Remember that inertia is the resistance an
object has to movement, it depends solely on
mass
 Rotational inertia measures the amount of
torque it takes to get an object rotating, in
other words it is the resistance of an object to
accelerate angularly
 It depends not only on the mass of the object,
but where the mass is relative to the hinge or
axis of rotation
 The rotational inertia is bigger if more mass is
located farther from the axis.
 Moment of Inertia variable is I
 In general I=mr2 (units are kg m2)

 If
these cylinders have the same mass, which
will reach the bottom of the ramp 1st?
 The
solid one! It has less moment of inertia
because its mass is evenly distributed and
the hollow one has it mass distributed
farther away from its rotational axis
 Which
of these two rods will be harder to
pick up? To spin?
 They
will be the same to pick up because
their mass is the same, but the one on the
left will be harder to spin because its mass is
located farther from axis of rotation.
2
cylind rical sh ell : I  M R
solid cy lin der : I 
1
MR
bicycle rim
2
filled can of coke
2
1
rod ab ou t cen ter : I 
ML
2
baton
12
rod ab ou t en d : I 
sph erical sh ell : I 
1
3
2
ML
MR
3
solid sph ere : I 
2
5
2
MR
2
baseball bat
2
basketball
boulder
 For
a given amount of torque applied to an
object, its rotational inertia determines its
rotational acceleration  the smaller the
rotational inertia, the bigger the rotational
acceleration
Fn e t  m a
  I
Same torque,
different
rotational inertia
Big rotational
inertia
Small rotational
inertia
spins
slow
spins
fast
Treat the spindle as a solid cylinder.
a) What is the moment of Inertia of the spindle?
b) If the tension in the string is 10N, what is the angular
acceleration of the wheel?
c) What is the acceleration of the bucket?
d) What is the mass of the bucket?
e) How far has the bucket dropped after 2.5 sec?
a) 0.9 kgm2 b) 6.7 rad/s2 c) 4 m/s2 d) 1.7 kg
e) 12.5 m
A cylindrical space station of
(R=12, M=3400 kg) has moment
of inertia 0.75 MR2. Retrorockets are fired tangentially
at the surface of space station
and provide impulse of 2.9x104 N·s.
a) What is the angular velocity of the space station
after the rockets have finished firing?
b) What is the centripetal acceleration at the edge
of the space station?
a) w= 0.948 rad/s
b) a=10.8 rad/s2
A 4 m beam with a 30 kg mass is free to rotate on a hinge.
It is attached to a wall with a horizontal cable. The cable is
then cut, find the initial angular acceleration of the beam.
I
1
ml
2
3
Fy
+

.
90o-θ θ
θ
mg
Fx
   I
θ = 35o
mg cos(  )
L
2
3  10  cos(35 )
o
 
24
 2
 3.1 rad / s
 

1
mL 
3
3g cos(  )
2L
2
vt  rw
2
K  1 mv ,
2
2
K  1 m ( rw )
2
2
2
K  1 mr w ,
2
K rot  1
2
Iw
2
I  mr
2
A 4 m beam with a 30 kg mass is free to rotate on a hinge.
It is attached to a wall with a horizontal cable. The cable is
then cut, find the angular velocity when the beam is
horizontal.
1
I
ml
2
3
Ei  E
f
Ug  KR
.
θ=
θ = 35o
35o
h 
L

