1. Copyright © 2010 Pearson Education, Inc. Lecture Outline Chapter 11 Physics, 4 th Edition James S. Walker

2. Copyright © 2010 Pearson Education, Inc. Chapter 11 Rotational Dynamics and Static Equilibrium

3. Copyright © 2010 Pearson Education, Inc. Units of Chapter 11 Torque Torque and Angular Acceleration Zero Torque and Static Equilibrium Center of Mass and Balance Dynamic Applications of Torque Angular Momentum

4. Copyright © 2010 Pearson Education, Inc. Units of Chapter 11 Conservation of Angular Momentum Rotational Work and Power The Vector Nature of Rotational Motion

5. Copyright © 2010 Pearson Education, Inc. 11-1 Torque From experience, we know that the same force will be much more effective at rotating an object such as a nut or a door if our hand is not too close to the axis. This is why we have long-handled wrenches, and why doorknobs are not next to hinges. Applying a torque (a) When a wrench is used to loosen a nut, less force is required if is applied far from the nut. (b) Similarly, less force is required to open a revolving door if it is applied far from the axis of rotation.

6. Copyright © 2010 Pearson Education, Inc. 11-1 Torque We define a quantity called torque: The torque increases as the force increases, and also as the distance increases.

7. Copyright © 2010 Pearson Education, Inc. 11-1 Torque Only the tangential component of force causes a torque: Only the tangential component of a force causes a torque. (a). A radial force causes no rotation. In this case, the force F is opposed by an equal and opposite force exerted by the axle of the merry-go-around. The merry-go-around does not rotate. (b) A force applied at an angle Ɵ with respect to the radial direction. The radial component of this force, F cos θ, cause no rotation; the tangential component, F sin θ, can cause a rotation. Because it is the tangential component alone that causes rotation, we define the torque to have a magnitude of r(F sin θ).

8. Copyright © 2010 Pearson Education, Inc. 11-1 Torque This leads to a more general definition of torque:

9. Copyright © 2010 Pearson Education, Inc. 11-1 Torque If the torque causes a counterclockwise angular acceleration, it is positive; if it causes a clockwise angular acceleration, it is negative. Sign Convention for Torque By convention, if a torque ד acts alone, then r > 0 if the torque causes a counterclockwise angular acceleration r < 0 if the torque causes a clockwise angular acceleration

10. Copyright © 2010 Pearson Education, Inc. 11-2 Torque and Angular Acceleration Newton’s second law: If we consider a mass m rotating around an axis a distance r away, we can reformat Newton’s second law to read: Or equivalently, α = a/r; α = a/r = F/mr Finally, multiplying both numerator and denominator by r gives

11. Copyright © 2010 Pearson Education, Inc. 11-2 Torque and Angular Acceleration Once again, we have analogies between linear and angular motion:

12. Copyright © 2010 Pearson Education, Inc. 11-2 Torque and Angular Acceleration A light rope wrapped around a disk-shaped pulley is pulled tangentially with a force of 0.53 N. Find the angular acceleration of the pulley, given that its mass is 1.3 kg and its radius is 0.11 m. The torque applied to the disk is ד = r x F = (0.11 m)(0.53 N) = 5.8 x 10 -2 N.m Since the pulley is a disk, its moment of inertia is given by I = ½ mr 2 I = ½ mr 2 = ½(1.3 kg)(0.11 m) 2 = 7.9 x 10 -3 kg.m 2 Thus, the angular acceleration of the pulley is α = ד / I = 5.8 x 10 -2 N.m 7.9 x 10 -3 kg.m 2 α = 7.3 rad /s 2

13. Copyright © 2010 Pearson Education, Inc. 11-3 Zero Torque and Static Equilibrium Static equilibrium occurs when an object is at rest – neither rotating nor translating.

14. Copyright © 2010 Pearson Education, Inc. 11-3 Zero Torque and Static Equilibrium If the net torque is zero, it doesn’t matter which axis we consider rotation to be around; we are free to choose the one that makes our calculations easiest.

15. Copyright © 2010 Pearson Education, Inc. 11-3 Zero Torque and Static Equilibrium A child of mass m is supported on a light plank by his parent, who exert the forces F 1 and F 2 as indicated. Find the forces required to keep the plank in static equilibrium. Use the right end of the plank as the axis of rotation.. Set the net force acting on the plank equal to zero. F 1 + F 2 – mg = 0 Set the net torque acting on the plank equal to zero. -F 1 (L) + mg(1/4 L) = 0 F 1 = 1/4mg F 2 = mg – 1/4mg = 3/4mg F 2 = ¾ mg

16. Copyright © 2010 Pearson Education, Inc. 11-3 Zero Torque and Static Equilibrium A 5.0 m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board, as shown below; the other is 1.50 m away. Find the forces exerted by the pillars when a 90.0-kg diver stands at the far end of the board. Set the net force acting on the diving board equal to zero. ∑F y = F 1,y + F 2,y – mg = 0 Set the net torque acting on the diving board equal to zero. Torque about the F 1 -F 2,y (d) - mg( L) = 0 F 2,y = mg (L / d) F 2,y = (90.kg)(9.81 m/s2)(5.0 m/ 1.50 m) F 2,y = 2940 N F 1,y = mg – F 2,y = (0 kg) (9.81 m/s 2 ) – 2940 N F 1,y = -2060 N

