2. Physics: Lecture 11 (Ch. 10-11 Halliday) Direction and the right hand rule Rotational dynamics and torque Work and energy with example More about rolling

3. Rotational v.s. Linear Kinematics AngularLinear And for a point at a distance R from the rotation axis: x =  R  v =  R  a =  R

4. Rotational Dynamics: What makes it spin? Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the (tangential) direction at some instant: – a t =  r Now use Newton’s 2nd Law in the  direction: – F t = ma t = m  r rF t = mr 2  r aaaa  F m r^ ^^ ^ FF l Multiply by r :

5. Rotational Dynamics: What makes it spin? rF  = mr 2  use Define torque:  = rF . –  is the tangential force F  times the lever arm r. Torque has a direction: – + z if it tries to make the system spin CCW. – - z if it tries to make the system spin CW. r aaaa  F m r^ ^ FF

7. Rotational Dynamics: What makes it spin? So for a collection of many particles arranged in a rigid configuration: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3  FF4FF4 FF1FF1 FF3FF3 FF2FF2 ii I Since the particles are connected rigidly, they all have the same .

8. Rotational Dynamics: What makes it spin?   NET = I  l This is the rotational analogue of F NET = ma l Torque is the rotational analogue of force: è The amount of “twist” provided by a force. Moment of inertia I is the rotational analogue of mass. Moment of inertia I is the rotational analogue of mass.  If I is big, more torque is required to achieve a given angular acceleration. l Torque has units of kg m 2 /s 2 = (kg m/s 2 ) m = Nm.

9. Torque = r F sin  = r sin  F  = r p F l Equivalent definitions! r p = “distance of closest approach” or torque arm.  = rF  Recall the definition of torque: r  rprp F  FF FrFr 

10. Torque  = r Fsin  So if  = 0 o, then  = 0 And if  = 90 o, then  = maximum r F r F

11. In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. (a) (a) case 1 (b) (b) case 2 (c) (c) same L L FF axis case 1case 2

12. QuestionSolution Torque = F x (distance of closest approach) L FF case 1case 2 L Torque is the same! è The applied force is the same. è The distance of closest approach is the same.

13. Torque and the Right Hand Rule: The right hand rule can tell you the direction of torque: – Point your hand along the direction from the axis to the point where the force is applied. – Curl your fingers in the direction of the force. – Your thumb will point in the direction of the torque. r F x y z 

14. The Cross Product We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. – The cross product of two vectors is a third vector: A B C A X B = C C The length of C is given by: C = AB sin  C AB The direction of C is perpendicular to the plane defined by A and B, and in the direction defined by the right hand rule. ABC 

15. The Cross Product Cartesian components of the cross product: C A B C = A X B C X = A Y B Z - B Y A Z C Y = A Z B X - B Z A X C Z = A X B Y - B X A Y A B C Note: B X A = - A X B

16. Torque Vector Fig. 11-10 (a) A force F, lying in an x-y plane, acts on a particle at point A. (b) This force produces a torque  = r x F on the particle with respect to the origin O. By the right-hand rule for vector (cross) products, the torque vector points in the positive direction of z. Its magnitude is given by in (b) and by in (c).

17. Sample problem Calculations: Because we want the torques with respect to the origin O, the vector required for each cross product is the given position vector r. To determine the angle  between the direction of r and the direction of each force, we shift the force vectors of Fig.a, each in turn, so that their tails are at the origin. Figures b, c, and d, which are direct views of the xz plane, show the shifted force vectors F 1, F 2, and F 3. respectively. In Fig. d, the angle between the directions of and is 90°. Now, we find the magnitudes of the torques to be

18. Torque & the Cross Product: So we can define torque as:  rF  = r X F = rF sin   X = r Y F Z - F Y r Z = y F Z - F Y z  Y = r Z F X - F Z r X = z F X - F Z x  Z = r X F Y - F X r Y = x F Y - F X y r F x y z 

19. Comment on  = I  When we write  = I  we are really talking about the z component of a more general vector equation. (Recall that we normally choose the z- axis to be the the rotation axis.)  z = I z  z We usually omit the z subscript for simplicity. z zz zz IzIz

20. Example To loosen a stuck nut, a (stupid) man pulls at an angle of 45 o on the end of a 50 cm wrench with a force of 200 N. – What is the magnitude of the torque on the nut? – If the nut suddenly turns freely, what is the angular acceleration of the wrench? (The wrench has a mass of 3 kg, and its shape is that of a thin rod). L = 0.5 m F = 200 N 45 o

21. Example L = 0.5m F = 200 N 45 o  Torque  = LFsin  = (0.5 m)(200 N)(sin 45) = 70.7 Nm Wrench w/ bolts If the nut turns freely,  = I   We know  and we want , so we need to figure out I.  = 283 rad/s 2 So  =   / I = (70.7 Nm) / (0.25 kgm 2 )

22. Work F Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d  : Fdr – dW = F. dr = FR d  cos(  ) = FR d  cos(90-  ) = FR d  sin(  ) = FR sin(  ) d   dW =  d  We can integrate this to find: W =  Analogue of W = F  r W will be negative if  and  have opposite signs! R F dr = R d  dd axis  

23. Work & Kinetic Energy: Recall the Work/Kinetic Energy Theorem:  K = W NET This is true in general, and hence applies to rotational motion as well as linear motion. So for an object that rotates about a fixed axis:

24. Connection with CM motion... So for a solid object which rotates about its center or mass and whose CM is moving:  V CM We will use this formula more in the future.

