb
32.2 lb/s
ω = =n
k
m
64.343 rad/s=
2
0.9765 sn
n
π
τ
ω
= =
With the initial conditions: (0) 15 in. 1.25 ft, (0) 0x x= = =&
( )1.25 ft sin 0mx φ= +
(0) 0 cos(0 )
2
m nx x
π
ω φ φ= = + ⇒ =&
1.25 ftmx =
Then
( ) (1.25 ft)sin 64.343
2
x t t
π 
= + 
 
(1.5) (1.25 ft)sin 64.343(1.5 s) 0.80137 ft
2
x
π 
= + = − 
 
( )(1.5) (1.25 ft)(64.343)cos 64.343 1.5 s 61.726 ft/s
2
x
π 
= + = − 
 
&
In 1.5 s, the block completes
1.5 s
15.361 cycles
0.09765 s/cycle
=

So, in one cycle, the block travels
4(1.25 ft) 5 ft=
Fifteen cycles take
15(0.09765 s/cycle) 1.46477 s=
Thus, the total distance traveled is
15(5 ft) 1.25 ft (1.25 0.80137)ft 77.1 ft+ + − =
Total 77.1 ft= !
(b)
2
(1.5) (1.25 ft)(64.343 rad/s) sin (64.343 rad/s)(1.5 s)
2
x
π 
= − + 
 
&&
2
3317.68 ft/s= 2
3320 ft/sx =&& !

2
2
10 lb
0.31056 lb s /ft
32.2 ft/s
m = = ⋅
With the given properties:
2
50 lb/ft
12.6886 rad/s
0.31056 lb s /ft
n
k
m
ω = = =
⋅
From free fall of the collar
( )0 2 2 1.5 ft 3 9.82853 ft/sv gh g g= = = =
The free-fall time is thus:
( )
1
2 1.5 ft2 3
0.30523 s
y
t
g g g
= = = =
Now to simplify the analysis we measure the displacement from the
position of static displacement of the spring, under the weight of the
collar:
Note that the static deflection is:
10 lb
0.2 ft
50 lb/ft
st
W
k
δ = = =
Then 0,mx kx+ = where x is measured positively up from the
position of static deflection. The solution is:
( )sin ,m nx x tω φ= + with velocity
( )cosm n nx x tω ω φ= +
Now to determine and ,mx φ impose the conditions at impact and
count the time from there.
Thus:
At impact:
0, 0.2 ft and 9.82853 ft/sstt x vδ= = = = − (down)
or 0.2 ft sinmx φ=
( )9.82853 ft/s 12.6886 rad/s cosmx φ− =

Solving for andmx φ
0.800 ftmx = −
0.25268 radφ = −
So, from time of impact, the ‘time of flight’ is the time necessary for the
collar to come to rest on its downward motion. Thus, is the time such2t
that
( )2 20 12.6886
2
x t t
π
φ= ⇒ + =
or 212.6886 0.25268
2
t
π
− =
Hence, 2 0.14371 st =
(a) Thus, the period of the motion is
( ) ( )1 22 2 0.30523 s 0.14371 st tτ = + = +
0.89788 s= 0.898 sτ =
(b) After 0.4 seconds, the velocity is
( ) ( )10.4 cos 0.4m n nx x tω ω φ⎡ ⎤= − +⎣ ⎦
( )( ) ( )( )12.6886 rad/s 0.8 ft cos 12.6886 rad/s 0.4 0.30523 s⎡= − −⎣
]0.25268 rad−
5.91ft/s= − ( )0.4 5.91 ft/sv =

sin , cos ,m n m n n n
g
t t
l
θ θ ω θ θ ω ω ω= = =&
9.81
2.8592 rad/s,
1.2
nω = =
0.18
0:
1.2
m nt θ θ ω= = =&
0.052462 radiansmθ∴ =
At 1.5 s,t = ( )( )0.052462 sin 2.8592 1.5θ =
(a) 0.047826θ = − radians 2.74= − ° !
(b) ( )( ) ( )( )1.2 0.052462 2.8592 cos 2.8592 1.5v =
74.0 mm/sv = !
( )( ) ( )( )2
1.2 0.052462 2.8592 sin 2.8592 1.5a =
2
469 mm/sa = !

(a)
( )0 sin 0 0.75 ftmx x φ= + =
( )0 0 cos 0 ,
2
m nx x
π
ω φ φ= = + ⇒ =&
0.75 ftmx∴ =
When the collar just leaves the spring, its acceleration is
g (downward) and 0.v =
Now
( )0.75 ft cos
2
n nx t
π
ω ω
 
= + 
 
&
( ) ( )0 0 0.75 ft cos ,
2 2 2
n n nx v t t
π π π
ω ω ω
 
= = = + ⇒ + = 
 
&
And
( ) 2
0.75 ft sin
2
n na g t
π
ω ω
 
= − = − + 
 
or ( ) 2
0.75ft ng ω− = −
2
32.2 ft/s
6.5524 rad/s
0.75 ft
nω = =
Then
( )22
2
10 lb
, 6.5524 rad/s
32.2 ft/s
n n
k
k m
m
ω ω= ⇒ = =
13.333 lb/ft=
13.33 lb/ftk = !
(b) 6.5524 rad/snω =

At 1.6 s:t =
( ) ( )( )0.75 ft sin 6.5524 rad/s 1.6 s 0.36727 ft
2
x
π 
= + = − 
 
0.367 ftx = − above equilibrium !
( )( ) ( )( )0.75ft 6.5524 rad/s cos 6.5524 rad/s 1.6 s 4.2848ft/s
2
v x
π 
= = + = 
 
&
4.28 ft/sv = !
( )( ) ( )( )2 2
0.75 ft 6.5524 rad/s sin 6.5524 rad/s 1.6 s 15.768 ft/s
2
a x
π 
= = − + = 
 
&&
2
15.77 ft/sa = !

Determine the constant k of a single spring equivalent to the three springs shown.
Springs 1 and 2:
1 1 1
1 2
1 2
, and
P P P
k k k
δ δ δ= + = +
′
Hence
1 2
1 2
k k
k
k k
′ =
+
Where k′is the spring constant of a single spring equivalent of springs 1 and 2.
Springs k′and 3 Deflection in each spring is the same
So 1 2 1 2 3, and , ,P P P P k P k P kδ δ δ′= + = = =
Now 3k k kδ δ δ′= +
1 2
3 3
1 2
k k
k k k k
k k
′= + = +
+
or
( )( )
( ) ( )
33.5 kN/m 2.1 kN/m
2.8 kN/m 4.11 kN/m 4.11 10 N/m
3.5 kN/m 2.1 kN/m
k = + = = ×
+
(a)
3
2 2
0.361 s
4.11 10 N/m
13.6 kg
n
k
m
π π
τ = = =
×
1 1
2.77 Hz
0.3614 s
n
n
f
τ
= = =

(b) ( )sinm nx x tω φ= +
And 44 mm 0.044 mmx = =
( )( )2 2 2.77 Hz 17.384 rad/sn nfω π π= = =
And
( ) ( )0.044 m sin 17.4 rad/sx t φ = + 
( )( ) ( )0.044 m 17.4 rad/s cos 17.4 rad/sx t φ = + &
Then ( )( )max 0.044 m 17.4 rad/s 0.766 m/sv = =
( )( ) ( )2
0.044 m 17.4 rad/s sin 17.4 rad/sx t φ = − + &&
Then ( )( )2 2
max 0.044 m 17.4 rad/s 13.3 m/sa = =
max 0.765 m/sv =
2
max 13.31 m/sa =

(a) First, calculate the spring constant
( ) ( ) ( ) ( )24 kN/m 12 kN/m 12 kN/m 48 kN/mP δ δ δ δ= + + =
48 kN/mk∴ =
Then
3
40 10 N/m
30.984 rad/s
50 kg
n
k
m
ω
×
= = =
2
0.20279 sn
n
π
τ
ω
= =
1
4.9312 Hzn
n
f
τ
= =
0.203 snτ = !
4.93 Hznf = !
(b) Now
And, since 0 0.060 mx =
( ) ( )0.060 m sin 30.984 rad/sx t φ = + 
( )( ) ( )0.060 m 30.984 rad/s cos 30.984 rad/sx t φ = + &
( )( ) ( )2
0.060 m 30.984 rad/s sin 30.984 rad/sx t φ = − + &&
Hence max 1.859 m/sv = !
2
max 57.6 m/sa = !

Equivalent spring constant
2 2 4k k k k′ = + =
(Deflection of each spring is the same.)
( )1
2
2
6.8 s
10 lb
32.2 ft/s
π
τ = =n
k
2
2
10 6.8
32.2
k π 
⇒ =  
 
0.2651 lb/ftk =
( )2
2
2
4
6 lb
32.2 ft/s
π
τ =n
k
( )( )2
2
4 0.2651 lb/ft 32.2 ft/s
6 lb
π
=
2.633 s=
2.63 snτ =

Equivalent Springs
Series: 1 2
1 2
s
k k
k
k k
=
+
Parallel: 1 2pk k k= +
2 2 2 2
;s p
s ps pk k
m m
π π π π
τ τ
ω ω
= = = =
( )
( )
( )
2 22
1 21 2
1 2 1 2
1 2
5
2
τ
τ
  ++ 
= = = =       
+
ps
p s
k k kk k
k kk k k
k k
( )( ) 2 2
1 2 1 1 2 26.25 2k k k k k k= + +
( ) ( )2 2 2
2 2 2
1
4.25 4.25 4
2
k k k
k
−
=
m
1
2
2.125 3.516
k
k
= m
1
2
1
4 or
4
k
k
=

For a static load P the total elongation of the spring is:
1 2 3
P P P
k k k
δ = + +
1 1 1
8 kN/m 12 kN/m 16 kN/m
P
 
= + + 
 
6 4 3 13
48 48
δ
+ + 
= = 
 
P P
48
3.6923 kN/m
13
P
k
δ
= = =
3
3.6923 10 N/m
( ) 12.153 rad/s
25 kg
k
a
m
ω
×
= = =
2 2
0.517 s
12.153
π π
τ
ω
= = = 0.517 sτ = !
12.153
1.9342 Hz
2 2
f
ω
π π
= = = 1.934 Hzf = !
(b) For 30 mm = 0.03 mmx =
( )( )0.03 m 12.153 rad/sm mv x ω= = 0.365 m/smv = !
( )( )22
0.03 m 12.153 rad/sm ma x ω= = 2
4.43 m/sma = !

