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Chapter 8: Rotational Motion

Chapter 8: Rotational Motion. Radians. 1) 1 radian = angle subtended by an arc (l) whose length is equal to the radius (r) 2) q = l r 360 0 = 2 p radians = 1 rev Radians are dimensionless. l. r. q. Convert the following: 20 o to radians 20 o to revolutions

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Chapter 8: Rotational Motion

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  1. Chapter 8: Rotational Motion

  2. Radians 1) 1 radian = angle subtended by an arc (l) whose length is equal to the radius (r) 2) q = l r • 3600 = 2p radians = 1 rev • Radians are dimensionless l r q

  3. Convert the following: • 20o to radians • 20o to revolutions • 5000o to radians • 0.75 rev to radians • 0.40 radians to degrees

  4. A bird can only see objects that subtend an angle of 3 X 10-4 rad. How many degrees is that? 3 X 10-4 rad 360o = 0.017o 2p rad

  5. How small an object can the bird distinguish flying at a height of 100 m? q = l r l = q r l = (3 X 10-4 rad)(100m) l = 0.03 m = 3 cm q r l (approx.)

  6. How at what height would the bird be able to just distinguish a rabbit that is 30 cm long (and tasty)? (ANS: 1000 m) q r l (approx.)

  7. A tiny laser beam is aimed from the earth to the moon (3.8 X 108 m). The beam needs to have a diameter of 2.50 m on the moon. What is the angle that the beam can have? (6.6 X 10-9 radians)

  8. Convert: • 0.0200 rev/s to radians/s (0.126 rad/s) • 30.0o/s to radians/s (0.534 rad/s) • 1.40 rad/s to rev/s (0.223 rev/s) • 3000 rpm to radians/s (314 rad/s)

  9. The Mighty Thor swings his hammer at 400 rev/min. Express this in radians/s. 400 rev 1 min 2p rad 1 min 60 s 1 rev = 13.3p rad/s or 41.9 rad/s

  10. Angular Velocity v = Dxw = Dq Dt Dt Angular Velocity – radians an object rotates per second • All points on an object rotate at the EXACT same angular velocity • All points on an object DO NOT rotate with the same linear speed. (the farther out the faster)

  11. Merry-Go-Round Example wa = wb (both rotate through 3600 in the same time period) va < vb (Point b travels a longer distance around the circle) vb va a b

  12. If a person at point b flies off the merry-go-round, will they travel in a curve or straight line? b

  13. Angular Acceleration a = Dva = Dw Dt Dt • The angular acceleration of all points on a circle is the same. • All points on an object DO NOT experience the same linear acceleration. (the farther out the more acceleration)

  14. Frequency and Period Frequency = Revolutions per second Period = Time for one complete revolution f = w 2p T = 1 f

  15. Converting between Angular and Linear Quantities atan Linear = Radius X Angular v = rw atan = ra Note the use of atan to differentiate from centripetal acceleration, ac or ar: ar

  16. One child rides a merry-go-round (2 revolutions per minute) on an inside lion 2.0 m from the center. A second child rides an outside horse, 3.0 m from the center. • Calculate the angular velocity in rad/s. (0.209 rad/s) • Calculate the frequency and the period. (0.0333 Hz, 30 s) • Calculate the linear velocity of each rider. (0.419 m/s, 0.628 m/s) • Who has a higher angular velocity?

  17. A clock has a seconds hand that is 7.00 cm long. • Calculate the angular velocity of the second hand in radians/s. (0.105 rad/s) • Calculate the frequency and period. (0.0167 Hz, 60 s) • Calculate the angular velocity in degrees/s. (6o/s) • Calculate the linear velocity at the end of the seconds hand. (7.35 X 10-3 m/s)

  18. A baseball bat is found to have a linear acceleration of 13.9 m/s2 at the “sweet spot.” The sweet spot is at 59 cm from the handle. • Calculate the angular acceleration in rad/s2 (a = 23.6 rad/s2) • Convert the angular acceleration to rev/s2 (3.75 rev/s2)

  19. A golf club has an angular acceleration of 20.0 rad/s2 and is 114 cm long. • Calculate the tangential acceleration of the club. (22.8 m/s2) • Convert the angular acceleration to rev/s2. (3.18 rev/s2) • Calculate the force the club could give to a 45.93 g golf ball. (1.05 N) • Calculate the velocity of the ball if the club is in contact with the ball for 50 cm. (4.77 m/s)

