2. Physics 151: Lecture 23, Pg 1 Physics 151: Lecture 23 Today’s Agenda l Topics çMore on Rolling Motion çCh. 11.1 Angular MomentumCh. 11.3-5

3. Physics 151: Lecture 23, Pg 2 Example : Rolling Motion l A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? M  h M v ? Cylinder has radius R M M M M M

4. Physics 151: Lecture 23, Pg 3 Lecture 22, ACT 4a Rolling Motion l A race !! Two cylinders are rolled down a ramp. They have the same radius but different masses, M 1 > M 2. Which wins the race to the bottom ? A) Cylinder 1 B) Cylinder 2 C) It will be a tie M1  h M? M2

5. Physics 151: Lecture 23, Pg 4 Lecture 22, ACT 4b Rolling Motion l A race !! Two cylinders are rolled down a ramp. They have the same moment of inertia but different radius, R 1 > R 2. Which wins the race to the bottom ? A) Cylinder 1 B) Cylinder 2 C) It will be a tie R1  h M? R2 animation

6. Physics 151: Lecture 23, Pg 5 Lecture 22, ACT 4c Rolling Motion l A race !! A cylinder and a hoop are rolled down a ramp. They have the same mass and the same radius. Which wins the race to the bottom ? A) Cylinder B) Hoop C) It will be a tie M1  h M? M2 animation

7. Physics 151: Lecture 23, Pg 6 Remember our roller coaster. Perhaps now we can get the ball to go around the circle without anyone dying. Note: Radius of loop = R Radius of ball = r

8. Physics 151: Lecture 23, Pg 7 How high do we have to start the ball ? 1 2 -> The rolling motion added an extra 2/10 R to the height) h h = 2.7 R = (2R + 1/2R) + 2/10 R

9. Physics 151: Lecture 23, Pg 8 Angular Momentum: Definitions & Derivations l We have shown that for a system of particles Momentum is conserved if l What is the rotational version of this ?? See text: 11.3 F  The rotational analogue of force F is torque  p l Define the rotational analogue of momentum p to be angular momentum p=mv Animation

10. Physics 151: Lecture 23, Pg 9 Definitions & Derivations... L l First consider the rate of change of L: So(so what...?) See text: 11.3

11. Physics 151: Lecture 23, Pg 10 Definitions & Derivations... l Recall that l Which finally gives us: l Analogue of !!   EXT See text: 11.3

12. Physics 151: Lecture 23, Pg 11 What does it mean? l where and l In the absence of external torques Total angular momentum is conserved See text: 11.5

13. Physics 151: Lecture 23, Pg 12 i j Angular momentum of a rigid body about a fixed axis: l Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momenta of each particle: rr1rr1 rr3rr3 rr2rr2 m2m2 m1m1 m3m3  vv2vv2 vv1vv1 vv3vv3 L We see that L is in the z direction. Using v i =  r i, we get I (since r i, v i, are perpendicular) Analogue of p = mv !!  See text: 11.4

14. Physics 151: Lecture 23, Pg 13 Lecture 23, ACT 2 Angular momentum In the figure, a 1.6-kg weight swings in a vertical circle at the end of a string having negligible weight. The string is 2 m long. If the weight is released with zero initial velocity from a horizontal position, its angular momentum (in kg · m2/s) at the lowest point of its path relative to the center of the circle is approximately a. 40 b. 10 c. 30 d. 20 e. 50

15. Physics 151: Lecture 23, Pg 14 Angular momentum of a rigid body about a fixed axis: In general, for an object rotating about a fixed (z) axis we can write L Z = I  The direction of L Z is given by the right hand rule (same as  ). We will omit the ” Z ” subscript for simplicity, and write L = I   z See text: 11.4

16. Physics 151: Lecture 23, Pg 15 Lecture 23, ACT 2 Angular momentum l Two different spinning disks have the same angular momentum, but disk 1 has more kinetic energy than disk 2. çWhich one has the biggest moment of inertia ? (a) disk 1 (b) disk 2 (c) not enough info I 1 < I 2  disk 2  disk 1 K 1 > K 2

17. Physics 151: Lecture 23, Pg 16 Example: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity  0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity  F. What is  F ? 00 z FF z

18. Physics 151: Lecture 23, Pg 17 Example: Two Disks l First realize that there are no external torques acting on the two-disk system. çAngular momentum will be conserved ! l Initially, the total angular momentum is due only to the disk on the bottom: 00 z 2 1

19. Physics 151: Lecture 23, Pg 18 Example: Two Disks l First realize that there are no external torques acting on the two-disk system. çAngular momentum will be conserved ! l Finally, the total angular momentum is due to both disks spinning: FF z 2 1

20. Physics 151: Lecture 23, Pg 19 Example: Two Disks l Since L INI = L FIN 00 z FF z L INI L FIN An inelastic collision, since E is not conserved (friction) !

21. Physics 151: Lecture 23, Pg 20 Example: Two Disks l Let’s use conservation of energy principle: 00 z FF z E INI E FIN 1/2 I  0 2 = 1/2 (I + I)  F 2  F 2 = 1/2  0 2  F =  0 / 2 1/2 E INI = E FIN

22. Physics 151: Lecture 23, Pg 21 l Using conservation of angular momentum: L INI = L FIN we got a different answer ! Example: Two Disks  F ’ =  0 / 2 1/2 Conservation of energy ! Conservation of momentum !  F ’ >  F Which one is correct ?  F =  0 / 2

23. Physics 151: Lecture 23, Pg 22 Example: Two Disks Is the system conservative ? Are there any non-conservative forces involved ? In order for top disc to turn when in contact with the bottom one there has to be friction ! (non-conservative force !) So, we can not use the conservation of energy here. correct answer:  F =  0 /2 We can calculate work being done due to this friction ! FF z W =  E = 1/2 I  0 2 - 1/2 (I+I) (  0 /2) 2 = 1/2 I  0 2 (1 - 2/4) = 1/4 I  0 2 = 1/8 MR 2  0 2 This is 1/2 of initial Energy !

24. Physics 151: Lecture 23, Pg 23 Lecture 23, ACT 2 Angular momentum

25. Physics 151: Lecture 23, Pg 24 Recap of today’s lecture l Chapter 11.1-5, çRolling Motion çAngular Momentum l For next time: Read Ch. 11.1-11.