PS 11 GeneralPhysics I for the
Life Sciences
ROTATIONAL MOTION
DR. BENJAMIN CHAN
ASSOCIATE PROFESSOR
PHYSICS DEPARTMENT
FEBRUARY 2014
Questions and Problems for Contemplation
 Chapter 8
 Questions: 1, 4, 8, 10, 15, 18, 22, 24
 Problems: 1, 2, 4, 8, 10, 15, 22, 27, 30, 38, 45, 52, 55, 61, 64
 General Problems: 72, 80, 81
Describing Rotational Motion
 Angular displacement q
 Let O be the axis of rotation
 How far the object has rotated


Only 2 directions possible: clockwise(-) and
counter-clockwise(+)
Measured in radians

1 radian (rad) is the angle subtended by an
arc whose length is equal to the radius of
motion
θ
l
r
Distance traveled
 Arc length traversed
l  rq
 For one complete revolution
l  2 r
 q can be expressed in revolutions
1 rev  2  rad
Example: Bike Wheel
 A bike wheel rotates 4.5 revolutions. How many
radians has it rotated?
θ  4 . 5 revs
2 rad
rev
 28 rad
 If the wheel has a diameter of 45 cm, what is the
distance traveled by a point on the rim of the
wheel?
l
d
2
q 
45 cm
2
28 rad  630 cm
Example: Bird of Prey
 A bird’s eye can distinguish objects
that subtend an angle no smaller
than 3  10-4 rad. How many
degrees is this?
θ  3  10
4
rad
360 
2  rad
 0 . 017 
 How small an object can the bird
just distinguish when flying at a
height of 100m?

For small angles (<15), arc length and
chord length are nearly the same
4
l  r q  (100 m )( 3  10 rad )  3 cm
Angular Velocity w
 Average w
w 
Dq
Dt
 Instantaneous w
 Dt must be very small
 Velocity v of a point on a rotating wheel
v  rw
 Changes direction as vector turns
 Increases in proportion to distance from the axis of
rotation
Angular Acceleration a
 Average a
a 
Dw
Dt
 Instantaneous a
 Make Dt as small as possible
 Tangential acceleration
a tan  r a
 Radial acceleration
aR 
v
2
r

(rw )
r
2
 rw
2
Review of Linear and Angular Quantities
 Frequency = number of complete revolutions per second = f
 w = 2f
 Period = time required to complete one revolution = T = 1/f
Equations of Motion
 Zero angular acceleration
 a = 0, w = constant
 Uniform circular motion
 q = wt + qo
 Linear velocity is not constant
 Magnitude is constant: v = wr
 Direction is changing
 Acceleration is not constant
 atan= 0 but aR= rw2 = constant (centripetal)
 Direction is changing
Example: Earth’s Rotation
 How fast is the earth’s equator turning?
 w = 2/T = (2 rad)/84,600s = 7.27 x 10-5 rad/s
 v = rw = (6,380 km)(7.27 x 10-5 rad/s) = 464 m/s
 How will your speed change as you go to the North
or South pole?
v = (r cos f)w = (464 cos f) m/s
 f = 14.5°, v = 449 m/s
 f = 30°, v = 402 m/s
 f = 60°, v = 232 m/s
 f = 90°, v = 0 m/s

The Coriolis Effect
 As you go from the equator towards the N pole, you
are moving faster than the ground you are moving
into: veer to your right (earth rotates west to east)
 As you go from the N pole towards the equator, you
are moving slower than the ground you are moving
into: veer to your right

Clockwise flow!
 To or from the S pole: veer to the left!
 Counterclockwise flow!
Example: Hard Drive
 The platter of the hard drive of a computer rotates
at 7200 rpm. What is the angular velocity of the
platter?
w  2 f  2
7200 rev 1 min
min
60 sec
 2
120 rev
s
 754
rad
s
 If the reading head of the drive is 3.00 cm from the
axis of rotation, how fast is the disk moving right
under the head?
v  r w  ( 3  10
2
m )( 754 rad / s )  22 . 6 m / s
Example (continued)
 If a single bit requires 0.50 mm of length along the
direction of motion, how many bits per second can
the writing head write when it is 3.00 cm from the
axis?

The number of bits passing the head per second is
22 . 6 m / s
0 . 50  10
6
m / bit
or 45 megabits/s (Mbps)
 45  10 bits / s
6
Constant Angular Acceleration
 a = constant
 w = wo + at
 q = qo + wot + ½ at2
 Eliminate t between w and q
 w2 = wo2 + 2aq
Total Acceleration
 atotal = atan + aR
 atan
 Constant magnitude, changing
direction
 aR
 Variable magnitude, variable
direction
Example: Centrifuge
 A centrifuge motor is accelerated from rest to 20,000
rpm in 30s. Determine its angular acceleration and
how many revolutions it makes while it is
accelerating.
 Solution

