very important lecture notes for dynamics!

1. Chapter Three
Kinetics of particles

2. Kinetics of particles
• It is the study of the relations existing between
the forces acting on body, the mass of the body,
and the motion of the body.
• It is the study of the relation between unbalance
d forces and the resulting motion.

3. • Newton ’s first law and third law are sufficient fo
r studying bodies at rest (statics) or bodies in mo
tion with no acceleration.

4. • When a body accelerates ( change in velocity mag
nitude or direction) Newton ’s second law is requi
red to relate the motion of the body to the force
s acting on it.

5. Force, mass and acceleration
• Newton ’s Second Law: If the resultant force acti
ng on a particle is not zero the particle will have a
n acceleration proportional to the magnitude of r
esultant and in the direction of the resultant.

6. • The basic relation between force and acceleratio
n is found in Newton's second law of motion and it
s verification is entirely experimental.

7. • Consider a particle subjected to constant forces
• We conclude that the constant is a measure of so
me property of the particle that does not change.
const
a
F
a
F
a
F



 ...
2
2
1
1

8. • This property is the inertia of the particle which is its resistanc
e to rate of change of velocity.
• The mass m is used as a quantitative measure of inertia, and the
refore the experimental relation becomes,
F=ma
• This relation provides a complete formulation of Newton's secon
d law; it expresses not only that the magnitude F and a are prop
ortional but also that the vector F and a have the same direction
.

9. Equation of motion and solution of problems
• When a particle of mass m acted upon by several forces. The Ne
wton’s second law can be expressed by the equation
• To determine the acceleration we must use the analysis used in k
inematics, i.e
• Rectilinear motion
• Curvilinear motion


ma
F

10. Rectilinear Motion
• If we choose the x-direction, as the direction
of the rectilinear motion of a particle of mass
m, the acceleration in the y and z direction wi
ll be zero, i.e






0
0
z
y
x
x
F
F
ma
F

11. • Generally,
• Where the acceleration and resultant force are given
by
Z
Z
y
y
x
x
ma
F
ma
F
ma
F






2
2
2
z
y
x
z
y
x
a
a
a
a
k
a
j
a
i
a
a






  2
2
2
)
(
)
( 










z
y
x
z
y
x
F
F
F
F
k
F
j
F
i
F
F

12. Curvilinear motion
• In applying Newton's second law, we shall make us
e of the three coordinate descriptions of acceler
ation in curvilinear motion.

13. Rectangular coordinates
Where and




y
y
x
x
ma
F
ma
F


 x
ax


 y
ay

14. Normal and tangential coordinate
• Where




t
t
n
n
ma
F
ma
F




 v
a
v
a t
n ,
2
2




15. Polar coordinates
• Where and





 ma
F
ma
F r
r
2




 
r
r
a
r





 
 r
r
a
n 2

16. Examples

17. Example 1
• Block A has a mass of 30kg and block B has a mass of
15kg. The coefficient of friction between all plane sur
faces of contact are and . Knowing
that Ѳ=300 and that the magnitude of the force P appl
ied to block A is 250N, determine
a) The acceleration of block A ,and
b) The tension in the cord
15
.
0

s
 10
.
0

k


18. Example 2
• A small vehicle enters the top A of the circular path with a horizont
al velocity vo and gathers speed as it moves down the path.
• Determine an expression for the angle β to the position where the v
ehicle leaves the path and becomes a projectile. Evaluate your expre
ssion for vo=0. Neglect friction and treat the vehicle as a particle

20. Example 3
• The slotted arm revolves in the horizontal about the fixed vertical axis
through point O. the 2 Kg slider C is drawn towards O at the constant r
ate of 50mm/s by pulling the cord S. At the instant for which r=225mm
, the arm has a counterclockwise angular velocity ω=6rad/s and is slowin
g down at the rate of 2rad/s2 .for this instant determine the tension T
in the cord and the magnitude N of the force exerted on the slider by t
he sides of the smooth radial slot. Indicate which side, A or B, of the sl
ot contacts the slider. (problem 3/69)

21. work and kinetic energy

22. • The method of work and energy directly relates force, mass,
velocity, and displacement.
• We apply this method:
 When intervals of motion are involved where the change in veloci
ty or the corresponding displacement of the particle is required.
• Integration of the forces with respect to the displacement o
f the particle leads to the equation of work and energy.

