Dynamics and Vibrations
Dynamics of Rigid Bodies
Mohammad I. Kilani
Center of Mass (Center of Gravity) for System of Particles
 The center of mass of the system of particles is
defined as

m
r
 ii
n

rG 
i 1
n
m
i 1

m
r
 ii
n

i 1
m
1 n 
  mi ri
m i 1
[1]
n
i
xG 
 The velocity and acceleration of the center of
mass can be found by differentiating the above
equation with respect to time once and twice,
n
to obtain

 
rG  vG 
 m r
i 1
n
i i
m
i 1
1 n 
  mi vi
m i 1
i
r
m
 ii
n
r  v  a 
G
G
G
[2]
i 1
n
m
i 1
i

m
a
 ii
n

i 1
m
1 n 
  mi ai
m i 1
[3]
m x
i 1
n
i i
m
i 1
i
n
, yG 
m y
i 1
n
i
m
i 1
i
n
i
, zG 
m z
i 1
n
i i
m
i 1
i
System of Particles under External Force
 For each particle in the system, Newton’s law
states that:
 

f i  Fi  mi ai
 Summing over all the particles in the system,
we obtain:
n

i 1
 n 

f i   Fi  mi ai
i 1
 Noting that the first term on the left hand side
of the above equation is zero (internal actionreaction forces cancel one another), we obtain:
n

 
F

m
 i  i ai
i 1
[4]

f i : Internal force on particle i

Fi : External force on particle i
System of Particles under External Force
 Using equations [3] and [4], we obtain
n

 


F

m
a

m
a

m
v
 i  ii
G
G
[5]
i 1
 Note that the linear momentum for a particle was
defined as


Li  mi vi
 The linear momentum for a system of particles can
then be defined as




L   Li   mi vi  mvG
 Differentiating with respect to time, we obtain:



L   mi vi  mvG
[6]
 Using the results of Equations [5] and [6], we have
 
F
 L
[7]
Angular Moment and Angular Momentum for a Particle
around a fixed point
 Starting with Newton’s law for a particle and taking the moment of
both sides of the equation, we obtain
 

f i  Fi  mi vi

  


ri  f i  Fi  M o i  ri  mvi

   


ri  f i  ri  Fi  M o i  ri  mvi
[8]


 Define the angular momentum for a particle as around a point O as:



H Oi  ri  mi vi
 The time derivative of the angular momentum is then


 
 
 
 

H Oi  ri  mi vi  ri  mi vi  vi  mi vi  ri  mi vi  ri  mi vi
 Using equation [8] and [9], we have


M Oi  H Oi
[9]
Angular Moment and Angular Momentum for a Particle
around a fixed point
 The angular momentum for a system of particles is
defined as:




H O   H Oi   ri  mi vi
 Differentiating with respect to time




H O   H Oi   ri  mi vi
 Using the results of equation [5] and summing
over all the particles in the system, we have

   


r

f

r

F

M

r

m
v
 i i i i  oi  i i
 
  
 ri  fi   ri  Fi H O


 
[10]
 ri  Fi  M o i  H O








Center of Mass (Center of Gravity) for System of Particles
 Recall that from equations [1], [2] and [3], we had:
 1 n  
1 n 
1 n 
rG   mi ri , vG   mi vi , aG   mi ai
m i 1
m i 1
m i 1
 If the reference point was taken to be the center of
mass, and letting ρi denote the position vector of
particle i from the center of mass, the above
equation becomes
 1 n
 1 n 
1 n
0   mi  i   mi  i   mi i
m i 1
m i 1
m i 1
 or
n
n


  0
m


m


m

 i i  i i  i i
n
i 1
i 1
i 1
[11]
Angular Momentum for a System of Particles
 The angular momentum for a system of particles
around the center of mass relative to a fixed
reference or to a moving reference, are defined,
respectively, as:
Relative to a fixed reference point O



H G   H Gi    i  mi vi
[12]
Relative to a moving reference point P

 
H G P    i  mi vi  vP 



H G P    i  mi vi    i  mi vP

 
H G P    i  mi vi  vP    i mi

 
H G P    i  mi vi H G
[13]