  20. Angular Kinematics v=vo + at w=wo + at x = vot + ½ at2 q = wot + ½ at2 v2 = vo2 + 2ax w2 = wo2 + 2aq

  21. A bike wheel starts at 2.0 rad/s. The cyclist accelerates at 3.5 rad/s2 for the next 2.0 s. • Calculate the wheel’s new angular speed (9.0 rad/s) • Calculate the number of revolutions. (1.75)

  22. A DVD rotates from rest to 31.4 rad/s in 0.892 s. • Calculate the angular acceleration. (35.2 rad/s2 ) • How many revolutions did it make? (2.23) • If the radius of the disc is 4.45 cm, find the linear speed of a point on the outside edge of the disc. (1.40 m/s)

  23. A bicycle slows from vo = 8.4 m/s to rest over a distance of 115 m. The diameter of each wheel is 68.0 cm. • Calculate the angular velocity of the wheels before braking starts. (24.7 rad/s) • How many revolutions did each wheel undergo? (HINT: calculate the circumference of the circle first) (53.8 rev) • Calculate the angular acceleration. (-0.903 rad/s2) • Calculate the time it took the bike to stop (27.4 s)

  24. A spinning bike tire of radius 33.0 cm has an angular velocity of 50.0 rad/s. Twenty seconds later, its angular speed is 150.0 rad/s. • Calculate the angular acceleration. (5.00 rad/s2) • Calculate the angular displacement over the 20 s. (2000 radians) • Calculate the revolutions travelled in the 20 s (318 rev) • Calculate the linear distance the tire travelled in 20 s. (660 m)

  25. A pottery wheel turning with an angular speed of 30.0 rev/s is brought to rest in 60.0 revolutions. • Calculate the radians that the wheel travelled (377 rad) • Calculate the angular acceleration. (-47.1 rad/s2) • Calculate the time required to stop. (4.00 s) • If the radius of the wheel is 12.0 cm, calculate the linear distance the outside of the wheel travelled. (45.2 m)

  26. A game show wheel with a 90 cm radius is initially turning at 3.0 rev/s. A point on the outside of the wheel travels 147 meters before stopping. • Calculate how many revolutions it when through. (26 rev) • Calculate the angular deceleration. (1.09 rad/s2) • Calculate how long it took to stop. (17.3 s) • Calculate the initial linear speed and linear acceleration on the outside of the wheel (17.0 m/s, 0.98 m/s2) • Calculate the initial centripetal acceleration on a point at the edge of the wheel. (319 m/s2)

  27. Friction and Rolling Wheels Rolling uses static friction • A new part of the wheel/tire is coming in contact with the road every instant B A

  28. Braking uses kinetic friction Point A gets drug across the surface A

  29. Torque Torque – tendency of a force to rotate a body about some axis (the force is always perpendicular to the lever arm) t= Fr r pivot F

  30. Torque Sign Conventions Counter-clockwise Torque is positive Clockwise Torque is negative

  31. Torque: Example 1 A wrench is 20.0 cm long and a 200.0 N force is applied perpendicularly to the end. Calculate the torque. t = Fr t = (200.0 N)(0.20 m) t = 40.0 m-N 20.0 cm 200.0 N

  32. Torque: Example 2 Suppose that same 200.0 N force is now applied at a 60o angle as shown. Calculate the Torque. Is it greater or less? 20.0 cm 200.0 N 60o

  33. 200.0 N First we need to resolve the Force vector into x and y components Only Fy has any effect on the torque (perpendicular) Fy = Fsinq = (200.0 N)(sin 60o) = 173.2 N t = Fr = (173.2 N)(0.20 m) = 34.6 m-N Fy 60o Fx

  34. Torque: Example 3 The biceps muscle exerts a 700 N vertical force. Calculate the torque about the elbow. • = Fr = (700 N)(0.050 m) • = 35 m-N

  35. A force of 200 N acts tangentially on the rim of a wheel 25 cm in radius. • Calculate the torque. • Calculate the torque if the force makes an angle of 40o to a spoke of the wheel. • If the wheel is mounted vertically, draw a free body diagram of the wheel if the force is the one in (a).

