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  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?

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 = radians / time

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 = r  V and  Linear (m/s) Angular (rad/s) V  d / t  / t 2  r f / t 2  f / t v = r 

a = 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?

 = - 4.4 rad/s2 f = i +  t 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 +  t 0 = 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 = distance x force  = r x F Easy!

Which one is easier to turn? 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 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 ID = ICM + MD2 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