1. Ying Yi PhD Chapter 9 Rotational Dynamics 1 PHYS I @ HCC

2. Outline PHYS I @ HCC 2 Torque Center of Gravity Two conditions for equilibrium Newton’s second law for rotational motion

3. Torque There are three factors that determine the effectiveness of the force in opening the door: The magnitude of the force The position of the application of the force The angle at which the force is applied 3 PHYS I @ HCC

4. Definition of Torque 4 PHYS I @ HCC

5. 5 Lever Arm

6. PHYS I @ HCC 6 Torque and Axis The value of the torque depends on the chosen axis of rotation Torques can be computed around any axis There doesn’t have to be a physical rotation axis present Once a point is chosen, it must be used consistently throughout a given problem

7. Multiple Torques When two or more torques are acting on an object, the torques are added As vectors If the net torque is zero, the object’s rate of rotation doesn’t change 7 PHYS I @ HCC

8. Example 9.1 Torques PHYS I @ HCC 8

9. Group Problem: Battle of the revolving Door PHYS I @ HCC 9 Two disgruntled businesspeople are trying to use a revolving door, as in Figure 8.3. The woman on the left exerts a force of 625 N perpendicular to the door and 1.20 m from the hub’s center, while the man on the right exerts a force of 8.5×10 2 N perpendicular to the door and 0.800 m from the hub’s center. Find the net torque on the revolving door.

10. Example: The swinging door PHYS I @ HCC 10 (a) A man applies a force of F=3.00×10 2 N at an angle of 60.0º to the door of Figure, 2.00 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?

11. Center of Gravity The force of gravity acting on an object must be considered Definition of center of Gravity: The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. 11 PHYS I @ HCC

12. Calculating the Center of Gravity 12 PHYS I @ HCC

13. Coordinates of the Center of Gravity The coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object 13 PHYS I @ HCC

14. Center of Gravity of a Uniform Object The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry Often, the center of gravity of such an object is the geometric center of the object 14 PHYS I @ HCC

15. In class experiment: Center of gravity PHYS I @ HCC 15

16. Experimentally Determining the Center of Gravity The wrench is hung freely from two different pivots The intersection of the lines indicates the center of gravity A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is acting upward through the object’s center of gravity 16 PHYS I @ HCC

17. Example 9.6 Center of Gravity PHYS I @ HCC 17

18. Torque and Equilibrium First Condition of Equilibrium The net external force must be zero This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium This is a statement of translational equilibrium 18 PHYS I @ HCC

19. Torque and Equilibrium, cont To ensure mechanical equilibrium, you need to ensure rotational equilibrium as well as translational The Second Condition of Equilibrium states The net external torque must be zero 19 PHYS I @ HCC

20. Selecting an Axis The value of depends on the axis of rotation You can choose any location for calculating torques It’s usually best to choose an axis that will make at least one torque equal to zero This will simplify the torque equation 20 PHYS I @ HCC

21. Example 9.5 bodybuilding PHYS I @ HCC 21

22. Group Problem: Balancing act PHYS I @ HCC 22 A woman of mass m=55.0 kg sits on the left end of a seesaw a plank of length L=4.00 m, pivoted in the middle as in figure 8.8. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M=75.0 kg sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of m pl =12.0 kg. (c) repeat part (a), but this time compute the torques about an axis through the left end of the plank.

23. Notes About Equilibrium A zero net torque does not mean the absence of rotational motion An object that rotates at uniform angular velocity can be under the influence of a zero net torque This is analogous to the translational situation where a zero net force does not mean the object is not in motion 23 PHYS I @ HCC

24. Solving Equilibrium Problems Diagram the system Include coordinates and choose a rotation axis Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on the object For systems containing more than one object, draw a separate free body diagram for each object 24 PHYS I @ HCC

25. Problem Solving, cont. Apply the Second Condition of Equilibrium This will yield a single equation, often with one unknown which can be solved immediately Apply the First Condition of Equilibrium This will give you two more equations Solve the resulting simultaneous equations for all of the unknowns Solving by substitution is generally easiest 25 PHYS I @ HCC

26. Group Problem: Walking a horizontal beam PHYS I @ HCC 26 A uniform horizontal beam 5.00 m long and weighing 3.00×10 2 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0º with the horizontal. If a person weighing 6.00×10 2 N stands 1.50 m from the wall, find the magnitude of the tension T in the cable and the force R exerted by the wall on the beam.

27. PHYS I @ HCC 27 Torque and Angular Acceleration When a rigid object is subject to a net torque ( Στ ≠ 0), it undergoes an angular acceleration The angular acceleration is directly proportional to the net torque Moment of Inertia I

28. PHYS I @ HCC 28 Moment of Inertia The angular acceleration is inversely proportional to the analogy of the mass in a rotating system This mass analog is called the moment of inertia, I, of the object SI units are kg m 2

29. PHYS I @ HCC 29 Newton’s Second Law for a Rotating Object The angular acceleration is directly proportional to the net torque The angular acceleration is inversely proportional to the moment of inertia of the object

30. PHYS I @ HCC 30 More About Moment of Inertia There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object The moment of inertia also depends upon the location of the axis of rotation

31. PHYS I @ HCC 31 Moment of Inertia of a Uniform Ring Imagine the hoop is divided into a number of small segments, m 1 … These segments are equidistant from the axis

32. PHYS I @ HCC 32 Other Moments of Inertia

33. Example 9.10 Torque of an Electric saw motor PHYS I @ HCC 33 The motor in an electric saw brings the circular blade from rest up to the rated angular velocity of 80.0 rev/s in 240.0 rev. One type of blade has a moment of inertia of 1.41×10 -3 kgm 2. What net torque must the motor apply to the blade?

34. Angular momentum PHYS I @ HCC 34

35. Application 1: figure skating PHYS I @ HCC 35

36. Application 2: Figure Diving PHYS I @ HCC 36

37. Homework 3, 9, 13, 15, 23, 25, 29 PHYS I @ HCC 37