  36. Torque: Example 4 Two wheels, of radii r1 = 30 cm and r2 =50 cm are connected as shown. Calculate the net torque on this compound wheel when two 50 N force act as shown. 50 N 30o r2 Note that Fx will pull the wheel r1 50 N

  37. First find the horizontal component of the top force: Fx = (50 N)(cos 30o) = 43 N The top force is pulling clockwise (-) and the bottom force pulls counterclockwise (+) St = F1r1 – F2r2 St = (50N)(0.30m) – (43 N)(0.50 m) = -6.5 m-N

  38. Two children push on a merry-go-round as shown. Calculate the net torque on the merry-go-round if the radius is 2.0 m. (+1800 m N)

  39. A 60.0 cm diameter wheel is pulled by a 500.0 N force.  The force acts at an angle of 65.0o with respect to the spoke.  Assume that a frictional force of 300.0 N opposes this force at a radius of 2.00 cm.  Calculate the net torque on the wheel. (130.0 Nm)

  40. Moment of Inertia (I) • Measure of Rotational Inertia • An objects resistance to a change in angular velocity • Would it be harder to push a child on a playground merry-go-round or a carousel?

  41. Deriving I Consider pushing a mass around in a circle (like the child on a merry-go-round) F = ma t = Fr a = ra t = mrar F = mra t = mr2a F r m

  42. I = moment of inertia • I = mr2 • More properly I = Smr2 = m1r12 + m2r22 +…. St= Ia Would it be harder (require more torque) to twirl a barbell in the middle (pt. M) or the end (Pt. E) E M

  43. Moment of Inertia: Example 1 Calculate the moment of inertia (I) for the barbell when rotated about point M. We will assume the barbell is 1.0 m long, and that each weight is a point mass of 45.4 kg. I = Smr2 = (45.4 kg)(0.50 m)2 + (45.4 kg)(0.50 m)2 I = 22.7 kg-m2 M

  44. Moment of Inertia: Example 2 Now calculate I assuming Mr. Fredericks uses his massive musculature to twirl the barbells from point E. I = Smr2 = (45.4 kg)(0 m)2 + (45.4 kg)(1 m)2 I = 45.4 kg-m2 E

  45. Moment of Inertia: Example 3 Calculate I for Bouncing Boy (75 kg, radius = 1.2 m). Use the formulas from the book. I = 2/5 MR2 I = (2)(75 kg)(1.2 m)2 5 I = 43.2 kg-m2

  46. Calculate the moment of inertia of an 8.00 kg solid wheel with a radius of 25.0 cm.

  47. A wheel has a moment of inertia of 0.50 kg m2. • Calculate the torque (St= Ia) is required to give it an acceleration of 3 rad/s2. (1.5 mN) • Calculate its angular speed (from rest) after 5.00 s. (15 rad/s) • Calculate the number of revolutions it goes through in 5.00 s. (37.5 rad, 5.97 rev)

  48. A 15.0 N force is applied to a cord wrapped around a pulley of radius 33.0 cm. The pulley reaches an angular speed (w) of 30.0 rad/s in 3.00 s. • Calculate the angular acceleration. (10.0 rad/s2) • Calculate the torque (4.95 m-N) • Calculate the moment of inertia of the pulley. (0.495 kg-m2) 33.0 cm 15.0 N

  49. A 25.0 kg wheel has a radius of 40.0 cm. A 1.20 kg mass is hung on the end of the wheel by a string and falls freely. • Calculate the moment of inertia of the wheel. (2 kg m2) • Calculate the torque on the pulley. (4.70 mN) • Calculate the angular acceleration of the wheel. (2.35 rad/s2) • Calculate the tangential acceleration of the wheel (and bucket) (0.94 m/s2) • Calculate the tension in the string. (10.6 N)

  50. A 15.0 N force is applied to a cord wrapped around a pulley of radius 33.0 cm. The pulley reaches an angular speed (w) of 30.0 rad/s in 3.00 s. Since this is a real pulley, there is a frictional torque (tfr= 1.10 m-N) opposing rotation. • Calculate net torque on the pulley. (3.85 m-N) • Calculate the moment of inertia of the pulley. (0.385 kg-m2) 33.0 cm 15.0 N

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