For equilibrium
Equal stretch
12 N 16 N
,A C B
k k
δ δ δ= = =
12 N ,A CT T kx= = + 16 NBT kx= +
( ) ( ) ( )4.08 9.81 2 12 16mx kx kx= × − + − +&&
3 0mx kx+ =&&
( )
3
0.0125 cos
k
x t
m
=
(a) Then ( )( )16 0.0125 1 0BT k= + − =
1.280 kN/mk = !
(b)
( )
( )
2 2 23 12803
941.76 rad /s
40 9.81
n
k
m
ω = = =
30.688 rad/s 4.88 Hzn fω = ⇒ = !
(c) ( )
3
12 1280 0.0125 cosA
k
T t
m
= +
3
Minimum when cos 1
k
t
m
= = −
( ) ( )min
12 1280 0.0125 4AT = − = −
Max Comp: 4 N !

Springs in parallel, eq 1 22k k k= +
12 16,000 2 2
10
0.2
n
k
m
π π
ω π
τ
+
= = = =
After removal
12 2
8
0.25
n
k
m
π
ω π= = =
Elimination: 2 12 16,000
100
k
m
π
+
=
2 12
64
k
m
π =
(a) 2 2 16,000
100 64 ,
m
π π∴ = × 45.0 kgm = !
(b) 2 1
1
2
64 , 14,222 N m
45.0316
k
kπ = =
1 14.22 kN/mk = !

The equivalent spring constant for springs in series is:
1 2
1 2
e
k k
k
k k
=
+
1 2For andk k
( )
1 2
1 2
2 2π π
τ = =
+
e
A A
k k k
m m k k
1For alonek
1
2π
τ′ =
A
k
m
(a)
2
2 1 2k k k
τ
τ
 
= + ′ 
0.2 s
0.16667
0.12 s
τ
τ
= =
′
And with 2 3.5 kN/mk =
( )( )2
13.5 kN/m 1.6667 3.5 kN/mk= +
or 1 6.22 kN/mk =
(b)
1
2π
τ′ =
A
k
m
A
A
W
m
g
=
( )
( )
2
1 1
2
2
A
k
m
τ
π
=
And with 3
1 6.22 kN/m 6.22 10 N/mk = = ×
( ) ( )
( )
2 3
2
2
0.12 s 6.22 10 N/m
2.2688 Ns /m 2.2688 kg
2
Am
π
×
= = =
2.27 kgAm =

The equivalent spring constant, accounting for the springs in series, is
eq
2
k
k k= +
( )( ) ( )( )
eq
22
2 2
0.4 s
3 /2
100 lb 32.2 ft/s100 lb 32.2 ft/s
k k
π π
τ = = =
It follows that 510.849 lb/ftk =
511 lb/ftk = !
After the changes
eq
2
0.4 s
120/32.2
k
π
τ = =
Thus eq 919.528 lb/ftk =
Since
eq
A
A
kk
k k
k k
= +
+
or
( ) ( )
( )
eq
eq
510.849 919.528 510.849
2 2 510.849 919.528
A
k k k
k
k k
− −
= =
− −
2043.4 lb/ft=
2040 lb/ftAk = !

Initially
2
1.6 s
A
k
m
π
τ = =
After the 14 lb weight is added to A,
2
2.1 s
A B
k
m m
π
τ′ = =
+
(a) A B
A
m m
m
τ
τ
′ +
=
2
2.1
1.6
A B
A
m m
m
+ 
= 
 
( )( )1.7227 A A Bm m m= + Note 2
14 lb
32.2 ft/s
Bm =
2
14 lb
1.3838 1.3838
32.2 ft/s
A Bm m
 
= =  
 
( )2
2
14 lb
32.2 ft/s 1.3838 19.373 lb
32.2 ft/s
A AW m
 
= = = 
 
19.37 lbAW = !
(b)
2
A
k
m
π
τ =
( )
( )
( )
2
2
2 Am
k π
τ
=
( )
( )
2
2
2
19.373 lb
32.2 ft/s
2
1.6 s
k π
 
 
 =
9.278 lb/ftk =
9.28 lb/ftk = !

, 2 2 2s s
d d
x x F kx k x
h h
= = =
( ) ( )eff
: 2A AM M Fd mgx mx hΣ = Σ − = − &&
2
d
k x d mg x mxh
h
 
− = − 
 
&&
2
2
2
0
kd g
x x
hmh
 
+ − = 
 
&&
22
2 2
2
2 2
,
kd g k d g
h m h hmh
ω ω
   
= − = −   
  
(1)
Data: 0.3 m; 0.7 m; 20 kgd h m= = =
(a) For 1 s:τ =
2
;
π
τ
ω
= 2 22
1 s = ; 4
π
ω π
ω
=
Eq. (1):
2
2 2 0.3 9.81
4
20 0.7 0.7
k
π
 
= − 
 
2912.4 N/mk = 2.91 kN/mk =
(b) For Infinite:τ =
2
;
π
τ
ω
= 0ω =
Eq. (1):
2
2 2 0.3 9.81
0
20 0.7 0.7
k
ω
 
= = − 
 
763.0 N/mk = 763 N/mk =

(a) The equation of motion is:
( )( )( ) ( )( ): 0.75 0.75 2 1.2 m 1.2AM kθ θΣ − = &&
1.125 1.44k mθ θ− = &&
1.44 1.125 0m kθ θ+ =&&
( )3
1.125 1.35 10
0
1.44m
θ θ
×
+ =&&
1054.69
0
m
θ θ+ =&&
Then 2 1054.69 2
, and 4n n
n
s
m
π
ω τ
ω
= = =
2
2 1054.69 1054.69 2
or
4 4m m
π π 
= =  
 
427.45 kgm =
427 kgm =
(b)
( )sinm ntθ θ ω φ= +
At 0t =
0.06
0.03636 rad
1.65
m
m
θ = =

And 0θ =&&
Then 0.03636 sinmθ φ=
( )0 cosm n ntθ ω ω φ= +
Thus , 0.0363 rad
2
m
π
φ θ= =
Now
0.06
0.03636 rad
1.65
m
m
θ = =
1054.69
1.5708 rad/s
427.45
nω = =
Then
0.06
0.03636 rad
1.65
m
m
θ = =
( )( ) ( )0.03636 rad 1.5708 rad/s cos 1.5708 rad/s
2
t
π
θ
 
= + 
 
&
( )( )max 0.03636 rad 1.5708 rad/s 0.05712 rad/sθ = =&
( ) ( )( )max
1.2 0.05712 rad/s 0.06854 m/scv m= =
( )max
68.5 mm/scv =

(a)
( )( )( )
( )
6 2 2 2 3 4
3 3
48 30 10 lb/in. 144 in. /ft 2 10 ft48
15 ft
P EI
k
Lδ
−
× ×
= = =
122880 lb/ft= 5
1.229 10 lb/ftk = ×
(b)
( )2
122880 lb/ft
51.360 rad/s
1500 lb 32.2 ft/s
n
k
m
ω = = =
or
1 51.36
8.17 Hz
2 2 2
n
n
k
f
m
ω
π π π
= = = =

(a) , ,e
PL AE
P k P
AE L
δ δ δ
 
= = =  
 
( )2
6
3
0.32 in.
29 10 psi
4
129.57 10 lb/in.
18 in.
e
AE
k
L
π 
   ×   = = = ×
3
129.6 10 lb/in.ek = × !
(b)
( )
( )
( )
3
2
129.57 10 lb/in.
16 lb
32.2 ft/s
81.272 Hz
2 2
e
n
k
mf
π π
×
= = =
81.3 Hznf = !

st
W
k
δ
=
W
m
g
=
2 st
W
n W
stg
k g
m
δ
ω
δ
= = =
1
2 2
n
n
st
g
F
ω
π π δ
= = !

(a) ( )( )2
0.10 kg 9.81 m/s 0.981 Nmg − =
1
2
2
00
0.981 N
4
4
F mg x x
 
= = ⇒ =  
 
0.06015 m=
0 60.1 mmx = !
(b) At 0,x ( )
1 1
2 2
0
0
4
2 0.06015
2x
dF
x
dx
− − 
= = 
 
8.1549 N/m=
0
8.1549 N/me
x
dF
k
dx
 
= = 
 
8.1549 N/m
0.10 kg
1.4372 Hz
2 2π π
= = =
e
n
k
mF
1.437 HznF = !

Using the Binomial Theorem we write
1
2
2
2 2
1
1 sin sin
2
1 sin sin
2
m
m
θ
φ
θ
φ
−
  
= −   
    
−  
 
2 21
1 sin sin
2 2
mθ
φ= + + ⋅ ⋅ ⋅ ⋅
Neglecting terms of order higher than 2 and setting ( )2 1
sin 1 cos2 ,
2
φ φ= − we have
( )2 2
0
1 1
4 1 sin 1 cos2
2 2 2
m
n
l
d
g
π θ
τ φ φ
  
= + −  
  
∫
2 2 2
0
1 1
4 1 sin sin cos2
4 2 4 2
m ml
d
g
π θ θ
φ φ
 
= + − 
 
∫
2
2 2
0
1 1
4 sin sin sin 2
4 2 8 2
m ml
g
π
θ θ
φ φ φ
  
= + −  
  
21
4 sin 0
2 4 2 2
ml
g
π θ π  
= + +  
  
21
2 1 sin
4 2
m
n
l
g
θ
τ π
 
= + 
 
!

For small oscillations: 0 2
l
g
τ π=
We want ( )01.01 1.01 2
l
g
τ τ π= =
Using the formula of Prob. 19.34, we write
2
0 0
1
1 sin 1.01
4 2
mθ
τ τ τ
 
= + = 
 
( )2
sin 4 1.01 1 0.04; sin 0.2; 11.54
2 2 2
m m mθ θ θ
= − = = = °
23.1mθ = ° !