Angular Momentum for a System of Particles
 Differentiate Equation [12] w.r.t time

HG

HG

HG

HG

HG

HG



  H Gi    i  mi vi
[12]




   i  mi vi    i  mi ai





   i  mi vG   i    i  mi ai






 vG   mi  i    i  mi  i    i  mi ai


   i  mi ai

  MG

Angular Momentum for a System of Particles
 Differentiate Equation [12] w.r.t time

HG

HG

HG

HG

HG

HG



  H Gi    i  mi vi
[12]




   i  mi vi    i  mi ai





   i  mi vG   i    i  mi ai






 vG   mi  i    i  mi  i    i  mi ai


   i  mi ai

  MG

Mass Moment of Inertia
I   r dm
2
I  I G  md 2
I G  mkG  ,
2
kG : Radius of gyration
Rectilinear Translation
 F  ma 
 F  ma 
M  0
x
G x
y
G y
G
1
2
T  mvG 
2
Curvelinear Translation
 F  ma 
 F  ma 
M  0
n
G n
t
G t
G
1
2
T  mvG 
2
Rotation about a Fixed Axis
2


F

m
a

m

rG
 n
G n
 F  ma   mr
 M  I  or  M
t
G t
G
G
G
1
1
2
T  mvG   I G 2
2
2
1
or T  I O 2
2
O
 I O
General Plane Motion
 F  ma 
 F  ma 
M  I 
 Μ  sum
x
G x
y
G y
G
k P
G
or
 M  Μ 
P
of moments of

I G and maG  around P
1
1
2
mvG   I G 2
2
2
1
or T  I IC  2
2
T
k P
General Plane Motion
 F  ma 
 F  ma 
M  I 
 Μ  sum
x
G x
y
G y
G
k P
G
or
 M  Μ 
P
of moments of

