2. Lecture Outline Chapter 8 College Physics, 7 th Edition Wilson / Buffa / Lou © 2010 Pearson Education, Inc.

3. Chapter 8 Rotational Motion and Equilibrium © 2010 Pearson Education, Inc.

4. Units of Chapter 8 Rigid Bodies, Translations, and Rotations Torque, Equilibrium, and Stability Rotational Dynamics Rotational Work and Kinetic Energy Angular Momentum © 2010 Pearson Education, Inc.

5. 8.1 Rigid Bodies, Translations, and Rotations A rigid body is an object or a system of particles in which the distances between particles are fixed (remain constant). In other words, a rigid body must be solid (but not all solid bodies are rigid). Raise your hand if you can think of an exception! © 2010 Pearson Education, Inc.

6. 8.1 Rigid Bodies, Translations, and Rotations A rigid body may have translational motion, rotational motion, or a combination. It may roll with or without slipping. © 2010 Pearson Education, Inc.

7. 8.1 Rigid Bodies, Translations, and Rotations For an object that is rolling without slipping, © 2010 Pearson Education, Inc.

8. 8.2 Torque, Equilibrium, and Stability It takes a force to start an object rotating; that force is more effective the farther it is from the axis of rotation, and the closer it is to being perpendicular to the line to that axis. © 2010 Pearson Education, Inc.

9. 8.2 Torque, Equilibrium, and Stability The perpendicular distance from the line of force to the axis of rotation is called the lever arm. The product of the force and the lever arm is called the torque. © 2010 Pearson Education, Inc.

10. 8.2 Torque, Equilibrium, and Stability Torque is a vector (it produces an angular acceleration), and its direction is along the axis of rotation, with the sign given by the right-hand rule. © 2010 Pearson Education, Inc.

11. 8.2 Torque, Equilibrium, and Stability In order for an object to be in equilibrium, the net force on it must be zero, and the net torque on it must be zero as well. © 2010 Pearson Education, Inc.

12. 8.2 Torque, Equilibrium, and Stability The left stick and the triangle are in equilibrium; they will neither translate nor rotate. The stick on the right has no net force on it, so its center of mass will not move; the torque on it is not zero, so it will rotate. © 2010 Pearson Education, Inc.

13. 8.2 Torque, Equilibrium, and Stability For an object to be stable, there must be no net torque on it around any axis. The axis used in calculation may be chosen for convenience when there is no motion. © 2010 Pearson Education, Inc.

14. 8.2 Torque, Equilibrium, and Stability If an object is in stable equilibrium, any displacement from the equilibrium position will create a torque that tends to restore the object to equilibrium. Otherwise the equilibrium is unstable. © 2010 Pearson Education, Inc.

15. 8.2 Torque, Equilibrium, and Stability Whether equilibrium is stable or unstable depends on the width of the base of support. © 2010 Pearson Education, Inc.

16. 8.3 Rotational Dynamics The net torque on an object causes its angular acceleration. For a point particle, the relationship between the torque, the force, and the angular acceleration is relatively simple. © 2010 Pearson Education, Inc.

17. 8.3 Rotational Dynamics We can consider an extended object to be a lot of near-point objects stuck together. Then the net torque is: The quantity inside the parentheses is called the moment of inertia, I. © 2010 Pearson Education, Inc.

18. 8.3 Rotational Dynamics The moments of inertia of certain symmetrical shapes can be calculated. Here is a sample: © 2010 Pearson Education, Inc.

19. 8.3 Rotational Dynamics The parallel-axis theorem relates the moment of inertia about any axis to the moment of inertia about a parallel axis that goes through the center of mass. © 2010 Pearson Education, Inc.

20. Example 1: A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias. R I = mR 2 Hoop R I = ½mR 2 Disk I = 0.120 kg m 2 I = 0.0600 kg m 2

21. 8.4 Rotational Work and Kinetic Energy The work done by a torque: As usual, the power is the rate at which work is done: © 2010 Pearson Education, Inc.

22. 8.4 Rotational Work and Kinetic Energy The work–energy theorem still holds—the net work done is equal to the change in the kinetic energy. This gives us the form of the rotational kinetic energy. © 2010 Pearson Education, Inc.

23. 8.4 Rotational Work and Kinetic Energy There is a strict analogy between linear and rotational dynamic quantities. © 2010 Pearson Education, Inc.

24. Example 2: What is the rotational kinetic energy of the device shown if it rotates at a constant speed of 600 rpm? 3 kg 2 kg 1 kg 1 m 2 m 3 m  First: I =  mR 2 I = (3 kg)(1 m) 2 + (2 kg)(3 m) 2 + (1 kg)(2 m) 2 I = 25 kg m 2  = 600  rpm = 62.8 rad/s K = ½Iw 2 = ½(25 kg m 2 )(62.8 rad/s) 2 K = 49,300 J

25. Example 3: What is the linear accel- eration of the falling 2-kg mass? Apply Newton’s 2nd law to rotating disk: R = 50 cm 6 kg 2 kg +a T T mg  I  TR = (½MR 2 )  T = ½MR  and T = ½Ma T = ½MR( ) ; aRaR Apply Newton’s 2nd law to falling mass: mg - T = ma mg - = ma T (2 kg)(9.8 m/s 2 ) - ½(6 kg) a = (2 kg) a 19.6 N - (3 kg) a = (2 kg) a a = 3.92 m/s 2 a =  R;  = but aRaR ½Ma R = 50 cm 6 kg 2 kg a = ? M

