Rotational Motion
Hope you learned polar coordinates
Rotation


The motion of these objects
cannot be easily described
by using linear equations of
motion.
These involve a body that
rotates about an axis that is
stationary in some inertial
frame of reference.
Purely Rotational Motion


No Translational
Motion
Rotation of a rigid
body about an axis of
rotation
◦ Rigid body – object with
definite shape. It will not
deform.
 The tire is not exactly a
rigid body but work with
me
◦ Axis of Rotation – The
line that passes through
a point where all other
points move around in
circles
Purely Rotational Motion
O
P
Velocity of a point
0.5m diameter tire
 P is on the rim


O
Rotating at 500 rpm
P
Velocity of a Point
d
v
t
500rotation
2r
min
v


min
rotation 60 s
500rotation 2 (.25) min
v


min
rotation 60 s
v  13.1 m s
Velocity of a point
0.5m diameter tire
 Q halfway to the rim


O
Rotating at 500 rpm
Q
P
Velocity of a Point
d
v
t
500rotation
2r
min
v


min
rotation 60s
500rotation 2 (.125) min
v


min
rotation 60s
v  6.54 m s
Angular Quantities
Different points on a
rotating rigid body
have different linear
velocities
 This makes
calculations clunky
 Better to use angular
velocities

Linear Displacement
P'

r
O
P
Angular Displacement
P'

r'

O
l

r
P
Angular Displacement
Angle measured in
radians
 1 Radian – angle
subtended by an arc
whose length is equal
to the radius

l

r
360  2rad
Angular Velocity
P' 

r'

O

r
P 0
Angular Velocity
    0
P'

r'

O


r
P
Angular Velocity


t

Instantaneous
angular velocity d
P'

r'

O



r
P
dt
Angular Velocity
 rad


t
s




Convention:
Counterclockwise
rotation is positive
Clockwise rotation is
negative
All points in a rotating
object have the same
angular velocity !
Angular Velocity
Define f as the
frequency – number
of complete
revolutions per
second
 Period T – time to
complete one
revolution

  2f
 rev
f 

 Hz
2
s
1 2
T 
s
f

Angular Velocity Vector

Direction of  given
by right hand rule
 Useful for when the
axis of rotation
changes

Angular Acceleration

500 rpm at time t0
0

600 rpm at time t

Angular Acceleration
    0
 rad

 2
t
s

Instantaneous
acceleration
d

dt
Convention:
 All points in a
rotating object have
the same angular
acceleration !

Angular Acceleration Vector

Direction of  given
by right hand rule
 Useful for when the
axis of rotation
changes


 If  and  are in the
same direction,
objects rotation is
speeding up. If in
opposite direction
it’s slowing down



Relationship of linear and angular
l

r
l  r

Note: there is still a
radial component of
the acceleration
 

a  arad  atan
l

2
2
v
r
 r


v
r 
2
t
t
arad 

 r
r
r
v

atan 
r
 r
t
t
Example

You are asked to design
an airplane propeller to
turn at 2400 rpm. The
forward airspeed of the
plane is to be 75.0 m/s,
and the speed of the tips
of the propeller blades
through the air must not
exceed 270 m/s (a) What
is the maximum radius
the propeller can have?
(b) With this radius, what
is the acceleration of the
propeller tip
Example
rev  2rad  min 

  2400rpm   2400



min  rev  60 s 

  251 rad s
v fastest at the tip
vtip  v 2tan  v 2plane

vtip 
r 
270

m
2
s
r  1.03m
 v plane
2
r (251
rad
)   75.0
2
s
m

2
s
Example
r  1.03m
 

a  atan  arad

2

a  r  r 
2

a  0  r

2
a  (1.03)251

a  6.5 x10 4 m s 2
Kinematics of Angular Motion

If α is constant
 
a
t
t    0
  0  t
  0

ave 

2
t
20  t    0

2
t
2
1
   0   0 t  2 t
Kinematics of Angular Motion

If α is constant
  0  t

t

   0 t  1 2  t 2

  
   0
 2


  
2
2  20 (  0 )  (  0 )
2
1
    2
2
2
0
Energy in Rotational Motion


A rotating rigid body,
even with no
translational motion,
has kinetic energy
Imagine rigid body is
made up of many small
particles
1
1
2
mi vi  mi (rii ) 2
2
2

Where ri is
perpendicular distance
to axis of rotation
Energy in Rotational Motion

Total kinetic energy
is the sum of all the
KE of the particles
1
2
K   mi (rii )
2