sin(  )
mgh 
2
mg

Find ω when beam is
at lowest point
Answer: ω = 3.6 r/s
L
sin( ) 
2

w 
1
Iw
2
11
2
mL w
2
2
23
3 g sin( )
L
 1.9 r / s
2
Rolling Motion
Many rotational motion situations involve rolling objects.
Rolling without slipping involves both rotation
and translation so you need to account for
both rotational and translational kinetic
energy.
Friction between the rolling object and the
surface it rolls on is static, because the
rolling object’s contact point with the
surface is always instantaneously at rest.
this point on the wheel is
instantaneously at rest if the
wheel does not slip (slide)
The point of contact of the object with the surface is the axis of
rotation
w
In the past, everything was SLIDING. Now the object is
rolling and thus has MORE energy than normal. So let’s
assume the ball is like a thin spherical shell and was
released from a position 5 m above the ground.
Calculate the velocity at the bottom of the incline.
E before  E after
E before  E after
U g  KT  K R
U g  KT
2
2
m gh  1 m v  1 I w
2
2
v  Rw
 2
I sphere @ cm
3
2
m gh  1 m v
2
mR
2
m gh  1 m v
2
2
2
gh  1 v
2
v
2
m gh  1 m v  1 ( 2 m R )( 2 )
2
2 3
R
2
2
2
2
gh  1 v  1 v
2
3
v
6
5
gh 
6
5
(10)(5) 
7.7 m/s
v
2 gh 
2(10)(5)  10 m / s
If you HAD NOT
included the
rotational kinetic
energy, you see the
answer is very much
different.

A bowling ball with a radius of .3 m and mass of
4.6 kg is held by a wire. The wire is swung back
so that it is now .5 m off the floor. The string is
cut as the bowling ball makes contact with the
floor. The resulting distance it travels in 5
seconds is 6 m. What is its rotational kinetic
energy and speed?
19.7 J and 7.2 rad/s
Rigid body
L  Iw
L  m vr  m w r
2
Point particle
Analogy between L and p
Angular Momentum
Linear momentum
L = Iw
p = mv
t = DL
Ft = Dp
Conserved if no net
outside torques
Conserved if no net
outside forces
Rank the following from largest to smallest angular
momentum.
A 65-kg student sprints at 8.0 m/s and leaps
onto a 110-kg merry-go-round of radius 1.6
m. Treating the merry-go-round as a
uniform cylinder, find the resulting angular
velocity. Assume the student lands on the
merry-go-round while moving tangentially.
= 2.71 rad/s
Two twin ice skaters separated by 10 meters skate without friction in a
circle by holding onto opposite ends of a rope. They move around a
circle once every five seconds. By reeling in the rope, they approach
each other until they are separated by 2 meters.
a) What is the period of the new motion?
TF = T0/25 = 0.2 s
b) If each skater had a mass of 75 kg, what is the
work done by the skaters in pulling closer?
W = 7.11x105 J
The figure below shows two masses held together
by a thread on a rod that is rotating about its
center with angular velocity, ω. If the thread
breaks, what happens to the system's (a) angular
momentum and (b) angular speed. (Increase,
decrease or remains the same)
A 50 kg figure skater rotates with her arms out at
2 rev/s, this gives her a radius of .7m. She
then pulls her arms in which gives her a radius
of 0.25 m. Find the final speed in rev/s.
Four identical masses rotate
about a common axis, 1.2 meters
from the center. Each mass is
2.5 kg, and the system rotates at
2 rad/sec. The rods connecting
them are assumed to be
massless. Find the total angular
momentum of this system.
2.5 kg
1.2 m
90.5 kg m2/s
The four masses are then pulled toward the
center until their radii are 0.5 meters. This is
done in such a manner that no external torque
acts on the system. What is the new angular
speed of the system?
36.2 rad/s
Quantity
Linear
Position
x (or y)
Ɵ
Displacement
Δx
ΔƟ
Velocity
v=Δx/Δt
ω=ΔƟ/Δt=2π/T
v=rω
Acceleration
a=Δv/Δt
α=Δω/Δt
a=rα
1st kinematic
v f  v 0  at
1 2
D x  v o t  at
2
w f  w0   t
2nd kinematic
3rd kinematic
Centripetal
acceleration
v f  v0  2 a D x
2
2
ac 
v
Rotational
Connection
Δx=rΔƟ
1
D  w ot 
t
2
2
w f  w 0  2 D 
2
2
2
r
c  w r
2
Inertia
m
Kinetic Energy
KEtrans= ½ mv2
KErot= ½ Iω2
What causes
acceleration
Force
Torque
Newton’s 2nd Law
Fnet=ma
τnet=Iα
Momentum
p=mv
I
L=Iω or mωr2
τ=r*F