Eq. (19.20)
2
2
k l
g
τ π
π
 
=   
 
For 0.8 m:l =
2
0.8 m
2 2 1.7943 s
9.81 m/s
l
g
π π= =
Thus: ( )
2
1.7943 s
k
τ
π
=
Using Table 19.1:
(a) For small oscillations:
2
0; 1m
k
θ
π
= =
1.7943 sτ = !
(b) For
2
30 : 1.017m
k
θ
π
= ° =
( )( )1.017 1.7943 sτ = 1.825 sτ = !
(c) For
2
90 : 1.180m
k
θ
π
= ° =
( )( )1.180 1.7943 sτ = 2.12 sτ = !

From Equation 19.20:
2
2n
k l
g
τ π
π
  
=      
For 60 , 1.686m kθ = ° =
( ) ( )( )( )3 s 2 1.686 2
32.2
l
=
3
0.44484 0.19788
32.2 6.744 32.2
l l
= = ⇒ =
Thus 6.3718 ft 76.462 in.l = =
76.5 in.l = !

( ) ( )0.6 , 0.4A st B stA B
F k l F k lθ δ θ δ   = + = +   
(a) ( )eff
:C CM MΣ = Σ
( ) ( )0.6 0.6 0.1 0.4 0.4 0.1st st tA B
lk l lmg lk l I lmaθ δ θ δ α   − + + − + = +    (1)
Equilibrium position: 0θ =
( ) ( )0.6 0.1 0.4 0st stA B
lk lmg lkδ δ− + − = (2)
Substitute (2) into (1)
( ) ( )2 2
0.1 0.6 0.4 0tI lma l k l kα θ θ+ + + =
or 2
0.1 0.52 0tI lma l kα θ+ + =
But
2
, 0.1 , ,
12
t
ml
I a lα α θ= = = &&
So ( )
2
2 2
0.1 0.52 0
12
ml
l m l kθ θ
 
+ + = 
 
&&
or 2
0.09333 0.52 0l kθ θ+ =&&
2
5.5714 0l kθ θ+ =&&
( )
850 N/m
5.5714 0
9 kg
θ θ
 
+ = 
 
&&
2
526.19 s 0θ θ−
+ =&&

Thus 2
526.19 s 22.939 rad/s, 3.6508 Hz
2
n
n nf
ω
ω
π
−
= = = =
3.65 Hznf =
(b) ( ) ( )sin , cosm n m n nt tθ θ ω φ θ θ ω ω φ= + = +&
m m nθ θ ω=&
( ) ( )0.6A mm
x l θ= &&
( )( )( )0.0011 m/s 0.6 0.6 m 22.939 rad/smθ=
0.0001332 rad 0.0076mθ = = °
0.0076mθ = °

( )eff
:B BM MΣ = Σ
Diskcos
2 2 2
AB
L L L
mg kxr I m Iθ α α α
  
− = + +  
  
(1)
where where from statics
2
st st
L
x r mg k rθ δ δ= + =
into (1) and assuming small angles ( )cos 1θ ≈
2
L
mg θ 2
stkr krθ δ− −
2
Disk
2
AB
L
I m I α
  
 = + +    
so
2
2
Disk 0
4
AB
mL
I I krθ θ
 
+ + + =  
 
&&
so
2 2
2
2 2
2 2
Disk Disk
1 1
4 12 4 2
n
AB
kr kr
mL mL
I I mL M r
ω = =
+ + + +
(2)
Disk
: 0.9 m
7.5 kg
0.25 m into (2)
6 kg
5 kN/m
Data L
m
r
M
k
= 

= 

= 
= 
= 

( )( )
( )( ) ( )( ) ( )( )
2
2
2 2 2
5000 N/m 0.25 m
1 1 1
7.5 kg 0.9 m 7.5 kg 0.9 m 6 kg 0.25 m
12 4 2
nω =
+ +
2 2
141.243 rad/snω =
11.8846 rad/snω =
(a) so
2 2
11.8846n
π π
τ
ω
= = 0.529 sτ = !
(b) 0.020 mmx =
( )0.02 11.8846 0.2377 m/sm m nv x ω= = =
max 238 mm/sv = !

At equilibrium, ( )spring tension 20 lb sin15= °
( )
0
20 lb sin15
stretch
50 lb/ft
x
°
= =
20sin15PMΣ = °( ) 0
4
50
12
x−
4
12
x
  +    
2
4
12
1 20 4 20 4
16.667 0.310559
2 32.2 12 32.2 12
x
x x x
       
= + − =               
!!
!! !!
0.310559 16.667 0, 7.326 rad/snx x ω+ = =!!
(a) 0.858 sτ = !
(b) sin , cosn n nx A t x A tω ω ω= =!
( )0.5 7.326 3.66 in./snAω = = !

Kinematics
2
cos sin
Non-linear terms
: sin
2
A
x A
a l l
ml
F N ma
θ θ θ θ
θ θ
= +

+
Σ = − 

& &&
&&
2 2
sin cos sin sin
2 3 2
A A
kl ml l
M Nl ma
θ
θ θ θ θΣ = − + = −
&&
or, letting cos 1,θ = sin cos , sin 0, since 1θ θ θ θ θ= = <<
2 2
2 3
0,
3 2 2
n
ml kl k
m
θ θ ω+ = =&&
For 1.5 ft,l = 2
6 lb
, 23 lb/in. 276 lb/ft
32.2 ft/s
m k= = =
(a)
1 3
7.50 Hz
2 2
k
f
mπ
= = !
(b) 2 2 max
max maxn n
x
L
θ ω θ ω= − = −&&
( )
( )
2max
0.8
3 276
3 12
98.747 rad/s
62
2 1.5
32.2
k x
m L
 
−  
 = − = =
 
 
 
2
max 98.7 rad/sθ =&& !

(a)
( )eff
:A AM MΣ = Σ
( ) ( ) ( )1.2 0.825 2 0.75 0.825 1.2R S R t tG C
mg m g F I m a m aα+ − = + + (1)
Where ( )0.75S stF k θ δ= +
At equilibrium, 0θ =
And
( )0: 1.2 0.825 2 0.75 0A R stM mg m g kδΣ = + − = (2)
( ) ( ) ( ) ( )0.825 1.2 2 0.75 0.75R t tG C
I m a m a kα θ+ + = −
( ) ( ), 0.825 , 1.2t tG C
a aα θ α α= = =&&
( )( )22
21.2 kg 1.65 m
0.2725 kg m
12 12
ml
I = = = −
( ) ( ) ( ) ( ) ( )2 2 2
0.2725 0.825 1.2 1.2 2 0.75 450 0m θ θ + + + =  
&&
[ ]1.08925 1.44 506.25 0m θ θ+ + =&&
506.25
0
1.08925 1.44m
θ θ+ =
+
&&

Now ( )
( )
( )
2 2
22
2
2506.25 2
2 109.66
1.08925 1.44 0.6
n n
n
f
m
π π
ω π
τ
 
= = = = = 
+  
( )506.25 109.66 1.08925 1.44m= +
2.4495 kgm =
2.45 kgm =
(b) ( ) ( )max max
0.06
1.2 , 0.03636 rad
1.65 1.65
θ θ= = = =& B
C m
y
v
( )( )
1
22
0.03636 rad 109.66 s 0.38075 rad/sm m nθ θ ω −
= = =&
( ) ( )max
1.2 0.38075 rad/s 0.45691 m/sCv = =
( )max
457 mm/sCv =

2mx kx P= −&& Mx P F= −&&
21
2
Mr Frθ =&&
Eliminate , andP F θ
3
2 0
2
M
m x kx
 
+ + = 
 
&&
( )2 2 5000 N/m2
769.23
3 4 kg 9 kg
2
n
k
M
m
ω = = =
++
27.735 rad/snω =
(a)
2
0.22654 sn
n
π
τ
ω
= =
0.227 snτ = !
(b) ( )max 12 mm 333 mm/snv ω= = !

From Problem 19.43
( )2 2 3500 N/m2
538.46; 23.2048 rad/s
3 4 kg 9 kg
2
n n
k
M
m
ω ω= = = =
++
( )( ) ( )( )2 2
6 kg 9.81 m/s 4 kg 9.81 m/s 98.1 NN W Mg mg= = + = + =
2 21 1
2 2
x
Fr Mr Mr
r
θ
 
= =  
 
&&&&
( )1
22
6 kg
max , ; Amplitude
98.1 N
ω= = =&& &&m m n
F
x x A A
N
( )
( )2
1
6 kg
20.5 23.2048 rad/s
98.1 N
s Aµ = =
or 0.03036 mA =
30.4 mmA = !

2
2
2
4
2
2 3
0.021440 m
16
2
r r
r
b
r
r
π
π
π
  
−     = =
−
( )2
16 0.22546 kg
2
m tr t
π
ρ ρ
 
= − = 
 
( )
24 4 2
4 8 4
4 2
6 4 9 2 3
o
t r r r r
I r t r
ρ π π
ρ
π π
  
= − − + −  
   
( )4 136 9
0.009342
3 4
tr t
π
ρ ρ
 
= − = 
 
(a) 2
0,o n
o
tmgb
I mgb
I
ρ
θ θ ω+ = = =&& ( )( )( )0.22546 9.81 0.021440
tρ ( )0.009342
2
5.07602, 2.2530 rad/sn nω ω= =
2
2.79 s
n
π
τ
ω
= = !
(b)
2
2
2 2 2
9.81 m/s
, 1.9326 m
5.07602 rad /s
n
n
g g
l
l
ω
ω
= = = =
1.933 ml = !

Kinematics: / ,G O G Oa a a GO d= + =
0.3 m, 4 2 /3 0.180063 mr d r π= = =
( )
2
PM mgd I m r dθ θ θ Σ = − = + −  
&& &&
Where
2 2
2
32
2 9
mr mr
I
π
= −
2
n
m
ω =
gd
m 2
32
2
r m
−
2
2
9
r
m
π
+ ( )2
65.515
r d
=
 
− 
 
2
0.776264 s
n
π
τ
ω
= = 0.776 sτ = !

21
2
AI mr=
(a)
2
4
0
2 3
mr r
mgθ θ
π
 
+ = 
 
&&
or
8
0
3
g
r
θ θ
π
+ =&&
8
3
n
g
r
ω
π
=
2 2
6.82
8
3
n
n
r
gg
r
π π
τ
ω
π
= = = !
(b)
2
2 2 4
3
r
b r
π
 
= +  
 
2 2
2 2 21 4 4
2 3 3
B G
r r
I I mb mr m mr m
mπ
   
= + = − + +   
   
23
2
mr=
23
0
2
mr mgbθ θ+ =&&
or 2
2
3
gb
r
θ =&&
2
2
3
n
gb
r
ω =
2
2 2
7.38
16
2 1
9
3
n
n
r
g
g
r
π π
τ
ω
π
= = =
+
!