17. Copyright © 2010 Pearson Education, Inc. 11-3 Zero Torque and Static Equilibrium When forces have both vertical and horizontal components, in order to be in equilibrium an object must have no net torque, and no net force in either the x - or y -direction. A lamp in static equilibrium A wall mounted lamp of mass m is suspended from a light curved rod. The bottom of the rod is bolted to the wall. The rod is also connected to the wall by a Horizontal wire a vertical distance V above the bottom of the rod. ד∑ = T(V) – mg(H) = 0 Solve for the tension T, T = mg(H / V) ∑F y = f y – mg = 0 = => f y = mg ∑F x = f x – T = 0 = => f x = T = mg(H / V)

18. Copyright © 2010 Pearson Education, Inc. 11-3 Zero Torque and Static Equilibrium An 85-kg person stands on a lightweight ladder as shown. The floor is rough; hence, it exerts both a normal forces f 1, and a frictional force, f 2 on the ladder. The wall, on the other hand, is frictionless; it exerts only a normal force, f 3. Using the dimensions given in the figure, find the magnitudes of f 1, f 2 and f 3. Set the net torque acting on the ladder equal to zero. Use the bottom of the ladder as the axis: f 3 (a) - mg (b) = 0  f 3 = mg (b/a) = 150 N ∑F x = 0 = f 2 – f 3 = 0  f 2 = f 3 = 150 N ∑F y = 0  f 1 – mg = 0  f 1 = 830 N

19. Copyright © 2010 Pearson Education, Inc. 11-4 Center of Mass and Balance If an extended object is to be balanced, it must be supported through its center of mass. Zero torque and balance One section of a mobile. The rod is balanced when the net torque acting on it is zero. This is equivalent to having the center of mass directly under the suspension point

20. Copyright © 2010 Pearson Education, Inc. 11-4 Center of Mass and Balance This fact can be used to find the center of mass of an object – suspend it from different axes and trace a vertical line. The center of mass is where the lines meet.

21. Copyright © 2010 Pearson Education, Inc. 11-5 Dynamic Applications of Torque When dealing with systems that have both rotating parts and translating parts, we must be careful to account for all forces and torques correctly.

22. Copyright © 2010 Pearson Education, Inc. 11-6 Angular Momentum Using a bit of algebra, we find for a particle moving in a circle of radius r, L = Iω = (mr 2 ) (v/r) = rmv =  Note that mv is the linear momentum p; we find that the angular momentum of a point mass can be written in the following form :

23. Copyright © 2010 Pearson Education, Inc. 11-6 Angular Momentum For more general motion,

24. Copyright © 2010 Pearson Education, Inc. 11-6 Angular Momentum Looking at the rate at which angular momentum changes, ∆L = L f – L i = (∆ t) ד The final angular momentum of the object is L f = L i + (∆ t) ד

25. Copyright © 2010 Pearson Education, Inc. 11-7 Conservation of Angular Momentum If the net external torque on a system is zero, the angular momentum is conserved. The most interesting consequences occur in systems that are able to change shape:

26. Copyright © 2010 Pearson Education, Inc. 11-7 Conservation of Angular Momentum As the moment of inertia decreases, the angular speed increases, so the angular momentum does not change. Angular momentum is also conserved in rotational collisions:

27. Copyright © 2010 Pearson Education, Inc. 11-8 Rotational Work and Power A torque acting through an angular displacement does work, just as a force acting through a distance does. The work-energy theorem applies as usual, regardless whether the work is done by a force or a torque.

28. Copyright © 2010 Pearson Education, Inc. 11-8 Rotational Work and Power Power is the rate at which work is done, for rotational motion as well as for translational motion. Again, note the analogy to the linear form:

29. Copyright © 2010 Pearson Education, Inc. 11-8 Rotational Work and Power It takes a good deal of effort to make homemade ice cream. (a) If the torque required to turn the handle on an ice cream maker is 5.7 N.m, how much work is expended on each complete revolution of the handle? (b) How much power is required to turn the handle if each revolution is completed in 1.5 s? (b). Power is the work per unit time; that is P = W / ∆ t = (36 J) / (1.5 s) = 24 W = (5.7 N.m) (2 π rad) = 36 J a.

30. Copyright © 2010 Pearson Education, Inc. 11-9 The Vector Nature of Rotational Motion The direction of the angular velocity vector is along the axis of rotation. A right-hand rule gives the sign.

31. Copyright © 2010 Pearson Education, Inc. 11-9 The Vector Nature of Rotational Motion A similar right-hand rule gives the direction of the torque.

32. Copyright © 2010 Pearson Education, Inc. 11-9 The Vector Nature of Rotational Motion Conservation of angular momentum means that the total angular momentum around any axis must be constant. This is why gyroscopes are so stable.

33. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 11 A force applied so as to cause an angular acceleration is said to exert a torque. Torque due to a tangential force: Torque in general: Newton’s second law for rotation: In order for an object to be in static equilibrium, the total force and the total torque acting on the object must be zero. An object balances when it is supported at its center of mass.

34. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 11 In systems with both rotational and linear motion, Newton’s second law must be applied separately to each. Angular momentum: For tangential motion, In general, Newton’s second law: In systems with no external torque, angular momentum is conserved.

35. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 11 Work done by a torque: Power: Rotational quantities are vectors that point along the axis of rotation, with the direction given by the right-hand rule.