ω is the same for all
particles 1
K   mi (ri )
2
1
K   mi ri 2  2
2
2


Moment of Inertia
I   mi ri2
1 2
K  I
2


Moment of Inertia is
the rotational analogue
of mass
It is dependent on the
distribution of the
mass and the distance
from the axis of
rotation
Example

An engineer is designing a
machine part consisting of
three heavy disks linked by
lightweight struts. (a) What
is the moment of inertia of
this body about an axis
through the center of disk
A? (b) What is the moment
of inertia about an axis
through the center of disks
B and C? (c) If the body
rotates about an axis
through A perpendicular to
the plane of the diagram
with angular speed  =4.0
rad/s, what is its kinetic
energy?
Example

(a) I 
m r
2
i i
I  (0.3)(0)  (0.1)(0.5)  (0.2)(0.4)
2
I  0.057kg  m

(b) I 
m r
2
2
2
i i
I  (0.3)(0.4)  (0.1)(0)  (0.2)(0)
2
I  0.048kg  m
2
2
2
Example

(c) I  0.057kg  m 2
1 2 1
K  I  (0.057)( 4.0) 2
2
2
K  0.46 J
Moment of inertia is not an intrinsic
property, it depends on the axis of
rotation
Moment of Inertia Table
Parallel-Axis Theorem

Let the moment of inertia about an axis
that passes through the center of mass be
I cm

Then the moment of inertia through an
axis parallel to the original axis, but
displaced by some distance d is
I P  I cm  Md 2
Proof
I cm   mi ri
2
I cm   mi ( xi2  yi2 )
I p   mi (( xi  d ) 2  yi2 )
I p   mi ( x  2 xi d  d  y )
2
i
2
2
i
I p   mi ( x  y )  2d  mi xi  d
2
i
I p  I cm  0  Md
2
i
2
2
m
i
Example

A uniform rigid rod
has a mass of 3.6kg
and length 0.30m.
Find the moment of
inertia perpendicular
through one end?
Example
1
I cm  ML2
12
I p  I cm  Md 2
Ip
Ip
Ip
Ip
1
L
2
 ML  M  
12
2
1
2
 ML
3
1
2
 (3.6)(0.3)
3
2
 0.108kg  m
2
Brief Review
Angular
Linear
Acceleration
α
a
Velocity
ω
v
Displacement
Θ
x
1 2
I
2
1 2
mv
2
Equations
(constant acceleration)
Kinetic Energy
For a point at distance R away from the axis of rotation
x=R θ
v=R ω
a=R α
Dynamics of Rotational Motion
Torque, Angular Acceleration and Rotation of Rigid
Bodies
Rotation of Rigid Bodies
Suppose you have a
rigid body that is
pinned at point C
 Force F is applied to
point P
 The object will begin
to rotate about the
axis of rotation C (
to the screen)


F


r
Torque
Torque is the quantitative measure of the
tendency of the force to change the rotational
motion of a body. Rotational analog of force.
 Represented by the symbol τ

◦ Define: Torque is positive for counter-clockwise rotation and
negative for clockwise rotation



Vector form
Magnitude
Has SI unit N·m

  r F
  rF sin 
eq(1)
Torque – Simple Example
F
Ftan
Frad



P
r
O

l
Ftan = F sinθ = Force perpendicular to r
l = r sinθ = lever arm

To loosen a bolt at point
O, a force F is applied on
the wrench at point P.
Calculate for the
magnitude and direction
of τ.
Two Solutions to
  rF sin 
= rFtan = F l
From right hand rule
direction is outwards 
to the screen
Torque – Practical Application
Torque
The radial component of the force, FRad, has no
effect on the rotation of the object.
 The line of action of the force must not pass
through the axis of rotation (sinθ ≠ 0).
 Always define torque with respect to a specific
point. [Torque of F relative to point A]

A
Torque and Angular Acceleration


F
Ftan


r

1

Applying torque
changes the rotational
motion of a body, a
change in its
acceleration.
Assume the object is
made up of small
masses
For P1 apply Newton’s
Second law
F=ma
Ftan,1=ma=m1r1α
τ1=Ftan,1r=m1r12α
Torque and Angular Acceleration


F
Ftan




r
1

Torque going through point 1
τ1=m1r12α
Net Torque
τnet=Σ(miri2) α
τnet= I α
eq(2)
where I = Σ(miri2)
Moment of inertia, I, is the
rotational analog for mass and
it measures the resistance of
an object to a change in
angular acceleration.
Equation 2 is rotational analog
to Newton’s 2nd law of
motion. Note: valid only for
rigid bodies.
Effect of Internal Torque
Not so Simple Example

