1. Physics 215 – Fall 2014Lecture 10-21 Welcome back to Physics 215 Today’s agenda: Extended objects Torque Moment of inertia

2. Physics 215 – Fall 2014Lecture 10-22 Current homework assignment HW8: –Knight Textbook Ch.12: 8, 54, 58, 60, 62, 64 –Due Monday, Nov. 5 th in recitation

3. Physics 215 – Fall 2014Lecture 10-23 Motion of center of mass of a system: The center of mass of a system of point objects moves in the same way as a single object with the same total mass would move under the influence of the same net (external) force.

4. Physics 215 – Fall 2014Lecture 10-24 Throwing an extended object Demo: odd-shaped object – 1 point has simple motion = projectile motion - center of mass Total external force F ext a CM = F ext /M Translational motion of system looks like all mass is concentrated at CM

5. Physics 215 – Fall 2014Lecture 10-25 Equilibrium of extended object Clearly net force must be zero Also, if want object to behave as point at center of mass  ALL forces acting on object must pass through CM

6. Physics 215 – Fall 2014Lecture 10-26 A few properties of the center of mass of an extended object The weight of an entire object can be thought of as being exerted at a single point, the center of mass. One can locate the center of mass of any object by suspending it from two different points and drawing vertical lines through the support points. Equilibrium can be ensured if all forces pass through CM An object at rest on a table does not tip over if the center of mass is above the area where it is supported.

7. Physics 215 – Fall 2014Lecture 10-27 Conditions for equilibrium of an extended object For an extended object that remains at rest and does not rotate: The net force on the object has to be zero. The net torque on the object has to be zero.

8. Physics 215 – Fall 2014Lecture 10-28 Preliminary definition of torque: The torque on an object with respect to a given pivot point and due to a given force is defined as the product of the force exerted on the object and the moment arm. The moment arm is the perpendicular distance from the pivot point to the line of action of the force.

9. Physics 215 – Fall 2014Lecture 10-29 Computing torque d F  = |F|d = |F||r|sin   = (|F| sin  r| component of force at 90 0 to position vector times distance r O 

10. Physics 215 – Fall 2014Lecture 10-210 Definition of torque: where r is the vector from the reference point (generally either the pivot point or the center of mass) to the point of application of the force F. where  is the angle between the vectors r and F.

11. Physics 215 – Fall 2014Lecture 10-211 Vector (or “cross”) product of vectors The vector product is a way to combine two vectors to obtain a third vector that has some similarities with multiplying numbers. It is indicated by a cross (  ) between the two vectors. The magnitude of the vector cross product is given by: The direction of the vector A  B is perpendicular to the plane of vectors A and B and given by the right-hand rule.

12. Physics 215 – Fall 2014Lecture 10-212 Interpretation of torque Measures tendency of any force to cause rotation Torque is defined with respect to some origin – must talk about “torque of force about point X”, etc. Torques can cause clockwise (+) or anticlockwise rotation (-) about pivot point

13. Physics 215 – Fall 2014Lecture 10-213 Extended objects need extended free-body diagrams Point free-body diagrams allow finding net force since points of application do not matter. Extended free-body diagrams show point of application for each force and allow finding net torque.

14. Physics 215 – Fall 2014Lecture 10-214 1.Clockwise. 2.Counter-clockwise. 3.Not at all. 4.Not sure what will happen. A T-shaped board is supported such that its center of mass is to the right of and below the pivot point. Which way will it rotate once the support is removed?

15. Physics 215 – Fall 2014Lecture 10-215

16. Physics 215 – Fall 2014Lecture 10-216

17. Physics 215 – Fall 2014Lecture 10-217 Examples of stable and unstable rotational equilibrium Wire walker – no net torque when figure vertical. Small deviations lead to a net restoring torque  stable

18. Physics 215 – Fall 2014Lecture 10-218 Restoring torque dRdR dLdL Consider displacing anticlockwise  R increases  L decreases net torque causes clockwise rotation! e.g., wire walker

19. Physics 215 – Fall 2014Lecture 10-219 Rotations about fixed axis Every particle in body undergoes circular motion (not necessarily constant speed) with same time period v = (2  r)/T =  r. Quantity  is called angular velocity Similarly can define angular acceleration  t

20. Physics 215 – Fall 2014Lecture 10-220 Rotational Motion pivot  riri |v i | = r i  at 90 º to r i mimi * Particle i: FiFi * Newton’s 2 nd law: m i  v i /  t = F i T  component at 90 º to r i m i r i 2  /  t = F i T r i =  i * Finally, sum over all masses: * Substitute for v i and multiply by r i :  /  t)   m i r i 2 =   i =  net

21. Physics 215 – Fall 2014Lecture 10-221 Discussion  /  t)   m i r i 2 =  net  angular acceleration Moment of inertia, I I   =  net compare this with Newton’s 2 nd law M a = F

22. Physics 215 – Fall 2014Lecture 10-222 Moment of Inertia * I must be defined with respect to a particular axis

23. Physics 215 – Fall 2014Lecture 10-223 Moment of Inertia of Continuous Body r mm m0m0

24. Physics 215 – Fall 2014Lecture 10-224 Tabulated Results for Moments of Inertia of some rigid, uniform objects (from p.299 of University Physics, Young & Freedman)

25. Physics 215 – Fall 2014Lecture 10-225 Parallel-Axis Theorem CM D D *Smallest I will always be along axis passing through CM

26. Physics 215 – Fall 2014Lecture 10-226 Practical Comments on Calculation of Moment of Inertia for Complex Object 1.To find  for a complex object, split it into simple geometrical shapes that can be found in Table 9.2 2.Use Table 9.2 to get  CM  for each part about the axis parallel to the axis of rotation and going through the center-of-mass 3.If needed use parallel-axis theorem to get  for each part about the axis of rotation 4.Add up moments of inertia of all parts

27. Physics 215 – Fall 2014Lecture 10-227 Beam resting on pivot M = ? m M b = 2m N CM of beam Vertical equilibrium? Rotational equilibrium? r r x rmrm  F =   = M = N =

28. Physics 215 – Fall 2014Lecture 10-228 Suppose M replaced by M/2 ? vertical equilibrium? rotational dynamics? net torque? which way rotates? initial angular acceleration?  F =   =

29. Physics 215 – Fall 2014Lecture 10-229 Moment of Inertia? I =  m i r i 2 * depends on pivot position! I = * Hence 

30. Physics 215 – Fall 2014Lecture 10-230 Rotational Kinetic Energy K =  i (  m i v i 2 = (1/2)    i m i r i 2 Hence K = (1/2) I   This is the energy that a rigid body possesses by virtue of rotation

31. Physics 215 – Fall 2014Lecture 10-231 Spinning a cylinder Cable wrapped around cylinder. Pull off with constant force F. Suppose unwind a distance d of cable F What is final angular speed of cylinder? 2R Use work-KE theorem W = Fd = K f = (1/2) I  2 Mom. of inertia of cyl.? -- from table: (1/2)mR 2 –from table: (1/2)mR 2  = [2Fd/(mR 2 /2)] 1/2 = [4Fd/(mR 2 )] 1/2

32. Physics 215 – Fall 2014Lecture 10-232 Reading assignment Rotations Chapter 12 in textbook