I G and maG  around P
k P
Work Energy Principle
Example
 Determine the moment of inertia of
the cylinder shown about the z axis.
The density of the material, ρ, is
constant
Example
 Determine the moment of
inertia of the cylinder
shown about the z axis.
The density of the
material, ρ, is constant
Example
 Determine the moment of
inertia of the cylinder
shown about the z axis.
The density of the
material, ρ, is constant
Problem F 17-1
 The cart and its load have a total mass
of 100 kg. Determine the acceleration
of the cart and the normal reactions on
the pair of wheels at A and B. Neglect
the mass of the wheels.
Problem F 17-1
 The cart and its load have a total mass
of 100 kg. Determine the acceleration
of the cart and the normal reactions on
the pair of wheels at A and B. Neglect
the mass of the wheels.
mg
Problem F 17-1
mg
 The cart and its load have a total mass
of 100 kg. Determine the acceleration
of the cart and the normal reactions on
the pair of wheels at A and B. Neglect
the mass of the wheels.
m(aG)x
M
O
0
3
N A 0.6  0.4   maG x 0.5  100 0.3
5
4
 100 1.2   mg 0.4   0
5
N A  100  0.8  0.5  60  0.3  80  1.2  981 0.4  0
N A  96  392.4  18  40  430.4 N
F
y
0
3
N A  N B  mg  100   0
5
N B  mg  60  N A  981  430.4  610.6 N
Problem 17-35
The sports car has a mass of 1.5
Mg and a center of mass at G.
Determine the shortest time it
takes for it to reach a speed of 80
km/h, starting from rest, if the
engine only drives the rear wheels,
whereas the front wheels are free
rolling. The coefficient of static
friction between the wheels and
the road is μs . Neglect the mass of
the wheels for the calculation. If
driving power could be supplied to
all four wheels, what would be the
shortest time for the car to reach a
speed of 80 km/h?
Problem 17-35
The sports car has a mass of 1.5 Mg and
a center of mass at G. Determine the
shortest time it takes for it to reach a
speed of 80 , starting from rest, if the
engine only drives the rear wheels,
whereas the front wheels are free rolling.
The coefficient of static friction between
the wheels and the road is μs . Neglect
the mass of the wheels for the
calculation. If driving power could be
supplied to all four wheels, what would
be the shortest time for the car to reach
a speed of 80 km/h?
Problem 17-35
The sports car has a mass of 1.5 Mg and a center of mass at G. Determine the shortest time it takes for it to
reach a speed of 80 , starting from rest, if the engine only drives the rear wheels, whereas the front wheels
are free rolling. The coefficient of static friction between the wheels and the road is . Neglect the mass of the
wheels for the calculation. If driving power could be supplied to all four wheels, what would be the shortest
time for the car to reach a speed of 80 km/h?
Problem 17-35
The sports car has a mass of 1.5 Mg and a center of mass at G. Determine the shortest time it takes for it to
reach a speed of 80 , starting from rest, if the engine only drives the rear wheels, whereas the front wheels
are free rolling. The coefficient of static friction between the wheels and the road is . Neglect the mass of the
wheels for the calculation. If driving power could be supplied to all four wheels, what would be the shortest
time for the car to reach a speed of 80 km/h?
Problem 17-42
The uniform crate has a mass of 50 kg and rests
on the cart having an inclined surface.
Determine the smallest acceleration that will
cause the crate either to tip or slip relative to
the cart. What is the magnitude of this
acceleration? The coefficient of static friction
between the crate and the cart is μs = 0.5
Problem 17-42
The uniform crate has a mass of 50 kg and rests
on the cart having an inclined surface.
Determine the smallest acceleration that will
cause the crate either to tip or slip relative to
the cart. What is the magnitude of this
acceleration? The coefficient of static friction
between the crate and the cart is μs = 0.5
Problem 17-45
The handcart has a mass of 200 kg and
center of mass at G. Determine the largest
magnitude of force P that can be applied to
the handle so that the wheels at A or B
continue to maintain contact with the
ground. Neglect the mass of the wheels.
Problem 17-75
Determine the angular acceleration of
the 25-kg diving board and the
horizontal and vertical components of
reaction at the pin A the instant the
man jumps off. Assume that the board
is uniform and rigid, and that at the
instant he jumps off the spring is
compressed a maximum amount of 200
mm, ω = 0, and the board is horizontal.
Take k = 7 kN/m.
Problem 17-45
Determine the angular acceleration of
the 25-kg diving board and the
horizontal and vertical components of
reaction at the pin A the instant the
man jumps off. Assume that the board
is uniform and rigid, and that at the
instant he jumps off the spring is
compressed a maximum amount of 200
mm, ω = 0, and the board is horizontal.
Take k = 7 kN/m.
Problem 18-12
The spool has a mass of 60 kg and a radius of
gyration kG = 0.3 m. If it is released from rest,
determine how far its center descends down the
smooth plane before it attains an angular velocity
of ω = 6 rad/s. Neglect friction and the mass of
the cord which is wound around the central core.
[Note that IG = m(kG)2]
T1   U1 2 T2
1
1
1
1
2
2
mvG1  I G12  mg sin  s  mvG 2  I G22
2
2
2
2
1
1
1
1
2
2
m1ri   mkG2 12  mg sin  s  m2 ri   mkG2 22
2
2
2
2
1
1
2
2
2
0  0  609.81sin 30s   600.36  600.3 6
2
2
s  0.661 m
Problem 18-13
Solve Prob. 18-12 if the
coefficient of kinetic friction
between the spool and plane
at A is μk = 0.2.
Problem 18-13
Solve Prob. 18-12 if
the coefficient of
kinetic friction
between the spool
and plane at A is μk =
0.2.
T1   U1 2 T2
sG
sA
2

, s A  sG
0.3 0.5  0.3
3
F
y'
0
mg cos   N A  0
N A  mg cos 
N A  609.81 cos 30  509.7 N
1
1
1
1
2
2
mvG1  I G12  mg sin  sG    k N A s A   mvG 2  I G22
2
2
2
2
1
2  1
2
0  mg sin  sG   k mg cos   sG   m2 ri   mkG2 22
2
3  2
1
2  1
2
2
2
0  0  609.81sin 30sG   0.2609.81 cos 30 sG   600.36   60 0.3 6
2
3  2
s  0.859 m