23. Work of Force
   
dz
F
dy
F
dx
F
k
d
j
d
i
d
k
F
j
F
i
F
dU
Fds
dU
r
d
F
dU
z
y
x
z
y
x
z
y
x





















cos

24.  
 














2
1
2
1
2
1
2
1
2
1
2
1
2
1
cos
A
A
z
y
x
s
s
t
s
s
A
A
dz
F
dy
F
dx
F
U
ds
F
ds
F
U
r
d
F
U




25.  
x
F
U 

 
cos
2
1
Work of a constant force in rectilinear motion

26. y
W
y
y
W
U
Wy
Wy
Wdy
U
Wdy
dU
y
y













 
)
( 1
2
2
1
2
1
2
1
2
1
When y is negative (moves down), the work is posi
tive
Work of the Force of Gravity

27. Work of the Force Exerted by a Spring
  x
F
F
U
kx
kx
kxdx
U
kxdx
Fdx
dU
kx
F
x
x














 
2
1
2
1
2
2
2
1
2
1
2
1
2
1
2
1
2
1
If the spring returning to the undeforme
d position, then positive energy

28. Work of a gravitational Force
1
2
2
2
1
2
2
2
1
r
GMm
r
GMm
dr
r
GMm
U
dr
r
Mm
G
Fdr
dU
r
Mm
G
F
r
r












29. Kinetic Energy of a Particle
2
1
2
2
2
1
2
1
2
1
2
1
mv
mv
vdv
m
ds
F
mvdv
ds
F
ds
dv
mv
dt
ds
ds
dv
m
dt
dv
m
ma
F
v
v
s
s
t
t
t
t










2
2
1
1
1
2
2
1
T
U
T
T
T
U







30. Power and Efficiency
• Friction energy dissipated by heat an
d reduce kinetic energy
input
power
output
power
v
F
dt
r
d
F
dt
dU
power










31. Potential Energy
Wy
V
V
V
U
Wy
Wy
Wdy
U
g
g
g
y
y








 
2
1
2
1
2
1
2
1
)
(
)
(
2
1
When Vg2 >Vg1, potential energy increases, and U 1-2
is negative
Potential Energy of the body with respec
t to the force of gravity

32. r
WR
r
Mm
G
V
r
GMm
r
GMm
dr
r
GMm
U
g
r
r
2
1
2
2
2
1
2
1





 

R is from the center of the ear
th
it should be noted that the expression just obtained for the potential energy of a bod
y with respect to gravity is valid only as long as the weight of body can be assumed to
remain constant, i.e, as long as the displacements of the body are small compared with
the radius of the earth.

33. 2
2
1
2
1
2
2
2
1
2
1
2
1
)
(
)
(
2
1
2
1
2
1
kx
V
V
V
U
kx
kx
kxdx
U
e
e
e
x
x








 
Potential Energy of the body
with respect to the elastic fo
rce F

34. Only the initial and final deflection of
the spring are needed
Deflection of the spring is measured fr
om its undeformed position

35. Equation of work and energy
of
s
w
U
U
U
U
where
T
U
2
1
2
1
2
1
2
1
2
1










Where =work done by other
(non conservative/ path dependa
nt) forces
of
U 2
1

E
U
V
V
T
U
T
U
V
V
of
g
e
of
of
e
g


















2
1
2
1
2
1

36. Advantages and disadvantages of work and ener
gy method
• Advantage; 1. No need to calculate values between A1 and A
2. Only final stages are counted 2. All scalars so can be ad
ded easily 3. Forces that do no work are ignored
• Disadvantage; can not determine accelerations, can not deter
mine accelerations and forces that do no work

37. Example 1
• For the slider collar shown in the figure, if m=0.5kg, b=0.8
m, and h=1.5m, and if the velocity of the collar as it strike
s the base B is 4.70m/s after the release of the collar fro
m rest, calculate the work Q of friction. What happens to
the energy that is lost?