10-16. A 12.0 kg box resting
on the horizontal, frictionless
surface is attached to a 5.00
kg weight by a thin, light wire
that passes over a frictionless
pulley. The pulley has the
shape of a uniform solid disk
of mass 2.00 kg and a
diameter of 0.500m. After the
system is released, find:
◦ (a) tension in the wire on both
sides of the pulley
◦ (b) the acceleration of the box
◦ (c) the horizontal and vertical
components of the force that
the axle exerts on the pulley.
12 kg
5 kg
Problem 10-16





∑F=ma
For box 1
T1=m1a
For box 2
m2g-T2=m2a
For pulley (for convenience
let clockwise rotation be
positive)
∑τ=Iα
(T2-T1)R= m3R2α/2
(T2-T1)=m3a/2
Adding all three equations
m2g=(m1+m2+m3/2)a
(5)(9.8)=(12+5+2/2)a
a=2.72 m/s2
N
T1
12 kg
m1g
T2
5 kg
m2g
F
T1
m3g
T2
Problem 10-16
T1=m1a=(12)(2.72)
=32.6N
 T2=m2(g-a)=(5)(9.82.72)=35.4N
 Force on pulley
Fx=-T1=32.6N
Fy=-(T2+m3g)
=35.4+(2)(9.8)
=55.0N

T1
12 kg
m1g
T2
5 kg
m2g
F
T1
m3g
T2
Rigid Body Rotation about a Moving
Axis



We now combine rotational and
translational motion
Motion of a rigid body can always be divided
into translation of the center of mass and
rotation about the center of mass
1 2 1
K  mvcm  I cm 2
2
2
Rolling without Slipping

Linear velocity at the
rim has equal
magnitude but
opposite direction to
the velocity of the
center of mass.
vcm  R

Where R is the
radius of the car
vcm
vp
P
Rolling without Slipping


Velocity at point of contact is zero!
We use static friction
Example

A primitive yoyo is made by wrapping a
string around cylinder with mass 1kg and
radius 10cm. Holding one end of the
string stationary, the cylinder is released
from rest. The string unwinds without
slipping. Find the speed of the center of
mass of the cylinder after it has dropped a
distance 0.25m.
Example
vcm  R

at K
all times
U

K

U
0
0
1 2 1 2
mgh  0  0  mvcm  I
2
2
2
1 2 11
2  vcm 
mgh  mvcm   mR 

2
22
 R 
1 2 1 2
3 2
mgh  mvcm  mvcm  mvcm
2
4
4
4
4
vcm 
gh 
(9.8)(0.25)  1.81 m s
3
3
Example

A uniform rod of length
1.0m with weight 2.0 kg
is free to rotate around
a frictionless pin going
through one end.
(a)What is its angular
speed when it reaches
its lowest position? (b)
What is the tangential
speed of the center of
mass and the tangential
speed of the lowest
point when the rod is in
the vertical position?
Example
U 0  K0  U  K
  5.42 rad s
1 2 v  r  L   2.71 m
s
cm
mgh  0  0  I
2
2
m
v

R


L


5
.
42
s
 L 11 2 2
mg    mL 
 2 23

3g

 5.42 rad s
L
Conceptual Problem

Several objects roll without slipping down
a ramp of height H, all starting from rest
at the same moment. You have a hoop, a
basketball, a solid marble, a solid cylinder
and a greased box that slides without
friction. In what order will they reach the
bottom of the ramp?
Conceptual Problem
U 0  K 0  WNC  U  K
WNC  W f  0
U 0  K0  U  K
1 2 1 2
mgh  mv  I
2
2
1 2 1 v
mgh  mv  I  
2
2 R
2
Conceptual Problem
I  cmR 2

Where c is a coefficient dependent on the objects shape
1 2 1 v
mgh  mv  I  
2
2 R
2
1 2 1
2 v 
mgh  mv  cmR  
2
2
R
1 2 1
mgh  mv  cmv 2
2
2
2 gh  (1  c)v 2
2 gh
v
1 c
2
Conceptual Problem
2 gh
v
1 c
Larger c means lower v.
 Box c=0
 Solid Sphere c=2/5
 Solid Cylinder c=1/2
 Spherical Shell c=2/3
 Thin Cylinder c=1

Work and Power
Similar to linear, a
force applied on a
rigid object will cause
it to rotate a certain
distance θ.
 If torque applied is
constant