( )
( )
2
22 1.61 1.6
2 0.6
3 3.6 3.6 12
A
bm m
I b b
     
 = + +    
       
211
25
mb=
( )2 0.3
3.6 6
b b b
y
b
= =
13
0.6
6 30
b
AG b b= − =
2
2 211 13
0.64
25 30 36
B
b
I mb m b m b
  
= − + +  
   
223
25
mb=
24.04
6
b
GB =
(a)
2
11 13 2
0, 6.33
25 30
n
n
mb b b
mg
g
π
θ θ τ
ω
+ = = =&& !
(b)
2
23 24.04 2
0, 6.67
25 6
n
n
mb b
mg b
g
π
θ θ τ
ω
 
+ = = =  
 
&& !

( )2 21
12
I m b c= +
1
2
ta r cα θ= = &&
( )eff
1 1
: sin
2 2
A Α tM M mg c I ma cθ α
   
Σ = Σ − = +   
   
( )
2
2 21 1
sin
2 12 2
m
mgc b c m cθ θ θ
 
− = + +  
 
&& &&
2 2
sin 0
2
3 12
g c
c b
θ θ
 
 
+ = 
 +  
&&
2
2 2 2
6
2
4 1
3 12
c
g c g b
bc b c
b
ω
 
 
 = =
  + +  
   
(1)

(a) Since
2
,
π
τ
ω
= minimum τ occurs when 2
ω is maximum. We consider
c
b
 
 
 
as the variable.
( )
2
2
2
2
4 1 8
6
4 1
c c c
d b b bg
c b
d c
b
b
ω
        
+ −       
        = =
      +   
   
2 2
4 1 8 0
c c
b b
   
+ − =   
   
2
1
4
c
b
 
= 
 
0.5
c
b
= !
(b) For a simple pendulum of length c we have 2 8
.
c
ω = We equate the value of 2
ω to that of Eq. (1).
2
6
4 1
c
g gb
b cc
b
 
 
  =
  
+  
   
2 2 2
1
4 1;
2
     
= + =     
     
c c c
b
b b b
0.707
c
b
= !

Consider general pendulum of centroidal radius of gyration k.
( ) ( ) 2
0 eff
: sinM Μ mgr mr r mkθ θ θ0Σ = Σ = +&& &&
2 2
sin 0
gr
r k
θ θ
 
+ = 
+ 
&&
For small oscillations, sin ,θ θ≈ we have
2 2
2
2 2 2 2
2
0 ; ; 2
gr gr r k
grr k r k
π
θ θ ω τ π
ω
  +
+ = = = = 
+ + 
&&
Connecting rod suspended at A: 1.03 sτ =
160 mm 0.16 mar r= = =
Thus
( )
( )( )
2 2
0.16
1.03 2
9.81 0.16
k
π
+
=
( ) ( )( )
2
2 2 1.03
0.16 9.81 0.16 0.04218
2
k
π
 
+ = = 
 
2
0.04218 0.0256 0.01658 ; 0.12876 mk k= − = = 128.8 mmk = !

2
21 1 10
m 0.334722 m
2 2 12
I mr
 
= = = 
 
α θ= &&
(a) The disk is free to rotate and is in curvilinear translation.
Thus 0Iα =
Then ( )effB BM MΣ = Σ
sin , sintmgl lmaθ θ θ− = ≈
2
0ml mglθ θ+ =&&
And
2
2 2
26
12
32.2 ft/s
14.861 s
ft
n
g
l
ω −
= = =
1
3.8551 snω −
=
2
1.6299 sn
n
π
τ
ω
= =
1.630 snτ =

(b) When the disk is riveted at A, it rotates at an angular acceleration α
( )effB BM MΣ = Σ
21
sin , , sin
2
tmgl I lma I mrθ α θ θ− = + = ≈
2 21
0
2
mr ml mglθ θ
 
+ + = 
 
&&
Then
( )2
2 2
2 2
2
2
26
32.2 ft/s ft
12
13.838 s
10
ft
2 2612
ft
2 12
ω −
 
 
 = = =
    +           +    
 
  
n
gl
r
l
1
3.7199 snω −
=
2
1.6891 sn
n
π
τ
ω
= =
1.689 snτ =

( )0 0 eff
: sin tM M W r I ma rθ αΣ = Σ − = +
2 2
sinmgr mk mrθ θ θ− = +&& &&
2 2
sin 0
gr
r k
θ θ+ =
+
&& (1)
For a simple pendulum of length OA l=
0
g
l
θ θ+ =&& (2)
Comparing Equations (1) and (2)
2 2
r k
l
r
+
=
2
k
GA l r
r
= − = (Q.E.D) !

See Solution to Problem 19.52 for derivation of
2 2
sin 0
gr
r k
θ θ+ =
+
&&
For small oscillations sinθ θ≈ and
2 2 2
2 2
2n
n
r k k
r
gF rg
π π
τ π
ω
+
= = = +
For smallest nτ we must have
2
k
r
r
+ a minimum
2
2
2
1 0
k
d r
r k
dr r
 
+  
  = − =
2 2
r k=
r k= (Q.E.D) !

Same derivation as in Problem 19.52 with r replaced by .R
Thus 2
0
gR
R k
θ θ+ =
+
&&
Length of the equivalent simple pendulum is
2 2 2
R k k
L R
R R
+
= = +
( )
2
2
= − + =
k
L l r l
k
r
Thus the length of the equivalent simple pendulum is the same as in
Problem 19.52. It follows that the period is the same and that the new
center of oscillation is at O (Q.E.D)

2 2
2 2
5
4 4
l l
d l= + =
Ignore static terms: 2
2
C
l
M mg klθ θΣ = − −
2
2 2 2 5
2
12 4 4
ml
mgl kl ml ml
θ θ
θ θ θ
 
− − = + +   
 
&& &&
&&
2
3
2
2
5
3
mg
k
l
m
ω
+
=
1 6 9
2 5 10
n
k g
f
m lπ
= + !

( )eff
:
4 4 2
D D
kL r mrL mLr
M M mgr
θ
θ θ θΣ = Σ − − = +&& &&
3
0
4 2
mLr kLr
mgrθ θ
 
+ + = 
 
&&
2 4 2
3 3
n
g k
L m
ω = +
1 2 4
Hz
2 3 3
n
k g
f
m Lπ
= + !

2
2
2
1 16 lb 1 12.5 in.
12 12 12 in./ft32.2 ft/s
I ml
   
= = ⋅   
  
2
0.04493 lb ft s= ⋅ ⋅
α θ= !!
2
0.16667
12
ta α θ
 
= = 
 
!!
sinθ θ≈
( )effC CM MΣ = Σ
( ) ( )2 2
0.6875 0.16667 0.16667k mg I mθ θ θ θ− + = +!! !!
( )( )
16
0.04493 0.02778 0.47266 0.16667 16 0
32.2
kθ θ
  
 + + − =    
  
!!
[ ]0.05693 0.47266 2.6667 0kθ θ+ − =!! (1)
(a) For 50 lb/ft,k = 368.28 0θ θ+ =!!
368.28
3.007 Hz
2 2
n
nf
ω
π π
= = =
3.01Hznf = !

(b) For ,nτ → ∞ 0nω → oscillations will not occur
From Equation (1)
2 0.47266 2.6667
0
0.05873
n
k
ω
−
= =
5.642 lb/ftk =
5.64 lb/ftk = !

2 3
3
3 2 3
a a
b
 
= =  
 
4 4
23 3
96 96
O
a a
I t mbρ
 
= + +  
 
23 5
,
2 2 12
O
t a
m a I ma
ρ  
= =  
 
With
2
20 lb s 16
, ft, 14 lb/ft
32.2 ft 12
m a k
⋅ 
= = = 
 
So with sinθ θ≈ for small ,θ ( )effO ΟΜ ΜΣ = Σ yields
2
25
2 0
12 3
ma mga
kaθ θ
 
+ + = 
 
&&
2
2
12
2 0
5 3
mga
ka
ma
θ θ
 
+ + = 
 
&&
2 2
2
12 12 24
2
55 3 5 3
n
mga g k
ka
a mma
ω θ
     
= + = +     
    
2 212 32.2 24 14
141.655 s
16 2055 3
12 32.2
nω −
  
  
= + =  
     
   
11.90 rad/snω =
1.894 Hz
2
n
f
ω
π
= = !

Ignore static terms
( ) ( ) ( ) ( )( ) ( )( )2 2 2 21 1
50 0.6 50 0.9 1.2 0.3 1.2 0.6 1.8 0.9
3 3
BM gθ θ θ θ
 
Σ = − − + = + 
 
&&
54.9684
0.63 54.9684 0, 9.3408 rad/s
0.63
nθ θ ω+ = = =&&
0
15
cos cos
900
t tθ θ ω ω
 
= =  
 
At 2 2
00.7 s, cos 0.7 1.407 rad/sn nt θ ω θ ω= = − = −&&
2
1.407 rad/sα =

Determine location of the centroid G.
Let mass per unit lengthρ =
Then total mass ( ) ( )2 2m r r rρ π ρ π= + = +
About C 22
0 2
r
mgc r g r gπ ρ ρ
π
 
= + = 
 
2r
y
π
=
( ) 2
2 2r c rρ π ρ+ =
( )
2
2
r
c
π
=
+
( )0 0 eff tM M a c cα θ α θΣ = Σ = = =&& &&
sin sinnmgc I mcaθ α θ θ− = + ≈
( )2
00 0I mc mgc I mgcθ θ θ θ+ + = + =&& &&

But 2
0I mc I+ =
( ) ( )
( )2
2
0 0 0 s-circ linesemi circ line
2
12
r
I I I m r m= + = +
( )s-circ line 2
2
m
m r m r
r
ρπ ρ ρ
π
= = =
+
( )
2 2
2
0
2 2
3 2 3
r r mr
I r rρ π π
π
 ⋅  
= ⋅ + = +   +   
( ) ( )
2
2 2
0
2 3 2
mr r
mgπ θ θ
π π
 
