1 / 94

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

vanna-kirkland
Télécharger la présentation

Chapter 8: Rotational Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related