W  
Work and Power
Wtot  K
1
Wtot  I  2  02 
2

Work-energy theorem for rotation
W
P
 
t

Power for rotation
Example

An electric motor exerts a constant
torque of 10 N.m on a grindstone. The
moment of inertia of the grindstone
about its shaft is around 2.0 kg.m2. If the
system starts from rest, find the work
done by the motor in 8 seconds and the
kinetic energy at 8 s. What was the
average power delivered by the motor?
Example
  I
 10



5
rad
s2
I
2
1 2 1
  t  (5)(82 )  160rad
2
2
W    10(160)  1600 J
  0  t  0  5(8)  40 rad s
1 2 1
K  I  (2)( 40) 2  1600 J
2
2
W 1600
P

 200 J s  200W
t
8
Angular Momentum
Rotational analog to linear momentum

  r F



L  r  p  r  mv
 Magnitude
L  rp sin   rmv sin 
L  mvl


Where l is the perpendicular distance
Angular Momentum

For a slice of an
extended body
L   Li   ri mi vi sin 
L   ri mi vi sin 90
L   ri mi (ri )   mi ri 
2
L  I
Angular Momentum
We then compute the angular momentum
for other slices, computation gets
complicated for irregularly shaped
objects.
 If the body rotates about an axis of
symmetry


L  I
 If the axis of rotation is not a line of

symmetry, L is not parallel to the axis of
rotation

Example

Estimate the angular
momentum of a 6.0
kg bowling ball with a
radius of 12cm
spinning at 10 rev/s?
Example


L  I
10rev 2 62.8


s rev
s
2
2
L  mR 
5
2
L  (6)(0.12) 2 (62.8)
5
L  2.17 kg  m 2 / s
Conservation of Angular
Momentum

When the net external torque of a
system is zero, the total angular
momentum of the system is conserved
(constant).


L  I


L0  L


I 00  I
Example

A man on a turntable
holds two 5.0 kg
weights at arms length
away while he is
spinning at one
revolution per second.
What is his angular
velocity when he tucks
the weights nearer to
his body? The
dumbbells are 1.0m
away from the axis of
rotation initially and
0.20m at the end.
Example


L0  L


I 00  I

I0 
  0
I
 1rev 2
0 

 2
1s
1s
I 0   mi ri 2  5(1) 2  5(1) 2  10.0
I  5(0.2) 2  5(0.2) 2  0.4
 I0 
20
  0 
  50 rad s  25 rev s
I
0 .4
Example

A horizontal circular
platform rotates about a
vertical frictionless axle.
The platform has a mass
M=100.0 kg and a radius
of R=2.0m. A student of
m=60.0 kg walks from
the rim towards the
center. If the angular
speed when the student
is at the rim is 2.0 rad/s,
what is the angular
speed when the student
is at r=0.50m
Example

I0 
  0
I
I 0   mi ri 2  I disc  I student 
1
2
2
MR

mR
2
I 0  1 2 (100)( 2) 2  60(2) 2  440
I  1 2 MR 2  mr 2
I  2 (100)( 2)  60(0.5)  215
 440

(2)  4.1 rad s
215
1
2
2
Earths Rotation
Leap Seconds
The Earths Rotation
is Slowing down (0.5
billion years ago a
day was 22 hours
long)
 Tidal drag one of
main reasons

Leap Seconds
Glacial Rebound helps reduce the slowing
 But it won’t be there forever
 Source: http://what-if.xkcd.com/26/

Problem – Giancoli 8-20

A small rubber wheel is used to drive a large
pottery wheel, and they are mounted so that
their circular edges touch. The small wheel
has a radius of 2.0cm and accelerates at a
rate of 7.2rad/s2 and is in contact with the
pottery wheel (radius 25.0cm) without
slipping. Calculate (a) the angular
acceleration of the pottery wheel, and (b)
the time it takes the pottery wheel to reach
its required speed of 65rpm.
Young and Freedman 9.59

A thin uniform rod of mass M and length
L is bent at it’s center so that the two
segments are perpendicular to each other.
Find its moment of inertia about an axis
perpendicular to its plane and passing
through (a) the point where the two
segments meet and (b) the mid point of
the line connecting the two ends.
Serway 10-31

Find the net torque on the wheel about
the axle O if a= 10.0 cm and b=25.0 cm
Giancoli 8-65
An asteroid of mass 1.0x105 kg travelling
at a speed of 30 km/s relative to the
Earth, hits the Earth at the equator
tangentially, and in the direction of Earth’s
rotation. Use angular momentum to
estimate the percent change in the
angular speed of the Earth as a result of
the collision.
 ME=5.97x1024kg
 RE=6.38x106m