38. • The 0.8kg collar slides with negligible friction o
n the fixed rod in the vertical plane. If the coll
ar starts from rest at A under the action of 8N
horizontal force, calculate its velocity v as it hi
ts the stop at B.
Example 2

39. • The 6kg cylindrical collar is released from rest
in the position shown and drops onto the spring.
Calculate the velocity v of the cylinder when th
e spring has been compressed 50mm.
Example 3

40. • The two springs, each of stiffness k=1.2KN/m, are if equal
length and un-deformed when θ=0. If the mechanism is rel
eased from rest in the position θ=20, determine its angula
r velocity θ’ when θ=0. the mass m of each sphere is 3kg. T
reat the sphere as particles and neglect the masses of the
light rods and springs.
Example 4

41. Impulse and momentum

42. • Work and energy is obtained by integrating the
equation of F=ma with respect to the displacem
ent of the particle.
• Impulse and momentum can be generated by int
egrating the equation of motion (F=ma) with res
pect to time.

43. Linear impulse and momentum
• Consider a particle of a mass m which i
s subjected to several forces in space.
• The product of the mass & the velocity
is defined as the linear momentum.
)
1
..(
..........
)
( G
F
or
v
m
dt
d
v
m
F 
 
 


v
m
G


44. • Equation (1) states that the resultant of all for
ces acting on a particle equals its time rate of c
hange of linear momentum.
• It is valid as long as the mass m of the particle
is not changing with time.
• The scalar components of equation (1) are:


 

 z
z
y
y
x
x
G
F
G
F
G
F 

 ,
,

45. • The scalar component of angular momentum is
:-
)
(
)
(
),
(
)
(
)
(
)
(
y
v
x
v
m
H
x
v
z
v
m
H
z
v
y
v
m
H
v
v
v
z
y
x
k
j
i
m
H
k
y
v
x
v
m
j
x
v
z
v
m
i
z
v
y
v
m
v
m
r
H
x
y
z
z
x
y
y
z
x
z
y
x
o
x
y
z
x
y
z
o
















46. • If represents the resultant of all forces acting on the particles
P, the moment Mo about the origin O is the vector cross product.
• The moment about the fixed point O of all forces acting on M equals
the time rate of change of angular momentum of M about O.
F
....(*)
..........

 













o
o
o
o
H
M
v
m
r
v
m
v
v
m
r
v
m
r
H
v
m
r
F
r
M







47. • To obtain the effect of moment on the angular momentu
m of the particle over a finite period of time;
• The product of moment & time is angular impulse the total
angular impulse on M about a fixed point O equals the corr
esponding change in angular momentum of M about O.
1
1
2
2
1
2
1
2
2
1
& v
m
r
H
v
m
r
H
where
H
H
H
dt
M
o
o
o
o
o
t
t
o










 o
M

 

2
1
2
1
t
t
o
o
o H
dt
M
H

48. Conservation of angular momentum
• If the resultant moment about a fixed point O of all f
orces acting on a particle is zero during the interval o
f time, equation (*) requires that its angular momentu
m H0 about that point remains constant.
2
1
0 o
o
o H
H
or
H 



49. Impact

50. Impact
• Refers to the collision b/n two bodies a
nd is characterized by the generation of
relatively large contact forces that act
over a very short interval of time.

51. Direct central impact
• Consider the collinear motion of two spheres of masses m1
and m2 travelling with velocities V1 & V2. If V1 is greater th
an V2, collision occurs with the contact forces directed alo
ng the line of centers.

52. • In as much as the contact forces are equal
& opposite during impact; the linear momen
tum of the system remains unchanged.
'
' 2
2
1
1
2
2
1
1 v
m
v
m
v
m
v
m 



53. • For given masses & initial conditions, the momentum equati
on contains two unknowns, v1’ & v2’, an additional relationshi
p is required.
• The relationship must reflect the capacity of the contacti
ng bodies to recover from the impact & can be expressed
by the ratio e of the magnitude of the restoration impulse
to the magnitude of the deformation impulse. This ratio is
called the coefficient of restitution.

