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Physics 221 Chapter 10

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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) ?

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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

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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.

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Problem 2 What EXACTLY is a Radian? A. 55 0 B. 57 0 C. 59 0 D. 61 0

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Solution 2 What EXACTLY is a Radian? If each pie has an “arc” of r, then there must be 2 radians in a 360 0 circle. 2 radians = 360 0 6.28 radians = 360 0 1 radian = 57.3 0

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Angular Velocity Angular Velocity = radians / time = / t

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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?

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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

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

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a and Linear (m/s 2 ) Angular (rad/s 2 ) a ( V f - V i ) / t ( f - i ) / t a = r

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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

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Solution 4... CD player = ( f - i ) / t = (530 x 2 /60 - 0) / 3 = 18.5 rad/s 2 a = r a = 0.06 x 18.5 a = 1.1 m/s 2

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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

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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!

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Angular Analogs d v a

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Problem 6... Angular Analogs d = V i t + 1/2 a t 2 ?

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Solution 6... Angular Analogs d = V i t + 1/2 a t 2 = i t + 1/2 t 2

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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?

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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/s 2 f = i + t 0 = 34.8 - 4.4 t t = 7.9 s

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Torque Torque means the “turning effect” of a force. SAME force applied to both. Which one will turn easier?

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Torque Torque = distance x force = r x F Easy!

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Torque Which one is easier to turn?

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Torque... The Rest of the Story! = r F sin Easy!

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Problem 8... Inertia Experiment SAME force applied to m and M. Which one accelerates more?

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Solution 8... Inertia Experiment Since F = ma, the smaller mass will accelerate more

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Problem 9 Moment of Inertia Experiment SAME force applied to all. Which one will undergo the greatest angular acceleration?

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Solution 9 Moment of Inertia Experiment This one will undergo the greatest angular acceleration.

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What is Moment of Inertia? F = m a Force = mass x ( linear ) acceleration = I Torque = moment of inertia x angular acceleration

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I = mr 2 The moment of inertia of a particle of mass m spinning at a distance r is I = mr 2 For the same torque, the smaller the moment of inertia, the greater the angular acceleration.

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All about Sarah Hughes... Click me!

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Problem 10... Sarah Hughes Will her mass change when she pulls her arms in? Will her moment of inertia change?

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Solution 10... Sarah Hughes Mass does not change when she pulls her arms in but her moment of inertia decreases.

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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

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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 = MR 2 I = 1/2 MR 2 I = 2 /5 MR 2 hoop disc ball

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Problem 12... K.E. of Rotation What is the formula for the kinetic energy of rotation? A. 1/2 mv 2 B. 1/2 m 2 C. 1/2 I 2 D. I

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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

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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.

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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 r 2 dr I = (M/L)(L 3 / 3 ) I = 1/3 ML 2

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Problem 14... In the middle Suppose the rod spins about its C.M. One can use the Parallel Axis Theorem to calculate I CM I D = I CM + MD 2 D is the distance between the C.M. and the other axis of rotation

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Solution 14... In the middle I D = I CM + MD 2 1/3 ML 2 = I CM + M(L/2) 2 I CM = 1/3 ML 2 - 1/4 ML 2 I CM = 1/12 ML 2

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Problem 1 The race of the century! Will it be the hoop or the disc?

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Solution 1... Race of the Century Hoop Loses ! ! ! P.E. = K.E. (linear) + K.E. (angular) mgh = 1/2 mv 2 + 1/2 I 2 mgh = 1/2 mv 2 + 1/2 I (v/r) 2 For the disc, I = 1/2 mr 2 So mgh = 1/2 mv 2 + 1/2 (1/2 mr 2 )(v/r) 2 Disc v = (4/3 g h) 1/2 Similarly Hoop v = (g h) 1/2

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