+ + = + + 
&&
( )
( )
( )( )
2
2
2 2
3 3
2 9.81 m/s2
0.32 m
n
g
r
ω
π π
= =
+ +
2 2
16.0999 s 4.0125 rad/sn nω ω−
= =
( )sinm n Bt y rθ θ ω φ θ= + =
( ) ( ) ( )sin sinB m n B nm
y r t y tθ ω φ ω φ= + = +
At 0t = 0.03 m 0B By y= =&
( ) ( ) ( )0 0 cos 0
2
B B m
t y y
π
φ φ= = = + =&
( ) ( )0.03 sin 0 0.03 m
2
B B Bm m
y y y
π 
= = + = 
 
1
0.03sin 4.0125 s
2
B n ny t
π
ω ω − 
= + = 
 
( )( ) ( )0.03 cos 0.03 sin
2
B n n n ny t t
π
ω ω ω ω
 
= + = − 
 
&
( )( )2 2
0.03 sin 0.03 cos
2
B n n n ny t t
π
ω ω ω ω
 
= − + = 
 
&&
At 10 s:t = ( ) ( ) ( )( )2 2
0.03 4.0125 cos 4.0125 10 0.36437 m/sB Bt
a y= = = −&&
( ) ( ) ( )( )0.03 4.0125 sin 4.0125 10 0.07902 m/sB Bv y= = =&
( ) ( )
( )
11
2 22 222
2 2 20.07902
0.36437 0.365 m/s
0.32
B
B B t
v
a a
r
          = + = − + =             

( )221 1
0.250
2 2 32
m
I mr m α θ= = = = &&
0.650ta lα θ= = &&
(a) The disk is free to rotate and is in curvilinear translation.
Thus 0I α =
( )effB BM MΣ = Σ
sin tmgl lmaθ− =
2 2
0 n
g
ml mgl
l
θ θ ω+ = =&&
From 19.17, the solution to this Equation is
( )sinm ntθ θ ω φ= +
At 0,t = 2 rad, 0
180 90
π π
θ θ= ⋅ = =&
0 0 cos
2
m nt
π
θ ω φ φ= = =
sin 0
90 2
m
π π
θ
 
= + 
 
rad
90
m
π
θ =

Thus sin
90 2
nt
π π
θ ω
 
= + 
 
( ) maxmaxA m nv l lθ θ ω= =&
With 26 in. 2.1667 ft, rad,
90
m n
g
l
l
π
θ ω= = = =
( ) ( )
( )2
max
32.2 ft/s
2.1667 ft rad
90 2.1667 ft
Av
π 
=  
 
0.29156 ft/s=
3.4987 in./s=
( )max
3.50 in./sAv =
(b) For disk riveted at A ( I α included)
( )effB BM MΣ = Σ sin tmgl I lmaθ α− = +
2 21
0
2
mr ml mglθ θ
 
+ + = 
 
&
2
2
2
2
n
r
gl
l
ω =
+
sin
90 2
nt
π π
θ ω
 
= + 
 
(See (a))
( ) maxmaxA m nv l lθ θ ω= =&
( )
( )( )2
2 2
32.2 ft/s 2.1667 ft
2.1667 ft
90 1 10 in. 26 in.
2 12 in./ft 12 in./ft
rad
π 
=  
     
+   
   
0.28135 ft/s=
3.376 in./s=
( )max
3.38 in./sAv =

( )1
8 0.3
8 in. = ft, 2 ,
12 32.2
r m rπ
   
= =   
   
( )2
0.3
32.2
m r
 
=  
 
2
1 2
2 0.045767 ft
r
m
b
m m
= =
+
2 2 2
1 2
1
0.58813 lb s ft
3
= + = ⋅ ⋅AI m r m r
( )1 2 0AI m m gbθ θ+ + =&&
( )1 22
n
A
m m gb
I
ω
+
=
1.9105 rad/s,nω = 0 cos ntθ θ ω=
2
2 rad,
180
π
θ° = ° = 0 sinn ntω θ ω ω= −
2
0 cosn ntα θ ω ω= −
At 5 s,t = 0.00849 rad/sω = !
0.1264 rad/sα = !

( )eff
M MΣ = Σ
K Iθ θ− = &&
0
K
I
θ θ+ =&&
2
,n
K
I
ω = 2
I
K
τ π= (1)
For 50-mm-diameter sphere, 0.025 mr =
2 3 2 52 2 4 8
5 5 3 15
SI mr r r rπ ρ π ρ
 
= = = 
 
( ) ( )5 3 6 28
0.025 7850 kg/m 128.45 10 kg m
15
π −
= = × ⋅
Solve Eq. (1) For :I
2
2
;
4
K
I τ
π
=
2
R R
S S
I
I
τ
τ
 
=  
 
(2)
Data: 4.1 s, 6.2 s, 0.45 kgR S Rmτ τ= = =
Eq. (2):
2
4.1 s
0.4373
6.2 s
R
S
I
I
 
= = 
 
( )6 2
0.4373 0.4373 128.45 10 kg m−
= = × ⋅R SI I
6 2
56.17 10 kg m−
= × ⋅I
6 2
2 56.17 10 kg m
0.45 kg
R
R
R
I
k
m
−
× ⋅
= =
2 6 2
124.82 10 mk −
= ×
3
11.173 10 mk −
= × 11.17 mmRk = !

( )effG GM MΣ = Σ
K Iθ θ− = &&
0
K
I
θ θ+ =&&
2
n
K
I
ω = (1)
2
I
K
τ π= (2)
Data: 1.95 N m/radK = ⋅
3 kgm =
( )( )22 3 21 1
3 kg 0.5 m 62.5 10 kg m
12 12
I ml −
= = = × ⋅
(a) Eq. (2)
3 2
62.5 10 kg m
2
1.95 N m/rad
τ π
−
× ⋅
=
⋅
1.125 sτ = !
(b) Max velocity
Eq. (1) 2
3 2
1.95 N m/rad
31.2
62.5 10 kg m
ω −
⋅
= = =
× ⋅
n
K
I
5.586 rad/snω =
Simple harmonic motion
m m nω θ ω= 180 radmθ π= °=
( )( )rad 5.586 rad/s 17.548 rad/smω π= =
( ) ( ) ( )( )0.25 m 17.548 rad/s 4.387 m/sA mm
v AG ω= = =
( ) 4.39 m/sA m
v = !

Equivalent torsional spring constant
( )2 1 1 1, ,eT K T K T Kθ θ θ θ= = − =
( )2 1 2 1K K Kθ θ= +
2
1
1 2
K
K K
θ θ=
+
1eT K Kθ θ= =
2
1
1 2
e
K
K K
K K
θ θ=
+
1 2
1 2
e
K K
K
K K
=
+
Newton’s Law
( )effc cM MΣ = Σ
eK Jθ θ− = &&
21
2
J mr=
21
0
2
emr Kθ θ+ =&&
( )( )
2
2
4 40
2 2 32.2 12
2
3 1.52
2
3 1.5
n
n eK
mr
π π
τ π
ω
  
  
  = = =
 
 
+ 
5.2197 s=
5.22 snτ = !

2 21
4
ABI mr mb= +
[ ]2 0ABI k mgbθ θ+ + =&&
2
2
2
2
4
n
k mgb
r
m b
ω
+
=
 
+  
 
2
120 lb 8
, ft
1232.2 ft/s
m r
   
= =   
  
6.4
ft,
12
b
 
=  
 
150 lb ft/radk = ⋅
(a)
( ) ( )
2
2
2
120 6.4
2 150 32.2
32.2 12
246.93
8
120 6.412
32.2 4 12
nω
   
+    
   = =
  
       +       
 
  
15.7139 rad/snω =
2 2
0.3998 s
15.7139n
π π
τ
ω
= = =
0.400 sτ = !
(b) 0 0 max 0sin ; cos ;n n n nt tθ θ ω θ θ ω ω θ θ ω= = =& &
So max max 0 nv b bθ θ ω= =&
( )max
6.4 2
ft 15.7139 rad/s 0.2925 ft/s
12 180
v
π  
= =  
  
max 3.51 in./sv = !

( )effG GM MΣ = Σ 0K J J Kθ θ θ θ− = + =&& &&
Empty platform centroidal of platformPJ J=
( )( )
22
2 2
27 N m/rad 2.2 s2 2
4 4
n
n P
n
P
K
J
K
J
π π τ
τ
ω π π
⋅
= = = =
2
3.31 N m sPJ = ⋅ ⋅
Platform with object A centroidal ofAJ J A=
( )
( )2
2
2 2
4
n
n P A
n
P A
K
J J
K
J J
τπ π
τ
ω π
′
′ = = + =
′
+
( )( )
2
2
2
27 N m/rad 3.8 s
3.31 N m s
4
AJ
π
⋅
= − ⋅ ⋅
2
9.88 3.31 6.57 N m sAJ = − = ⋅ ⋅
2
6.57 N m sAJ = ⋅ ⋅ !

Geometry
3
b
l
θ
φ =
3 3
mg b
F
l
θ
=
Then
3
G G
b
M I Fθ θ
 
Σ = = −  
 
&&
or
2 2
0
12 3
mb mgb
L
θ θ+ =&&
(a) 2 4 2
,n
n
g l
l g
π
ω τ π
ω
= = = !

(b)
3mx F= −&&
mgx
l
= −
2
n
g
l
ω =
2
l
g
τ π= !

( )2 2 21
2
T m b c θ= + &
2 21
2
V kc θ=
( )
2
2
2 2n
kc
m b c
ω =
+
720 N/m, 1.5 kgk m= =
17.5271 rad/snω =
0
0.015 m
0.033333 rad
0.45 m
θ = =
0max 0.351 m/sD nv cω θ= =
max 0.351 m/sDv = !

From Problem 19.69
17.5271 rad/snω =
( )
0
0.25 m/s
0.5555 rad/s
0.45 m
θ = =&
0
max 0.01902 mD
n
c
x
θ
ω
= =
&
max 19.02 mmDx = !

2
2 2 21 1
2 2 2
B
l
T I ml mθ θ
  
= = +  
   
& &
2
1
2 2
l
V k
θ 
=  
 
2
2
2
8
5
8
n
kl
ml
ω =
5
k
m
=
1
Hz
2 5
n
k
f
mπ
= !

