Transcript Physics 121
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=2fr • 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 • 2rf/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 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