Physics 221
Chapter 10
Problem 1 . . . Angela’s new bike
• The radius of the wheel is 30 cm and the speed v=
5 m/s. What is the rpm (revolutions per minute) ?
Solution 1 . . . Angela’s rpm
•
•
•
•
•
r = radius
circumference = 2  r
f = revolutions per second
v = d/t
v=2fr
• 5 = (2 )(f)(0.3)
• f = 2.6 revolutions per second or 159 rpm
What is a Radian?
• A “radian” is about 60 degrees which is 1/6 of the
circle (360 degrees).
• To be EXACT, the “radian pie” has an arc equal to
the radius.
Problem 2
What EXACTLY is a Radian?
•
•
•
•
A. 550
B. 570
C. 590
D. 610
Solution 2
What EXACTLY is a Radian?
• If each pie has an “arc” of r, then there must be 2 
radians in a 3600 circle.
• 2  radians = 3600
• 6.28 radians = 3600
• 1 radian = 57.30
Angular Velocity
• Angular Velocity = radians / time
•= /t
Problem 3 . . . Angular Velocity
• The radius of the wheel is 30 cm. and the (linear)
velocity, v, is 5 m/s. What is the angular velocity?
Solution 3 . . . Angular Velocity
• We know from problem 1 that :
• f = 2.6 rev/s
• But 1 rev = 2  radians
• So
=/t
 =(2.6)(2 ) /(1 s)
 = 16.3 rad/s
V and

• Linear (m/s)
Angular (rad/s)
•
V

•
d/t
/t
• 2rf/t
2 f/t
v=r
a and 
• Linear (m/s2)
•
•
Angular (rad/s2)

a
( Vf - Vi ) / t
(
a=r
f - i ) / t
Problem 4 . . .Your CD player
• A 120 mm CD spins up at a uniform rate from rest
to 530 rpm in 3 seconds. Calculate its:
• (a) angular acceleration
• (b) linear acceleration
Solution 4 . . . CD player
•  = ( f - i ) / t
•  = (530 x 2  /60 - 0) / 3
•  = 18.5 rad/s2
• a=r
• a = 0.06 x 18.5
• a = 1.1 m/s2
Problem 5 . . . CD Music
• To make the music play at a uniform rate, it is
necessary to spin the CD at a constant linear
velocity (CLV). Compared to the angular velocity of
the CD when playing a song on the inner track, the
angular velocity when playing a song on the outer
track is
• A. more
• B. less
• C. same
Solution 5 . . . CD Music
v=r
• When r increases,  must decrease in order for v
to stay constant. Correct answer B
• Note: Think of track races. Runners on the
outside track travel a greater distance for the
same number of revolutions!
Angular Analogs
• d

• v

• a

Problem 6 . . . Angular Analogs
• d = Vi t + 1/2 a t2
?
Solution 6 . . . Angular Analogs
• d = Vi t + 1/2 a t2
 = i t + 1/2  t2
Problem 7 . . . Red Corvette
• The tires of a car make 65 revolutions as the car
reduces its speed uniformly from 100 km/h to 50
km/h. The tires have a diameter of 0.8 m. At this
rate, how much more time is required for it to stop?
Solution 7 . . . Corvette
• 100 km/h = 27.8 m/s = 69.5 rad/s since v = r
• Similarly 50 km/h = 34.8 rad/s
(  f ) 2 = (  i) 2 + 2

• (34.8)2 = (69.5)2 + (2)()(65)(6.28)
 = - 4.4 rad/s2
0 = 34.8 - 4.4 t
f
=
i + 
t = 7.9 s
t

Torque
• Torque means the “turning effect” of a force.
• SAME force applied to both. Which one will turn
easier?
Torque
Torque = distance x force
=rxF
Easy!
Torque
Which one is easier to turn?
Torque . . . The Rest of the Story!
 = r F sin 
Easy!

Problem 8 . . . Inertia Experiment
• SAME force applied to m and M. Which one
accelerates more?
Solution 8 . . . Inertia Experiment
• Since F = ma, the smaller mass will accelerate
more
Problem 9
Moment of Inertia Experiment
• SAME force applied to all. Which one will undergo
the greatest angular acceleration?
Solution 9
Moment of Inertia Experiment
• This one will undergo the greatest angular
acceleration.
What is Moment of Inertia?
F=ma
Force = mass x ( linear ) acceleration

=I

Torque = moment of inertia x angular acceleration
I = mr2
• The moment of inertia of a particle of mass m
spinning at a distance r is I = mr2
• For the same torque, the smaller the moment of
inertia, the greater the angular acceleration.
All about Sarah Hughes . . .
Click me!
Problem 10 . . . Sarah Hughes
• Will her mass change when she pulls her arms in?
• Will her moment of inertia change?
Solution 10 . . . Sarah Hughes
• Mass does not change when she pulls her arms in
but her moment of inertia decreases.
Problem 11 . . . Guessing Game
• A ball, hoop, and disc have the same mass.
Arrange in order of decreasing I
•
•
•
•
A. hoop, disc, ball
B. hoop, ball, disc
C. ball, disc, hoop
D. disc, hoop, ball
Solution 11 . . . Guessing Game
• I (moment of inertia) depends on the distribution of
mass. The farther the mass is from the axis of
rotation, the greater is the moment of inertia.
• I = MR2
• hoop
I = 1/2 MR2
I = 2 /5 MR2
disc
ball
Problem 12 . . . K.E. of Rotation
• What is the formula for the kinetic energy of
rotation?
• A. 1/2 mv2
• B. 1/2 m2
• C. 1/2 I2
• D. I

Solution 12 . . . K.E. of Rotation
• The analog of v is

• The analog of m is I
• The K.E. of rotation is
1/2 I2
Problem 13 . . . Long, thin rod
• Calculate the moment of inertia of a long thin rod
of mass M and length L rotating about an axis
perpendicular to the length and located at one end.
Solution 13 . . . Long, thin rod
• I = mr 2
• However, r is a variable so we need to integrate.
(ain’t that fun!)
• A small mass m of length dr must = M/L dr
• I = M/L  r2 dr
• I = (M/L)(L3 / 3 )
• I = 1/3 ML2
Problem 14 . . . In the middle
• Suppose the rod spins about its C.M. One can use
the Parallel Axis Theorem to calculate ICM
ID = ICM + MD2
• D is the distance between the C.M. and the other
axis of rotation
Solution 14 . . . In the middle
ID = ICM + MD2
• 1/3 ML2 = ICM + M(L/2)2
ICM = 1/3 ML2 - 1/4 ML2
ICM = 1/12 ML2
Problem 1
The race of the century!
Will it be the hoop or the disc?
Solution 1 . . . Race of the Century
Hoop Loses ! ! !
• P.E. = K.E. (linear) + K.E. (angular)
•
•
•
•
•
mgh = 1/2 mv2 + 1/2 I2
mgh = 1/2 mv2 + 1/2 I (v/r)2
For the disc, I = 1/2 mr2
So mgh = 1/2 mv2 + 1/2 (1/2 mr2)(v/r)2
Disc v = (4/3 g h)1/2
• Similarly Hoop v = (g h)1/2