54. • Fr – contact force during restoration period
• Fd – contact force during deformation period
 
 
 
 
 
 
 
 
2
.
..........
..........
'
'
1
.....
..........
'
'
2
2
2
2
2
2
1
1
1
1
1
1
particle
for
v
v
v
v
v
v
m
v
v
m
dt
F
dt
F
e
particle
for
v
v
v
v
v
v
m
v
v
m
dt
F
dt
F
e
o
o
o
o
t
t
d
t
t
r
o
o
o
o
t
t
d
t
t
r
o
o
o
o























55. • According to classical theory of impact, the value e
=1 means that the capacity of the two particles to r
ecover equal their tendency to deform.
• The value e=0, on the other hand describes inelasti
c or plastic impact where the particles cling togeth
er after collision & the loss of energy is a maximum.

56. b) Oblique central impact
• Here the initial and final velocities are not parallel.
• The spherical particles of mass m1 & m2 have initial velocit
ies v1 & v2 in the same plane & approach each other on a co
llision course.
• The direction of the velocity vector are measured from th
e direction tangent to the contacting surfaces.

58.  
    ,
cos
,
sin
,
cos
,
sin
2
2
2
2
2
2
1
1
1
1
1
1




v
v
v
v
v
v
v
v
t
n
t
n






There will be four unknown namely, (v1’)n, (v1’)t, (v2’)n, & (v2’)t
1) Momentum of the system is conserved in the n-direction,
2) & 3) The momentum for each particle is conserved in the t-directi
on since there is no impulse on either particle in the t- direction
t
t
t
t
v
m
v
m
v
m
v
m
)
'
(
)
(
)
'
(
)
(
2
2
2
2
1
1
1
1



59. 3) the coefficient of restitution, the velocity component in the n
- direction,
   
   
n
n
n
n
v
v
v
v
e
2
1
1
2 '
'




60. Example 1
• The 4kg cart, at rest at t=0, is acted on by a ho
rizontal force that varies with time t as shown.
Neglect friction and determine the velocity of
the cart at t=1Sec and t=3Sec

61. Example 2
• The loaded mine skip has a mass of 3Mg. The hoisting dru
m produces a tension T in the cable according to the time
schedule shown. If the skip is at rest againest A a when th
e drum is activated, determine the speed v of the skip whe
n t=6Sec. Friction loss may be neglected.

62. Example 3
• a particle with a mass of 4kg has a position vector in mete
rs given by ,where t is the time in seconds. For
t=3sec, determine the magnitude of the angular momentum
of the particle and the magnitude of the moment of all for
ces on the particle, both about the origins of the coordina
tes.
k
t
j
t
i
t
r 3
2
3
2




63. Example 4
• A particle of mass m moves with negligible friction
on a horizontal surface and is connected to a light s
pring fastened at O. At position A the particle has
the velocity VA=4m/s. Determine the velocity VB o
f the particle as it passes position B.

64. Example 5
• The assembly of two 5Kg spheres is rotating freely about the vertical a
xis at 40rev/min with θ=900. if the force F necessary to maintain the gi
ven position is increased to raise the base collar and reduce θ to 600, d
etermine the new angular velocity ω. Also determine the work U done b
y F in changing the configuration of the system. Assume that the mass
of the arms and the collars is negligible.

65. Example 6
• The magnitude and direction of the velocities of tw
o identical frictionless balls defore strike each oth
er are as shown in the figure. Assume e=0.90,
A) determine the magnitude and direction of the velocity of ea
ch ball after impact.
B) calculate the percentage loss of energy due to the impact.

66. Example 7
• Sphere A has a mass of 23Kg and a radius of 75mm while sphere
B has a mass of 4Kg and a radius of 50mm. If the spheres are tr
avelling initially along the parallel paths with the speeds shown, d
etermine the velocities of the spheres immediately after impact.
Specify the angles θA and θB with respect to the x-axis made b
y the rebound velocity vectors. The coefficient of restitution is
0.4 and friction is neglected.

67. Example 8
• The 2Kg sphere is projected horizontally with velocity of 10m/s
against the 10kg carriage that is backed up by the spring with a
stiffness of 1600N/m. The carriage is initially at rest with the s
pring uncompressed. If the coefficient of restitution is 0.6, calc
ulate the rebound velocity v’, the rebound angle, and the maximu
m travel δ of the carriage after impact.