Physics 221Chapter 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 f r

• 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 r f / t 2 f / t

v = r

a and

• Linear (m/s2) Angular (rad/s2)

• a • ( Vf - Vi ) / t ( f - i ) / t

a = r

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

f = i + t0 = 34.8 - 4.4 t

t = 7.9 s

Torque

• Torque means the “turning effect” of a force.

• SAME force applied to both. Which one will turn easier?

Torque

Torque = distance x force

= r x F

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 = m a 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 I = 1/2 MR2 I = 2 /5 MR2

• hoop 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 m2

• C. 1/2 I2

• 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 I2

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 I2

• 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