Assuming constant angular acceleration
a 
w  wo
t

2100 rad / s  0
30 s
 70 rad / s
2
Example continued
 Where the final angular velocity w is
w  2 f 
2  rad 20000 rev / min
rev
60 s / m
 2100 rad / s
 The angular displacement in 30s is then
q  0  (1 / 2 )( 70 rad / s )( 30 s )  3 . 15  10 rad
2
2
4
 We divide by 2 to convert to revolutions
3 . 15  10 rad
4
q 
2 rad / rev
 5 . 0  10 rev
3
Rolling Motion
 Translational + rotational
motion
 No Slipping

Static friction between object and
rolling surface
v  rw
Example: Bicycle
 A bicycle slows down uniformly from a velocity of
8.40 m/s to rest over a distance of 115 m. The
overall diameter of the tire is 68.0 cm. Determine
the initial angular velocity of the wheels.
wo 
vo
r

8 . 40 m / s
0 . 340 m
 24 . 7 rad / s
Example (continued)
 Determine the number of revolutions each wheel
undergoes before stopping.

The rim of the wheel turns 115m before stopping. Thus,
115 m
115 m

 53 . 8 rev
2 r
2 ( 0 . 340 m )
 Determine the angular acceleration of the wheel
w  wo
2
a 
2
2q

0  ( 24 . 7 rad / s )
2
2 ( 2  rad / rev )( 53 . 8 rev )
  0 . 902 rad / s
2
Example continued
 Determine the time it took the bicycle to stop
t
w  wo
a

0  24 . 7 rad / s
 0 . 902 rad / s
2
 27 . 4 s
 Note: when the bike tire completes one revolution,
the bike advances a distance equal to the outer
circumference of the tire (no slipping or sliding).
Announcements
 FINAL EXAM
 Wednesday, March 19
 7.30 - 10.30
 F-113
 Long Test 4
 Thursday, March 13
 6.00 – 7.30
 Room TBA c/o Paulo
Center of Mass
 You can reduce an object to a point and describe its
translational motion by considering the motion of
this point (called its center of mass)
Determining Center of Mass
 Consider masses m1, m2, m3, … with coordinates (x1,
y1), (x2, y2), (x3, y3), …
x cm 
x1 m 1  x 2 m 2  x 3 m 3  
m1  m 2  m 3  
mx
i

i
i
m
i
i
y cm 
y1m 1  y 2 m 2  y 3 m 3  
m1  m 2  m 3  

m
i
yi
i
m
i
i
CM for a Leg
 Determine the center of mass of a leg when a)
stretched out and b) bent at 90°. Assume the
person is 1.70 m tall.
 Solution

a) Straight leg
Essentially 1-D
 Measure distance from hip joint

x cm 
( 21 . 5 )( 9 . 6 )  ( 9 . 6 )( 33 . 9 )  ( 3 . 4 )( 50 . 3 )
21 . 5  9 . 6  3 . 4
 20 . 4 units
CM is 52.1-20.4 = 31.7 units from base of foot
 For a height of 172 cm, xcm = 54.5 cm above the bottom of the
foot

CM of Leg
 b) Bent leg
x cm 
y cm 

( 21 . 5 )( 9 . 6 )  ( 9 . 6 )( 23 . 6 )  ( 3 . 4 )( 23 . 6 )
21 . 5  9 . 6  3 . 4
( 3 . 4 )( 1 . 8 )  ( 9 . 6 )( 18 . 2 )  ( 21 . 5 )( 28 . 5 )
21 . 5  9 . 6  3 . 4
 14 . 9 units
 23 . 0 units
For a height of 172 cm
xcm = (172 cm)(0.149) = 25.6 cm
ycm = (172 cm)(0.23) = 39.6 cm
 Center of mass of bent leg is 39.6 cm
above the floor and 25.6 cm from the
hip joint!
CM Trajectory
Center of mass of
swimmer in flight
follows projectile
motion (parabolic)
trajectory
Center of mass of
wrench follows
constant velocity
trajectory
Torque
 What causes an object to rotate?
 Torque = force x lever arm
  r F  rF sin q
More Torque
 Units: Nm (Newton-meter)
 Reserve J for work and energy
 Torque is a vector quantity
 Direction determined by the right hand rule
Newton’s First law
 Translational Equilibrium
 All
forces cancel out: SF = 0
 Rotational Equilibrium
 Torques
must balance out: SG = 0
 When is an object in equilibrium?
Newton’s Second Law
 F = ma
 G = Ia
 I = moment of inertia
 a = angular acceleration
 Only two possible directions
 Counter-clockwise rotation
 Clockwise rotation
Moment of Inertia of Particles
 For a single moving object
with mass m
 = rF = rma = rmra =mr2a
 I = mr2