26. Example 4: The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power if the 2-kg mass is lifted 20 m in 4 s.  F Fd=W s s = 20 m 2 kg 6 kg Work =  = FR  Work = (19.6 N)(0.4 m)(50 rad) sRsR  = = = 50 rad 20 m 0.4 m Work = 392 J F = mg = (2 kg)(9.8 m/s 2 ); F = 19.6 N Power = = Work t 392 J 4s Power = 98 W

27. Example (a): Find velocity v of a disk if given its total kinetic energy E. Total energy: E = ½mv 2 + ½I  

28. Example (b) Find angular velocity  of a disk given its total kinetic energy E. Total energy: E = ½mv 2 + ½I  

29. Strategy for Problems Draw and label a sketch of the problem.Draw and label a sketch of the problem. List givens and state what is to be found. List givens and state what is to be found. Write formulas for finding the moments of inertia for each body that is in rotation. Linear or Rotational variable? Write formulas for finding the moments of inertia for each body that is in rotation. Linear or Rotational variable? Recall concepts involved (power, energy, work, conservation, etc.) and write an equation involving the unknown quantity. Recall concepts involved (power, energy, work, conservation, etc.) and write an equation involving the unknown quantity. Solve for the unknown quantity – algebra. Solve for the unknown quantity – algebra.

30. Example 6: Find the velocity of the 2-kg mass just before it strikes the floor. (use E when you can!) h = 10 m 6 kg 2 kg R = 50 cm mgh o ½    ½mv o 2 = mgh f ½  f  ½mv f 2 2.5v 2 = 196 m 2 /s 2 v = 8.85 m/s

31. Example 7: A hoop and a disk roll from the top of an incline. What are their speeds at the bottom if the initial height is 20 m? 20 m mgh o = ½mv 2 + ½I  2 Hoop: I = mR 2 v = 16.2 m/s mgh o = ½mv 2 + ½mv 2 ; mgh o = mv 2 v = 14 m/s Hoop: mgh o = ½mv 2 + ½I  2 Disk: I = ½mR 2 ;

32. 8.5 Angular Momentum Definition of angular momentum: In vector form (the direction is again given by the right-hand rule): © 2010 Pearson Education, Inc.

33. 8.5 Angular Momentum The rate of change of the angular momentum is the net torque: Angular momentum is conserved: In the absence of an external, unbalanced torque, the total (vector) angular momentum of a system is conserved (remains constant). © 2010 Pearson Education, Inc.

34. 8.5 Angular Momentum Internal forces can change a system’s moment of inertia; its angular speed will change as well. © 2010 Pearson Education, Inc.

35. 8.5 Angular Momentum This can be demonstrated in the classroom. (What purpose do the hand weights serve?) © 2010 Pearson Education, Inc.

36. 8.5 Angular Momentum The conservation of angular momentum means its direction cannot change in the absence of an external torque. This gives spinning objects remarkable stability. © 2010 Pearson Education, Inc.

37. 8.5 Angular Momentum An external torque on a rotating object causes it to precess. © 2010 Pearson Education, Inc.

38. Example 9: A sharp force of 200 N is applied to the edge of a wheel free to rotate. The force acts for 0.002 s. What is the final angular velocity? R 2 kg  F    0 rad/s R = 0.40 m F = 200 N  t = 0.002 s Applied torque  FR I = mR 2 = (2 kg)(0.4 m) 2 I = 0.32 kg m 2 Impulse = change in angular momentum  t = I  f   o 0 FR  t = I  f  f = 0.5 rad/s

39. Summary of Chapter 8 Pure translational motion: all parts of object have same velocity Pure rotational motion: center of mass does not move; all parts of object have same rotational velocity Rolling without slipping: Torque: An object in mechanical equilibrium has no net force and no net torque acting on it. © 2010 Pearson Education, Inc.

40. Summary of Chapter 8 Moment of inertia: Newton’s second law: Parallel-axis theorem: Rotational work, power, and kinetic energy: © 2010 Pearson Education, Inc.

41. Summary of Chapter 8 Angular momentum: Newton’s second law, again: In the absence of an external net torque, angular momentum is conserved. © 2010 Pearson Education, Inc.

42. Summary – Rotational Analogies QuantityLinearRotational DisplacementDisplacement x Radians  InertiaMass (kg) I (kg  m 2 ) ForceNewtons NTorque N·m Velocity v “ m/s ”  Rad/s Acceleration a “ m/s 2 ”  Rad/s 2 Momentum mv (kg m/s) I  (kg  m 2  rad/s)

43. Analogous Formulas Linear MotionRotational Motion F = ma  = I  K = ½mv 2 K = ½I  2 Work = Fx Work =  Power = Fv Power = I  Fx = ½mv f 2 - ½mv o 2  = ½I  f 2 - ½I  o 2

44. Summary of Formulas: I =  mR 2 mgh o ½    ½mv o 2 = mgh f ½  f  ½mv f 2 Height? Rotation ? velocity?Height? velocity?