25. Example: Disk & String A massless string is wrapped 10 times around a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning). – How fast is the disk spinning after the string has unwound? F R M

26. Disk & String... The work done is W = F x d – The displacement is 2  x 0.1m x 10 rev = 2  m F R M dF So W = (10 N)(2  ) = 62.8 J

27. Disk & String... W NET = W = 62.8 J =  K Recall that  I  for a disk about its central axis is given by: So  = 792.5 rad/s  R M Flywheel, pulley, & mass

28. Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest. Which disk has the biggest angular velocity after the pull ? (a) (a) disk 1 (b) (b) disk 2 (c) (c) same FF 11 22 d

29. Solution FF 11 22 d l The work done on both disks is the same! çW = Fd The change in kinetic energy of each will therefore also be the same since W =  K. But we know So since I 1 = I 2  1 =  2

30. Spinning Disk Demo: We can test this with our big flywheel. I m negligible in this case In this case, I = 1 kg - m 2 W = mgh = (2 kg)(9.81 m/s 2 )(1 m) = 19.6 J  = 6.26 rad/s ~ 1 rev/s

31. Applications of Newton’s Second Law for Rotation Examples

33. Radius sprocket r s = 7cm. Find ω after t = 5 s. 1. ω = αt 2. 3. 4 5.

34. Known L, M. Find: (a)α immediately following the release (b)The force F A 12 3 a. b. 1 2. Mg-F A =Ma cmy 3. a t = rα 4. a cmy = r cm α = (L/2)α 5. From 2, 3 and 4: 6. F A = (1/4)Mg

35. Nonslip Conditions A string wrapped around a rotating wheel must move with a tangential velocity v t that is equal to the tangential velocity of the rim of the wheel, provided that the string remains taut and does not slip. By differentiating the previous equation we get:

36. Problem Solving Guide

37. An object of mass m is tied to a light string wound around a pulley that has the moment of inertia I and the radius R. In non slipping conditions, find the tension in the string and the acceleration of the object. 1. Draw a free body diagram, drawing each force in its point of application. 2. The only force causing a torque is T: Falling weight & pulley...

38. 3. Draw a free body diagram for the suspended object and apply Newton’s second law: 4. Relate the linear and angular accelerations: 5. α from 2) and at from 3) in 4) gives: so 6. Substitute T into 3) and solve for at: Falling weight & pulley...

39. Using 1-D kinematics (Lecture 1) we can solve for the time required for the weight to fall a distance L: I m R T mg  a L where

40. The work done by a torque  acting through a displacement  is given by: l The power provided by a constant torque is therefore given by: Power

42. Review: Torque and Angular Acceleration  NET = I  This is the rotational analogue of F NET = ma Torque is the rotational analogue of force: Torque is the rotational analogue of force: – The amount of “twist” provided by a force. Moment of inertia I is the rotational analogue of mass Moment of inertia I is the rotational analogue of mass – If I is big, more torque is required to achieve a given angular acceleration.

43. Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other. Forces F 1 and F 2 are applied as shown. What is F 2 / F 1 if the angular acceleration of the wheels is the same? 1.A) 1 2.B) 2 3.C) 4 F1F1 F2F2

44. Solution We know but and so F1F1 F2F2 Since R 2 = 2 R 1

45. Rotation around a moving axis. A string is wound around a puck (disk) of mass M and radius R. The puck is initially lying at rest on a frictionless horizontal surface. The string is pulled with a force F and does not slip as it unwinds. – What length of string L has unwound after the puck has moved a distance D? F R M Top view

46. Rotation around a moving axis... The CM moves according to F = MA F MA l The distance moved by the CM is thus  R The disk will rotate about its CM according to  = I  l So the angular displacement is

47. Rotation around a moving axis... So we know both the distance moved by the CM and the angle of rotation about the CM as a function of time: D F F  Divide (b) by (a): (a)(b) L The length of string pulled out is L = R  :

48. Comments on CM acceleration: We just used  = I  for rotation about an axis through the CM even though the CM was accelerating! – The CM is not an inertial reference frame! Is this OK?? (After all, we can only use F = ma in an inertial reference frame). YES! YES! We can always write  = I  for an axis through the CM. – This is true even if the CM is accelerating. – We will prove this when we discuss angular momentum! F R MA 

49. Chapter 11 (Halliday) Rolling, Torque, and Angular Momentum

50. Rolling Objects Rolling Without Slipping

51. Rolling as Translational and Rotation Combined Although the center of the object moves in a straight line parallel to the surface, a point on the rim certainly does not. This motion can be studied by treating it as a combination of translation of the center of mass and rotation of the rest of the object around that center.