 For several objects rigidly
attached to each other
S = (Smiri2)a
 I = Smiri2

Changing Moment of Inertia
 Determine the change in the moment of inertia of a
particle as the radius of its orbit doubles
 Solution
Ii
If

mR
2
m (2 R )
2

mR
2
4 mR
2

1
4
 It increases by
I f  Ii
4 1
 100 % 
 100 %  300 %
Ii
1
Changing Your I
 Vertical axis of rotation
 Arms on the side
 R = 25 cm, M = 9.6 kg
I side  MR
2
 ( 9 . 6 kg )( 0 . 25 m )  0 . 60 kg  m
2
2
 Raise your arms in a crucifixion pose
 R = 57.5 cm, M = 9.6 kg
I cross  MR
2
 433% increase
 ( 9 . 6 kg )( 0 . 575 m )  3 . 20 kg  m
2
2
Moments of Inertia
for Various Objects
Rotational Kinetic Energy
KE 
(
1
2
2
mivi ) 
(
rotational
Total KE 
1
2
1
2
m i ri w ) 
2
KE 
2
1
2
I CM w 
2
Iw
1
2
1
2
(  m i ri )w
2
2
Mv
2
CM
2
Example: Ball Rolling Down an Inclined Plane
 Determine the speed of a solid sphere of mass
M and radius R when it reaches the bottom of
an inclined plane if it starts from rest at a
height H and rolls without slipping. Assume no
slipping occurs. Compare the result to an object
of the same mass sliding down a frictionless
inclined plane.
Solution
 Initial mechanical energy
 PE = MgH
 KEtrans = 0
 KErot = 0
 Final mechanical energy
 PE = 0
 KEtrans = ½ Mv2
 KErot = ½ Iw2
 Conservation of Energy
MgH = ½ (Mv2 + Iw2)
Solution (continued)
 I = (2/5)MR2 for a solid sphere rotating about an
axis through its center of mass
 w = v/R
 Thus
MgH = ½ Mv2 + ½(2/5)MR2(v/R)2
(1/2 + 1/5) v2 = gH
v = [(10/7)gH]1/2
 v does not depend on the mass and radius of the
sphere!!
Frictionless Incline
 Ball slides down the incline and does not roll
 Thus,
½ Mv2 = MgH
v = (2gH)1/2
 The speed is greater!
 None of the original PE is converted into rotational energy.
Work Done on a Rotating Body
 W = FDl = F rDq
 W = Dq
 Power
P = W/Dt
P = Dq/Dt = w
Angular Momentum L
 L = Iw
 Newton’s second law becomes

 Ia  I
 Thus,


Dw
Dt
DL
Dt

Iw  Iw o
Dt
Conservation of Angular Momentum
 If the net torque acting on a rotating
object is zero, then its angular
momentum remains constant.
0
DL
Dt
0  L f  Li
Li  L f
I iw i  I f w f
The Ice Skater
 How can the ice skater spin so fast?
wf 
Ii
If
wi
The Diver
 How can the diver make somersaults?
 Does she have to rotate initially?
 What trajectory does she follow?
The Hanging Wheel
 Why is the wheel standing up?
 Why does it turn around about the point of support?
Rotating Disk Demo
 What happens when you tilt the rotating disk?
HINT:
 
dL
dt
Drunk Driver Test/Tightrope Artist
 Follow the line walk
 Increase your moment of inertia to minimize
rotations
Quiz 8
1. A 4 kg mass sits at the origin, and a 10 kg mass sits at x = + 21 m.
Where is the center of mass on the x-axis?
(a) + 7 m (b) + 10.5 m (c) + 14 m (d) + 15 m
2. An object moving in a circular path experiences
(a) free fall.
(b) constant acceleration.
(c) linear acceleration.
(d) centripetal acceleration.
3. A boy and a girl are riding on a merry-go-round which is turning at
a constant rate. The boy is near the outer edge, and the girl is closer
to the center. Who has the greater angular velocity?
a) The boy
b) The girl
c) Both have the same non-zero angular velocity.
d) Both have zero angular velocity.
Quiz 8
4. A wheel starts at rest, and has an angular acceleration of 4 rad/s2.
Through what angle does it turn in 3 s?
a) 36 rad
b) 18 rad
c) 12 rad
d) 9 rad
5. A wheel of diameter 26 cm turns at 1500 rpm. How far will a point
on the outer rim move in 2 s?
a) 314 cm
b) 4084 cm
c) 8995.5 cm
d) 17990.8 cm
6. What is the centripetal acceleration of a point on the perimeter of a
bicycle wheel of diameter 70 cm when the bike is moving 8 m/s?
a) 91 m/s2
b) 183 m/s2
c) 206 m/s2
d) 266 m/s2
7. A bicycle is moving 4 m/s. What is the angular speed of a wheel if
its radius is 30 cm?
a) 0.36 rad/s b) 1.2 rad/s
c) 4.8 rad/s
d) 13.3 rad/s
Quiz 8
8. An ice skater is in a spin with his arms outstretched. If he pulls in
his arms, what happens to his kinetic energy?
a) It increases.
b) It decreases.
c) It remains constant but non-zero. d) It remains zero.
9. What is the quantity used to measure an object's resistance to
changes in rotation?
a) mass
b) moment of inertia
c) linear momentum
d) angular momentum
10. A wheel of moment of inertia of 5.00 kg-m2 starts from rest and
accelerates under a constant torque of 3.00 N-m for 8.00 s. What is
the wheel's rotational kinetic energy at the end of 8.00 s?
a) 57.6 J
b) 64.0 J
c) 78.8 J
d) 122 J