Equilibrium:
0
0
0
2
2
20 lb
1
ft
9
k y
T
T k y
y

=

+ =

∴ =

Vibration: 2
2 2
1 20
,
2
T v
g
 
=  
 
( ) ( )
2
2 1
1 1
1 1
144 144
2 2 2
x
V x
 
= +  
 
1 2:V T= 2 2
1
1 20
2 32.2
n xω
 
 
 
( ) ( ) 2
1
1 1
144
2 8
x
 
= + 
 
2
289.8, 17.0235 rad/sn nω ω= =
(a)
2
0.36909 s,
n
π
τ
ω
= = 0.369 sτ = !
(b)
1
ft 1.8915 ft/s
9
nm
v ω
 
= = 
 
1.892 ft/sm
v = !

Datum at 1
Position 1
2
1 0 1
1
0
2
mT J Vθ= =&
Position 2
2 20T V mgh= =
( ) 2
1 cos 2sin
2
m
mh r r
θ
θ= − =
2
2
mrθ
≈
2
2
2
m
V mgr
θ
=
Conservation of energy
2
2
1 1 2 2 0
1
0 0
2 2
m
mT V T V J mgr
θ
θ+ = + + = +&
2 2 2
0m n m n m mJ mgrθ ω θ ω θ θ= =&
( )22
02 2
2
0
44
n n
n
Jmgr
J mgr
ππ
ω τ
ω
= = =
( )( )2
2 2
0 2
4
n mgr
J J mr J mr
τ
π
= + + =
( )( )2
2
2
4
n mgr
J mr
τ
π
= −
( ) ( )( )( )
( )( )
2 2
2
2
1.26 s 38 kg 9.81 m/s 0.175 m
38 kg 0.175 m
4π
= −
2 2 2
2.62345 N m s 1.16375 N m s 1.4597 N m sJ = ⋅ ⋅ − ⋅ ⋅ = ⋅ ⋅
2
1.460 kg mJ = ⋅ !

Find nω as a function of c.
Datum at 2
Position 1 1 10T V mgh= =
( )1 1 cos mV mgc θ= −
2
2
1 cos 2sin
2 2
m m
m
θ θ
θ− = ≈
2
1
2
m
V mgc
θ
=
Position 2
2
2
1
2
C mT I θ= &
2 2 21
12
CI I mc ml mc= + = +
2
2 2
2 2
1
0
2 12
m
l
T m c Vθ
 
= + =  
 
&
2 2 2
2
1 1 2 2 0 0
2 12 2
m ml
T V T V mgc m c
θ θ 
+ = + + = + +  
 
&
m n mθ ω θ=&
2
2 2
12
n
l
gc m c ω
 
= +  
 
2
2
2
12
n
gc
l
c
ω =
 
+  
 

Maximum c, when
2
2 2
2
2
2
2
12
0 0
12
n
l
g c c g
d
dc l
c
ω
 
+ −  
 = = =
 
+  
 
2
2
0
12
l
c− =
12
l
c = !

Consider a general pendulum of centroidal radius of gyration .k
Datum at 1
Position 1
2
1 0
1
2
mT J θ= !
1 0V =
Position 2
2 20T V mgh= =
( ) 2
1 cos 2sin
2
m
mh r F
θ
θ= − =
2 2
2
2 2
m m
V mg F
θ θ
= ≈
1 1 2 2T V T V+ = +
2 2
0
1 1
0 0
2 2
m mJ mgrθ θ+ = +!
m n mθ ω θ=!
2 2 2
0 n m mJ mgrω θ θ=
2 0
0
2
2n n
n
mgr J
J mgr
π
ω τ π
ω
= = =
2 2 2
0J J mr mk mr= + = +

(a)
2 2
2n
k r
gr
τ π
+
=
For a rod suspended at A:
2 2
0.895 2 ,a
n a
a
k r
r r
gr
τ π
+
= = = (1)
For a rod suspended at B:
2 2
0.805 2 ,b
n b
b
k r
r r
gr
τ π
+
= = = (2)
10.5 in. 0.875 fta br r+ = = (3)
From (1) and (2)
( )22 2
2
0.895
4
a
a
gr
k r
π
+ = ( )1′
( )22 2
2
0.805
4
b
b
gr
k r
π
+ = ( )2′
Taking the difference ( ) ( )1 2′ ′− :
[ ]2 2
2
0.80125 0.648025
4
a b a b
g
r r r r
π
− = −
( )( ) [ ]2
0.80125 0.648025
4
a b a b a b
g
r r r r r r
π
− + = −
( )
( )
[ ]2
32.2
0.80125 0.648025
4 0.875
a b a br r r r
π
− = −
( ) 0.746685 0.60406a b a br r r r− = −
0.25332 0.39594 0.63979a b b ar r r r= ⇒ =
0.63979 0.875 0.533605 fta a ar r r+ = ⇒ =
6.4033 in.= 6.40 in.ar = !
(b)
2
2 20.895
2
a ak gr r
π
 
= − 
 
( )( )
( )
( )
2
22
2
0.895 s
32.2 ft/s 0.533605 ft 0.533605 ft
2π
= −
2 2 2
0.34863 ft 0.28473 ft 0.0639 ft= − =
0.25278 ft 3.0334 in.k = =
3.03 in.k = !

2
2
P
mr
I >
2 2
24 4
1 0.715 0.715 ft
3 3
r
d r r
π π
   
= + − = =   
   
2
2
24 2
2 3
O
mr r
I m md
π
  
 = − +     
0.65117 m=
2
21
,
2 2
OT I V mgd
θ
θ= =&
2 2 35.3564
0.65117
2
n
d
mg
m
ω = =
5.9461 rad/snω =
2
1.057 s
n
π
τ
ω
= = !

Since vibration takes place about the position of equilibrium, we shall neglect the effect of weight and the
static deflection.
Position of Max Displacement
Spring Elongations
2
B m
l
x θ= C mx lθ=
Position 1: 1 0T =
( )
2
22 2
1
1 1 1
2 2 2 2
B B C C B m C m
l
V k x k x k k lθ θ
 
= + = + 
 
2 2
1
1 1
2 4
B C mV k k l θ
 
= + 
 
Position 2: 2 0V =
2
2 2 2 2
2
1 1 1 1 1
2 2 2 12 2 2
m m m m
l
T I mv ml mω ω ω
   
= + = +   
   
2 2
2
1
6
mT ml ω=
Conservation of Energy
2 2 2 2
1 1 2 2
1 1 1
: 0
2 4 6
B C m mT V T V k k l mlθ ω
 
+ = + + + = 
 
For simple harmonic motion, ( )22 2 21 1 1
;
2 4 6
m n m B C m n mk k l mlω ω θ θ ω θ
 
= + = 
 
2 3 1
4
n B Ck k
m
ω
 
= + 
 
(1)

Note: Result is independent of length of the rod.
Data
6 lb,W = 3 lb/in. 36 lb/ft,k = = 5 lb/in. = 60 lb/ft.Ck =
2
2
3 36 lb/ft
60 lb/ft 1110.9
46 lb
32.2 ft/s
nω
 
= + =    
 
 
33.33 rad/snω =
33.33
2 2
n
f
ω
π π
= = 5.30 Hzf = !

2 2
1
2 2 4
l l
V k mg
θ θ 
= + 
 
2 2 2 2 2
2
1 1
2 4 2 2 4
L mr L
T m
r
θ θ   
= +      
   
& &
2 2
3
16
mL θ
=
&
Then
2
2
2
8 4
3
16
n
kL mgL
mL
ω
+
=
2 2
3
k g
m L
 
= + 
 
1 2 4
Hz
2 3 3
n
k g
f
m Lπ
= + !

(a) Position 1
( )2
1 1
1
0
2
st mT V k rδ θ= = +
Position 2
( )22 2
2 2
1 1 1
2 2 2
m m stT J mv V mgh kθ δ= + = +!
1 1 2 2T V T V+ = +
( ) ( )2 22 21 1 1 1
0
2 2 2 2
st m m m stk r J mv mgh kδ θ θ δ+ + = + + +!
2
stkδ 2 2 2 2 2
2 2st m m m m stk r kv J mv mgh kδ θ θ θ δ+ + = + + +! (1)
When the disk is in equilibrium
0 sinC stM mg r k rβ δΣ = = −
Also sin mh r βθ=
Thus 0stmgh k rδ− = (2)
2 2 2 2
m m mkr J mvθ θ= +!
m n m m m n mv r rθ ω θ θ ω θ= = =! !
( )2 2 2 2 2
m m nkr J mrθ θ ω= +

2
2
2n
kr
J mr
ω =
+
21
2
J mr=
2
2
2 2
2
1 3
2
n
kr k
mmr mr
ω = =
+
( )
2 2
0.71983 s
800 N/m2
3 7 kg
n
n
π π
τ
ω
= = =
0.720 snτ = !
(b)
m mv rθ= !
m m nθ θ ω=!
0.01 mm m n mv r rθ ω θ= =
( )
2
0.01 m 0.0873 m/s
0.720 s
mv
π 
= = 
 
!

sinθ θ≈
2
2
1 cos 2sin
2 2
m m
m
θ θ
θ− = ≈
Position1
2
2
1
1 1
2
2 2 2
m m
l
T J mθ θ
   
= +   
   
& &
1 0V =
Position 2
( )
2
2 2
2
0 1 cos
2 2 2
m m
l l
T V W kθ θ
 
= = − − +  
 
2 2
2
2
2 2 4
m
m
Wl kl
V
θ
θ= − +
1 1 2 2T V T V+ = +
( )
2 2 2
2 2 21 1
2 0 0
2 2 4 2 2 4
m
m m m
l wl kl
J m
θ
θ θ θ+ + = − +& &
21
12
m n m
W
J l
g
θ ω θ= =&

2
2 2 2 2
6 4 2 2
n m m
W W Wl kl
l
g g
ω θ θ
   −
+ = +    
   
2 62 2
55
12
n
W kl
g k
WlW
l
gg
ω
 −
+  −
 = = +
   
  
  
2
6 9.81 m/s 120 N/m
5 0.160 m 0.6 kg
 −
= +  
 
2
166.43 s−
=
12.901 rad/snω =
1
2.0532 s
2
n
nf
ω
π
−
= =
2.05 Hznf = !

