1. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Physics I 95.141 LECTURE 21 11/24/10

2. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Exam Prep Question The system to the right consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. a) (10 pts) What is the moment of inertia of this system? b) (10 pts) Assume a 10kg mass is attached to a massless cord, wrapped around the cylinder, and dropped from rest. What is the acceleration of the mass? c) (5pts) What is the angular acceleration of the cylinder/mass system? d) (10pts) Determine the Kinetic Energy, as a function of time, associated with i) the rotating system and ii) the falling mass. 30cm 15cm 2kg 10kg

3. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Exam Prep Question The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. a) (10 pts) What is the moment of inertia of this system? 30cm 15cm 2kg

4. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Exam Prep Question The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. b) (10 pts) Assume a 10kg mass is attached to a massless cord, wrapped around the cylinder, and dropped from rest. What is the acceleration of the mass? 30cm 15cm 2kg 10kg

5. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Exam Prep Question The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. c) (5pts) What is the angular acceleration of the rotating cylinder/mass system? 30cm 15cm 2kg 10kg

6. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Exam Prep Question The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. d) (10pts) Determine the Kinetic Energy, as a function of time, associated with i) the rotating system and ii) the falling mass. 30cm 15cm 2kg 10kg

7. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Review Example What is the vector cross product of the two vectors:

8. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Administrative Notes Exam III –Wednesday 12/1 –In Class, 9am-9:50am –Chapters 9-11 Practice Exams posted Practice problems posted by end of day Exam Review Scheduled for 11/29….subject to change. Will probably have to be shifted to Tuesday.

9. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Outline Vector Cross Products Conservation of Angular Momentum What do we know? –Units –Kinematic equations –Freely falling objects –Vectors –Kinematics + Vectors = Vector Kinematics –Relative motion –Projectile motion –Uniform circular motion –Newton’s Laws –Force of Gravity/Normal Force –Free Body Diagrams –Problem solving –Uniform Circular Motion –Newton’s Law of Universal Gravitation –Weightlessness –Kepler’s Laws –Work by Constant Force –Scalar Product of Vectors –Work done by varying Force –Work-Energy Theorem –Conservative, non-conservative Forces –Potential Energy –Mechanical Energy –Conservation of Energy –Dissipative Forces –Gravitational Potential Revisited –Power –Momentum and Force –Conservation of Momentum –Collisions –Impulse –Conservation of Momentum and Energy –Elastic and Inelastic Collisions2D, 3D Collisions –Center of Mass and translational motion –Angular quantities –Vector nature of angular quantities –Constant angular acceleration –Torque –Rotational Inertia –Moments of Inertia –Angular Momentum

10. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Review of Lecture 20 Introduced concept of Angular Momentum Conservation of Angular Momentum –With no external torques acting on a system, the angular momentum of the system is conserved. Vector Cross products

11. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Review of Angular Motion We know equations of motion for angular motion We know torques cause angular acceleration Objects can have rotational kinetic energy And angular momentum So why the cross product?

12. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Torque and the Cross Product When we first introduced torque as the product of the radius and the perpendicular component of the Force, we were only interested in the magnitude of the torque! Magnitude given by RFsinθ…same as cross product FROM LECTURE 19

13. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Torque and the Cross Product However, we since learned that And we know that angular acceleration points in direction of axis of rotation… so Torque must as well! Torque is cross product of R,F FROM LECTURE 19

14. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Angular Momentum of a Particle We have already defined angular momentum as But this definition is for objects rotating with some angular velocity and moment of inertia around an axis of rotation. More general, alternate, definition:

15. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Equivalence of our two definitions Suppose we have a mass rotating around an axis Use cross product m=2kg 2m y x

16. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Equivalence of our two definitions Suppose we have a mass rotating around an axis Use m=2kg 2m y x

17. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 So why use cross product? Cross products are messy…why would we ever use them, instead of the simpler Because the cross product allows us to determine the angular momentum of, or torque on, objects which are not necessarily moving with constant, or even circular motion!

18. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Example Calculate the angular momentum (about the origin) of the rock of mass m dropped from rest off the cliff. (0,0) d

19. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Example What Torque is exerted (about the origin) on the rock? (0,0) d

20. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Relationship of torque to angular momentum When we discussed linear momentum, we revised Newton’s 2 nd Law to state Similarly, we can write an expression for net torque in terms of angular momentum Double check with our falling rock:

21. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Newton’s 2 nd Law: Angular Form The vector sum of all of the torques acting on a particle, object, or system, is equal to the time rate of change of the angular momentum of the particle, object or system.

22. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Conservation of Angular Momentum We used the revised expression for conservation of linear momentum to argue that if there is no net external force on a system or object, then the momentum of the system or object is conserved. Similarly: –If the net external torque acting on a system of object is zero, then the angular momentum of that object will remain constant. L constant!

23. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Conservation of Momentum

24. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Does this make sense? What is happening? What do we need to know? –System= Container + Skinner

25. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Moments of Inertia Skinner=point massShipping Container 12m 2.5m Mass=3,500kg

26. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 σ Container Determine surface mass density σ Divide into 6 slabs Total Mass = 3500kgs 12m 2.5m

27. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 I Container Divide into 6 slabs Top & Bottom:

28. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Ends Use parallel axis theorem h=6m

29. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Sides Use parallel axis theorem

30. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Conservation of Momentum Skinner seems to be making one rotation every 2 seconds.

31. Department of Physics and Applied Physics 95.141, F2010, Lecture 21 Finally In the clip, it takes about 20 seconds to turn the container 90 degrees.