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Physics 221 Chapter 10

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

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Physics 221 Chapter 10

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  1. Physics 221Chapter 10

  2. 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) ?

  3. 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

  4. 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.

  5. Problem 2 What EXACTLY is a Radian? • A. 550 • B. 570 • C. 590 • D. 610

  6. 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

  7. Angular Velocity • Angular Velocity = radians / time •  = / t

  8. 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?

  9. 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

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

  11. a and  • Linear (m/s2) Angular (rad/s2) • a  • ( Vf - Vi ) / t ( f- i ) / t a = r 

  12. 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

  13. 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

  14. 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

  15. 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!

  16. Angular Analogs • d  • v  • a 

  17. Problem 6 . . . Angular Analogs • d = Vi t + 1/2 a t2 ?

  18. Solution 6 . . . Angular Analogs • d = Vi t + 1/2 a t2  = it + 1/2  t2

  19. 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?

  20. 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

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

  22. Torque Torque = distance x force  = r x F Easy!

  23. Torque Which one is easier to turn?

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

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

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

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

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

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

  30. 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.

  31. All about Sarah Hughes . . . Click me!

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

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

  34. 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

  35. 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

  36. 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 

  37. Solution 12 . . . K.E. of Rotation • The analog of v is  • The analog ofm is I • The K.E. of rotation is 1/2 I2

  38. 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.

  39. 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

  40. 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

  41. 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

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

  43. 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

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