( )
2
21
2 4
l
V k l mg
θ
θ
 
= +   
 
2
2 21 1
2 12 2
l
T ml m θ
  
= +  
   
&
2
2
2
2 4
6
n
kl mgl
ml
ω
+
=
2 3 3
2
n
k g
m l
ω = +
With 1500 N/m, 0.6 m, 10 kgk l m= = =
2 2 2
474.525 rad /snω =
2
0.288 s
n
π
τ
ω
= = !

Let m be the mass of rod and Cm be mass of each collar
Then
2 2 2 2
2 4 2
C
kl mgl l
V m g
θ θ θ 
= + +  
 
( )
2 221 1
2 3 2
C
l
T m m lθ θ
 
= +  
 
& &
2
2
22
2 4 2
6 2
C
n
C
mgl m glkl
m lml
ω
+ +
=
+
2 4 2
6 2
C
C
mg m gk
l l
mm
+ +
=
+
( )( )
( )
( )( )
( )
2 2
2
5 kg 9.81 m/s 2.5 kg 9.81 m/s
1500 N/m
2 4 0.6 m 2 0.6 m
5 kg 2.5 k
6 2
nω
 
 + +
 
 =
 + 
 
2
397.62 s−
=
19.4838 rad/snω =
2
0.322 s
n
π
τ
ω
= = !

2 2
1 20, 2
2 2 2
l
V V mgl mg
θ θ 
= = +   
 
2
mglθ=
2
2 2 2
2 1
1 1
0, 2
2 2 3
ml
T T ml θ θ
 
= = +  
 
& &
2 2
5
6
ml θ
=
&
2
2 2 2 2 25 6
: ,
6 5
n n n
ml g
mgl
l
θ ω θ ω ω= = =&
( )
( )
2 2
1.586 s
9.81 m/s6
0.75 m5
n
π π
τ
ω
= = = !

6.87 NA AW m g= =
4.91 NC CW m g= =
9.81 NAC ACW m g= =
Position 1 ( ) ( ) ( )
2 2 2 2
1
1 1 1 1
0.1 0.16 0.03
2 2 2 2
A m C m AC m AC mT m m m Iθ θ θ θ= + + +& & & &
( )21
0.26
12
AC ACI m=
So ( ) ( ) ( ) ( )( )
2 2 2 2 2
1
1 1
0.7 0.1 0.5 0.16 1 0.03 1 0.26
2 12
mT θ
 
= + + + 
 
&
( )2 21
0.02633 kg m
2
mθ= ⋅ &
1 0V =
Position 2
( )( ) ( )( ) ( )( )2 20, 0.1 1 cos 0.16 1 cos 0.03 1 cosA m C m AC mT V W W Wθ θ θ= = − − + − + −

With
2
2
1 cos 2sin
2 2
m m
m
θ θ
θ− = "
( )( ) ( )( ) ( )( )
2 2
2 6.87 0.1 4.91 0.16 9.81 0.03 0.3929
2 2
m m
V
θ θ
 = − + + = 
( ) ( )
2
2
1 1 2 2
1 1
: 0.02633 0 0 0.3929
2 2 2
m
mT V T V
θ
θ+ = + + = +&
m n mθ ω θ=&
So 2 20.3929
14.922 s
0.02633
nω −
= =
2 2
1.6266 s
14.922
n
n
π π
τ
ω
= = =
1.627 snτ = !

Position 1
( ) ( ) ( )( )
2
22 2 2
1 0
1 1 1 1 2
2
2 2 2 2 2 4
A C m mm m
m l
T m v m v I θ θ
 
= + + +   
 
& &
( ) ( )21
12 2
G A C mm m
m
I l v v lθ= = = &
2 2
2 2 2
1
7
24 8 6
m m m
ml ml
T ml mlθ θ θ
 
= + + =  
 
& & &
1
5
2 cos cos cos
2 2
l
V mgl mg mglβ β β= − − = −
Position 2
2 0T =
( ) ( )2 cos cos
2 2
m m
m l
V mgl gβ θ β θ= − − − −
( ) ( )cos cos
2 2
m m
m l
mgl β θ β θ− + − +
[ ]2
5
cos cos sin sin cos cos sin sin
4
m m m mV mgl β θ β θ β θ β θ= − + + −

2
5
cos cos
2
mV mgl β θ= −
( )
2
cos 1 small angles
2
m
m
θ
θ ≈ −
2
2
5
cos 1
2 2
m
V mgl
θ
β
 
= − − 
 
1 1 2 2T V T V+ = +
2
2 27 5 5
cos 0 cos 1
6 2 2 2
m
mml mgl mgl
θ
θ θ β
 
− = − −  
 
&
m n mθ ω θ=&
2 2 27 5
cos
6 4
n m ml gω θ βθ= ⋅
2 15
cos
14
n
g
l
ω β=
2
2 15 32.2 ft/s
cos40
25 in.14
12 in./ft
nω
 
 
 = °
 
 
 
2
12.686 s−
=
1
3.5617 snω −
=
1
0.56686 s
2
n
nf
ω
π
−
= =
0.567 Hznf = !

D = Disk
R = Rod
2ABl r=
For small oscillations: ( ) 21
1 cos
2
m mh r rθ θ= − =
Position 1: 1 0T =
2
1
1
2
R R mV W h m grθ= =
Position 2: 2 0V =
( )22 2
2
1 1 1
2 2 2
D m D D R mm
T I m v Iω ω= + +
( )
22 2 2 2 21 1 1 1 1
2
2 2 2 2 12
D m D m R mm r m r m rω ω ω
   
= + +   
   
2 21 3 1
2 2 3
D R mm m r ω
 
= + 
 
Conservation of Energy.
1 1 2 2 :T V T V+ = +
2 2 21 1 3 1
0
2 2 2 3
R m D R mm g m m rθ ω
 
+ = + 
 
But for simple harmonic motion: m n mω ω θ=
( )22 21 1 3 1
2 2 2 3
R m D R n mm gr m m rθ ω θ
 
= + 
 

2
3 1
2 3
R
n
D R
m g
rm m
ω
 
=  
 +
or 2
3 1
2 3
R
n
D R
W g
rW W
ω
 
=  
 +
(1)
Data:
4
3 lb, 5 lb, ft
12
R DW W r= = =
( ) ( )
2 3 32.2
34.094
3 1 4
5 3
2 3 12
nω
 
 
= = 
 +  
 
5.839 rad/snω =
2 2
5.839n
π π
τ
ω
= = 1.076 sτ = !

Position 1: 1 0,T = 2
1
1
2
mV kx=
Position 2: 2 0,V = 2 2 2
2
1 1 1
2
2 2 2
AB m m Disk mT m v I m vω
 
= + + 
 
2
2 2 2
2
1 1
2 2
m
AB m Disk Disk m
v
T m v m r m v
r
  
= + +  
  
( ) 2
2
1
3
2
AB Disk mT m m v= +
1 1 2 2:T V T V+ = + ( )2 21 1
0 3
2 2
m AB Disk mkx m m v+ = +
But for simple harmonic motion, :m n mv xω=
( )( )221 1
3
2 2
m AB Disk n mkx m m xω= +
2
3
n
AB Disk
k
m m
ω =
+
Note: Result is independent of r
Data: 5 kN/m, 9 kg, 6 kgAB Diskk m m= = =
( )
2 5000 N/m
185.185
9 kg 3 6 kg
nω = =
+
13.608 rad/snω =
13.608
2 2
n
f
ω
π π
= = 2.17 Hzf = !

( )1
4
1 cos
3
r
V mgh mg θ
π
 
= = − 
 
2
1 cos
2
mθ
θ− ≈
2
1 2
3
m
V mgr
θ
π
=
2 2 2 2
2
1 1 1 1
2 2 2 2
A A B BT I I mrω ω ω
 
= + +  
 
Where
2 2
1
2 8 4 256
A B
m r mr
I I
  
= = =  
  
and 4A Bω ω ω= =
2 2 2 2
2
2
5
16 4 16
mr mr mr
T
ω
ω
 
∴ = + =  
 
1 2
2
,
m
V T=
2
mgrθ 5
3
m
π
=
2 2 2
n mr ω θ
16
2 32
,
15
n
g
r
ω
π
=
1 32
2 15
n
g
f
rπ π
 
=  
 

Kinematics:
Data
0.15 , 0.05m m m AB m mv r v bω ω ω ω= = = = 10kgABm =
4kgDm =
1 1 2 2T V T V+ = + (1)
Where
( )
1
1
0
Position 1, Max. Displacement
cosAB m
T
V m g r b θ
= 

= − − 
( )0.10 cos 9.81cosAB m mm g θ θ= − = −
2 2 2
2
1 1 1
2
2 2 2
D m D m AB ABT m v I m vω
 
= + + 
 
( )( ) ( )( ) ( )( )2 22 21 1 1 1
2 4 0.15 4 0.15 10 0.05
2 2 2 2
m m mω ω ω
  
= + +  
  
2
0.1475 mω=
( ) ( )2 0.10 9.81AB ABV m g r b m g= − − = − = −
Into (1)
2
0 9.81cos 0.1475 9.81m mθ ω− = −
( )2
0.1475 9.81 1 cosm mω θ= −
But 21
1 cos
2
m mθ θ− ≈
2 2
0.1475 4.905m mω θ= (2)

m n mω ω θ=
Into (2)
( )2 2 2
0.1475 4.905n m mω θ θ=
So
2 4.905
33.254
0.1475
nω = =
5.7666 rad/snω =
0.9178 Hz
2
n
f
ω
π
= = 0.918 Hznf =

2
2 21 1
2 7
2 2 6
x
T mx my m
 
= + = 
 
&
& &
Equilibrium of BC:
3
0
2 2 2
D
mg l l
M ku
 
Σ = = ⋅ −   
 
Where 0, natural length
2 3
mg
u l l u
k
= = = +
( ) ( )
2 2
0
1
2 2
2 2
k
k
V k l x l x u = − − = + 
,gV mgy= − where
22
2 3
2 2
l l
l x y
  
= − + +       
And ( )
( )2 2
2 2
3 3 4
3 0,
2
l l x lx
y ly x lx y
− + − −
+ + − = =
So 21 1 8
H.O.T.
23 3 3
y x x
l
   
= + − +   
   
Then
2
2 4
2 2 constant
3 3 3
x mgx
V kx kux mg
l
= + + − +
2
4
2
3 3
7
6
n
mg
k
l
m
ω
+
=
1 12 8
Hz
2 7 7 3
n
k g
f
m lπ
= + !