52. The part of the spool that is moving slowest is blurred the least.

53. Rolling = Translation + Rotation Note: “com” – center of mass

54. The wheel of radius R is rotating without slipping on a flat surface. Point P on the wheel moves as shown with the linear velocity: where r is the perpendicular distance from P to the axis of rotation. The CM of the wheel moves with the speed: Note: The top of the wheel @ 2R moves @ v = 2v cm

55. By differentiating each side of the equation 9-26 we obtain the acceleration of the CM A falling yo-yo unwinding from a string with the top fixed, follows the same rules as the wheel.

56. As the wheel rotates through the angle Φ, the point of contact b/w the wheel and the surface moves a distance: If the wheel is rolling on a flat surface, the CM remains directly over the point of contact, so it also moves through s.

57. The Forces of Rolling: Friction and Rolling A wheel rolls horizontally without sliding while accelerating with linear acceleration a cm. A static frictional force f s acts on the wheel at P, opposing its tendency to slide. If the wheel does slide when the net force acts on it, the frictional force that acts at P is a kinetic frictional force, f k. The motion then is not smooth rolling, and the above relation does not apply to the motion.

58. The Kinetic Energy of Rolling If we view the rolling as pure rotation about an axis through P, then (  is the angular speed of the wheel and I P is the rotational inertia of the wheel about the axis through P). Using the parallel-axis theorem (I P = I cm +Mh 2 ), I P = I CM + MR 2 (M is the mass of the wheel, I cm is its rotational inertia about an axis through its center of mass, and R is the wheel’s radius, at a perpendicular distance h). Using the relation v cm =  R, we get: A rolling object, therefore has two types of kinetic energy: 1)a rotational kinetic energy due to its rotation about its center of mass (=½ I cm  2 ), 2) and a translational kinetic energy due to translation of its center of mass (=½ Mv 2 cm )

59. Example 1: A bowling ball with R = 11 cm, v i = 2m/s, and M = 7.2 Kg is rolling up an incline and stops momentarily at h, before rolling back. Find h. Solution Apply conservation of energy: Substitute: Solve for h: and

60. A cue stick hits a ball x above CM. Find x for which the ball rolls w/o slipping. Express your answer in terms of the radius of the sphere R. 1. Free body diagram, negligible friction. 2. Torque about CM: 3. 2 nd Law: and 4. Nonslip condition: 5. Substitute: 6. Solve for x:

61. Remark: Striking the ball at a point higher or lower than (2/5)R from CM, will result in the ball rolling and slipping (skidding) which is often desirable in the game of billiard.

62. A sphere, a cylinder, and a hoop with the same mass are released together from rest at the top of an incline. The sphere reaches the bottom first, followed by the cylinder and then the hoop. Objects of Different I Rolling Down The Plane

63. Rolling Motion Objects of different I rolling down an inclined plane: h v = 0  = 0 K = 0 R  K = -  U = Mgh v CM =  R M Roll objects down ramp

64. Rolling... If there is no slipping: v CM 2v In the lab reference frame v In the CM reference frame v Where v =  R 

65. Rolling... Use v CM =  R and I CM = cMR 2. So: The rolling speed is always lower than in the case of simple sliding since the kinetic energy is shared between CM motion and rotation. hoop: c = 1 disk: c = 1/2 sphere: c = 2/5 etc... c c c c

66. Direction of Rotation: In general, the rotation variables are vectors (have direction) If the plane of rotation is in the x-y plane, then the convention is – CCW rotation is in the + z direction – CW rotation is in the - z direction x y z x y z

67. Direction of Rotation: The Right Hand Rule To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector! We normally pick the z-axis to be the rotation axis as shown. –  =  z –  =  z –  =  z For simplicity we omit the subscripts unless explicitly needed. x y z x y z

68. Example: A flywheel spins with an initial angular velocity  0 = 500 rad/s. At t = 0 it starts to slow down at a rate of 0.5 rad/s 2. How long does it take to stop?   Realize that  = - 0.5 rad/s 2. Use to find t when  = 0 : l So in this case

69. Rotations A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is on the ramp? (a) (a) down the ramp (b) (b) into the page (c) (c) out of the page

70. Solution l When the ball is on the ramp, the linear acceleration a is always down the ramp (gravity). a l The angular acceleration is therefore counter-clockwise.  Using your right hand rule,  is out of the page!

71. Recap of today’s lecture More about rolling Direction and the right hand rule Rotational dynamics and torque Work and energy with example