Position 1
( )
22 2 2 2
1
1 1 1
2 2
2 2 2
D m D m r mT I m r c m rθ θ θ
   
= + − +   
   
& & &
For one disk ( )
2 2 2
2 2 2
2
1 4 16
0.31987
2 3 2 9
D D D D D D DA
r r r
I I m c m r m m m r
π π
  
= − = − = − =  
   
( )
2
2 2 24
1 0.3313
3
D D Dm r c m r m r
π
 
− = − = 
 
( ) 2 2 2
1 0.3199 0.3313 0.5D r mT m r m r θ = + + 
&
( ) 2 2
0.6512 0.5D r mm m r θ= + &
But ,D r
D r
W W
m m
g g
= =
So ( ) ( )
2
2 2
1
1 6
0.65117 6 0.5 4 0.045866
32.2 12
m mT θ θ
 
 = + =  
 
& &

Position 2 ( )
2
2 2
2
0, 2 1 cos , , 1 cos 1
2 3 2
m m
D m m
r
T V m gc c
θ θ
θ θ
π
 
= = − = − ≅ 
 
"
( )
2
2 2 2
2
4 4 4 6
2 6 1.27324
3 2 3 3 12
m
D D m m m
r
V m g W r
θ
θ θ θ
π π π
   
= = = =   
   
&
2 2
1 1 2 2 : 0.045866 0 0 1.27324m mT V T V θ θ+ = + + = +&
2 1.27324
: 27.747
0.045866
m n m nθ ω θ ω= = =&
2
5.2677, 1.192 sn n
n
π
ω τ
ω
= = =
1.192 snτ = !

With 2 2
4 lb 6 lb 40
, , 20 lb/ft, ft
1232.2 ft/s 32.2 ft/s
m M k l
 
= = = =  
 
2 2 2
1
2 2 2 2 2
l l
V k mg Mgl
θ θ θ    
= + +             
And for small :θ
( )
2
22 2 21 1 1 1 1
2 3 2 2 2
l
T ml M l Mr
r
θ
θ θ
    
= + +     
    
&
& &
So
2
2
2 2
8 4 2 23.05935
3
6 4
n
kl l l
mg Mg
l l
m M
ω
+ +
= =
+
Then
2
1.308 s
n
π
τ
ω
= = !

2 2
2
1 1
2 12
T ml θ
 
=  
 
&
Ignoring static terms:
2 2
1
1 1
2 2 2 2
a a
V k k
θ θ   
= +   
   
2
2
4
ka
θ
=
2
2 2 2 2
2 1
1
:
24 4
n m m
a
T V ml kω θ θ
 
= =   
 
2
2
2
6
n
ka
ml
ω =
6
2
n
a k
f
l mπ
=

2 2
m m m m
a a
AA BB l
l
θ α α θ′ ′= = = =
Position 1
( )1 10 1 coscT V mgy mgl α= = = −
For small angles
2 2
2
2
1 cos 2sin
2 2 8
m m
m m
a
l
α α
α θ− = ≈ =
2
2
1 2
8
m
a
V mgl
l
θ
 
=   
 
Position 2 2 2 2
2 2
1 1 1
0
2 2 12
m mT I ma Vθ θ
 
= = = 
 
& &
m n mθ ω θ=&
1 1 2 2T V T V+ = +
2
2 2 2
2
1
0
248
n m
a
mgl ma
l
ω θ
 
+ +  
 
2 3
n
g
l
ω =
1 3
2
n
g
f
lπ
= !

( ) 2
1 cos
2
m mn r rθ θ
1
= − = ( )m mv r r ω= −
Position 1: 2
1 1
1
0,
2
mT V Wh mgrθ= = =
Position 2: 2 0V =
( )22 2 2 2 2
2
1 1 1 1
2 2 2 2
m m m mT I mv mk m r rω ω ω= + = + −
Conservation of energy
1 1 2 2T V T V+ = +
( )22 2 21 1
0
2 2
m mmgr m k r rθ ω + = + −  
But for simple harmonic motion, m n mω ω θ=
( ) ( )222 21 1
2 2
m n mmgr m k r rθ ω θ = + −  
( )
2
22n
r
g
k r r
ω =
+ −
(1)
For half section of pipe
2 2
1
r
r r r r
π π
 
= − = − 
 
Parallel-Axis Theorem: 2
0I I mr= +
2
2 2
0
2
;
r
I mr mr I m
π
 
= = +  
 

2
2
4
1I mr
π
 
= − 
 
2 2
2
4
1k r
π
 
= − 
 
Eq. (1): 2
2
2 2
2
2
4 2
1 1
n
r
g
r r
πω
ππ
=
   
− + −   
   
2
2
2 2
2 2
44 4 4
21 1
n
r
g
g
r
r
π πω
πππ π
= =
  −− + − + 
 
( )
2
2
1
2 4 2 4 2
n
g g g g
r r r
πω
π π π
π
= = =
− − −
2
n
π
τ
ω
=
( )2
2
r
g
π
τ π
−
= !

( ) 2
sin sinm m mr rθ θ θ≈
( )
2
1 cos
2
m
mr r
θ
θ− ≈
Position 1 maximum deflection 1 0T =
2
1
2
m
mV Wy mgr
θ
= =
Position 2 ( )0θ =
2 2 2
2
1 1 1
2 2 12
m mT I mlθ θ
 
= =  
 
& &
m n mθ ω θ=&
2 2 2
2
1 1
2 12
n mT ml ω θ
 
=  
 
1 1 2 2T V T V+ = +
2 2 2 21 1 1
0
2 2 12
m n mmgr mlθ ω θ
 
+ =  
 
2
2
2
12 2
2
12
n n
n
gr l
grl
π
ω τ π
ω
= = =
3
n
l
gr
π
τ = !

This is not a damped vibration. However, the kinetic energy of the fluid
must be included
(a) Position 2 2 0T =
2
2
1
2
mV kx=
Position 1
2 2
1 spere fluid
1 1
2 4
s m m
r
T T T m v Vv
g
 
= + = +  
 
1 0V =
2 2 2
1 1 2 2
1 1 1
0 0
2 4 2
s m m m
r
T V T V m v Vv kx
g
 
+ = + + + = + 
 
m m m nv x x ω= =&
2 2 2 21 1 1
,
2 2 2
ω ω
  
+ = =        +  
 
s m n m n
s
r k
m V x kx
g r
m V
g
3
2
2
1 1 62.4 lb/ft 4 4 in.
0.15032 lb s /ft
2 2 3 12 in./ft32.2 ft/s
r
V
g
π
    
= = ⋅    
     
2 2
2
2
40 lb/ft
220.54 s
1 lb
0.15032 lb s /ft
32.2 ft/s
nω −
= =
 
+ ⋅ 
 
1
14.850 snω −
=
2
0.4231 sn
n
π
τ
ω
= =
0.423 snτ = !
(b) The acceleration does not change mass
0.423 snτ∴ = !

Eq. (19.33)
2
1
m
m
f
n
P
kx
ω
ω
=
 
−  
 
1450 N/m
10.607 s
4 kg
n
k
m
ω −
= = =
13 N
0.28889 m
450 N/m
mP
k
= =
2
0.28889 m
1
10.607
m
f
x
ω
=
 
−  
 
(a) 2
0.28889 m
5: 0.03714 m
5
1
10.607
f mxω = = =
 
−  
 
( )In Phase 37.1 mmmx = !
(b) 2
0.28889 m
10: 0.25984 m
10
1
10.607
f mxω = = =
 
−  
 
( )In Phase 260 mmmx = !

Eq. (19.33)
2
2
,
1
m
m n
f
n
P
kkx
m
ω
ω
ω
= =
 
−  
 
( ) 22
, orm m
m
m ff
P P
x k
x mk m ωω
= =
+−
(a) In phase
( )( )1
9 N
0.15 m 4 kg 5 s
k −
=
+
160.0 N/m= !
(b) Out of phase 0.15 mmx = −
( )( )1
9 N
40.0 N/m
0.15 m 4 kg 5 s
k −
= =
− +
!

Eq. (19.33) 2
2
1
m
m
f
n
P
kx
ω
ω
=
−
2m
st n
P k
k m
δ ω= =
2
2
3
1
st
st
f
n
δ
δ
ω
ω
≥
−
2
2
1
1
3
f
n
ω
ω
− ≤
2
2
2
3
f
n
ω
ω
>
Also 2
2
3
1
st
st
f
n
δ
δ
ω
ω
< −
−
2 2
2 2
1 4
1
3 3
f f
n n
ω ω
ω ω
− < − <
2
2
2 4 2 4
3 3 3 3
f
f
n
k k
m m
ω
ω
ω
< < < < !

2 2
sinB m fM mb kl P l tθ θ ωΣ = = − +&&
2 2
sinm fmb kl P l tθ θ ω+ =&&
2
2
42.4641 rad/s, sinn m f
kl
t
mb
ω θ θ ω= = =
2
2 2 2
0.14 in. 6.762
0.0175 rad
8 in. 1803.2
m
m
n f f
P l
mbθ
ω ω ω
±
= = ± = =
− −
Lower frequency: ( )2
6.762 0.0175 1803.2 , 37.64 rad/sf fω ω= − =
Upper frequency: ( )2
6.762 0.0175 1803.2 , 46.79 rad/sf fω ω= − − =
37.6 rad/s 46.8 rad/sfω< < !

Referring to the figure and solution for Problem 19.101
2
2
32
140
1 1 28.309412
40
32.2
n
kl
b m b b
ω
 
 
 = = =
 
 
 
Range
2
2 2
0.14
12
m
m
n f
P l
mb
b
θ
ω ω
 
 
 = ± =
−
( ) 22 2
3.00533 3.00533
801.422 225225n
bb ω
= =
−−
Lower frequency 2
3.00533 9.3499 2.625b b= −
2
2.625 3.00533 9.3499 0b b+ − =
1.3998 ftb =
Upper frequency 2
3.00533 9.3499 2.625b b= − +
2
2.625 3.00533 9.3499 0b b− − =
2.5446 ftb =
16.80 in. 30.5 in.b< < !

solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 19

solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 19

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solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 19