Physics 2210 Fall smartphysics Conservation of Angular Momentum 11/20/2015

Transcription

1 Physics 2210 Fall 2015 smartphysics Conservation of Angular Momentum 11/20/2015

2 Poll In the two cases shown above identical ladders are leaning against frictionless walls and are not sliding. In which case is the force of friction between the ladder and the ground the biggest? A. Case 1 B. Case 2 C. Same

3 Static Equilibrium: The net force is zero: F x = f 1 N 2 = 0 1 F y = N 1 + f 2 MM = 0 (2) The net torque is zero about the lower pivot point y N 2 f 2 x τ = N 2 L sin θ + f 2 L cos θ MM L cos θ = 0 (3) 2 Going from case (1) to case (2), θ increases, so we concentrate on equation (3), but noting that from (1), we have f 1 = N 2 (and L cancels) f 1 sin θ + f 2 cos θ = 1 MM cos θ 3 2 Note that as θ sin θ, cos θ But there is too many degrees of freedom in (3) (f 1 and f 2 ) to say what happens to f 1. N 1 L θ f 1 MM However, I forgot to use the fact that the wall is frictionless f 2 = 0!!! Now we have f 1 sin θ = 1 MM cos θ f 2 1 = 1 MM cot θ 2 As θ sin θ, cos θ cot θ, but MM does not change! Therefore as θ increases from case 1 to case 2, f 1 decreases!!! The friction force on the floor is greatest in case 1

4 Unit 19

5 Unit 19 Law of Conservation of Angular Momentum If the sum of the external torques on a system is zero, the angular momentum of the system is conserved (relative to some rotation axis).

6 Poll Consider the two collisions shown above. In both cases a solid disk of mass M, radius R, and initial angular velocity ω 0 is dropped onto an initially stationary second disk having the same radius. In Case 2 the mass of the bottom disk is twice as big as in Case 1. If there are no external torques acting on either system, in which case is the final kinetic energy of the system biggest? A. Case 1 B. Case 2 C. Same

7 Poll The angular momentum of a freely rotating disk around its center is L disk. You toss a heavy block horizontally onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. We will choose the initial angular momentum of the disk to be positive. What is the L total, the magnitude of the angular momentum of the disk-block system? A. L total > L disk B. L total = L disk C. L total < L disk

8 Poll The angular momentum of a freely rotating disk around its center is L disk. You toss a heavy block horizontally onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. We will choose the initial angular momentum of the disk to be positive. What is the magnitude of the final angular momentum of the disk-block system? A. L total > L disk B. L total = L disk C. L total < L disk

9 Unit 20

10 Precessing Gyropscope Demo Video from MIT

11 Demos In many systems one can change the moment-of-inertia of a system by changing the radial distribution of mass. To conserve angular momentum the angular speed must change to compensate

12 Example 19-2 (1/3) A small puck of mass m=100 g travels with initial speed v 0 =25m/s to the right. It strikes and sticks to a rod (initially at rest) that is free only to rotate about the pivot P at one end, in the plane of the page. The rod has mass M=1.4 kg and length l=1.2 m. The point of impact is at a distance of b=0.80 m from the pivot. Find the angular velocity ω of the stick after the collision. Solution: TOP VIEW OVER ICE In general, collisions are NOT elastic CANNOT assume energy is conserved We would normally say that total momentum (of the system of the stick and puck) is conserved, but only if there are no external forces acting But the pivot provides WHATEVER EXTERNAL FORCE NECESSARY to prevent the stick from detaching from the rotation axis: Total Momentum is DEFINITELY NOT conserved Fortunately, the force of the pivot acts at the rotation axis: so it exerts NO (EXTERNAL) TORQUE abut P Total Angular Momentum of the system (puck +stick) about P is CONSERVED m v 0 b M P l

13 Law of Conservation of Angular Momentum Definition (1) : Angular Momentum of a rotor (sign of L follows sign of ω) about P L Iω, L II Definition (2) : Angular Momentum of a particle, in linear motion, about P L mmr Here r is the lever-arm: perpendicular distance (of closest approach) from the line-of-motion to the rotation axis Sign of L: is positive if particle misses P to the right Negative if it misses P to the left I P ω L II Total angular momentum of a system (rotors and particles defined above) is the sum (with sign: they are vectors!!!) of the angular momenta of all components. In the absence of external torque about a pivot (rotation axis) P, the total angular momentum (sum of individual angular momenta) of a system is conserved. m v L mm P r Line-of- Motion

14 Example 19-1 (2/2) Puck m=100 g, v 0 =25 m/s, hits and sticks to rod: M=1.4 kg, l=1.2m at b=0.80 m. Find ω of the puck+rod after the collision. Solution by Conservation of Angular Momentum Angular Momentum (about P) is conserved Before: (stick at rest) L i = mm 0 r = mm 0 b After: rod and putty become joined, with total moment-of-inertia: I (rrd+pppp) = I rrr + mb 2 And the system now rotates at angular velocity ω L f = I (rrr+ppcc) ω = Ml mb 2 ω Total Angular Momentum is conserved: L f = L i, Ml mb 2 ω = mm 0 b mm 0 b ω = Ml mb = 0.10 kg 25 m s 0.80 m kg 1.2 m kg 0.80 m 2 = 2.72 rad/s Note the change in kinetic energy K i = 1 2 mv 0 2 = kg 25 m s 2 = J K f = 1 2 I (rrr+pppp)ω 2 = kg 1.2 m kg 0.80 m rad/s 2 = 2.72 J

15 Example 19-2 (1/4) A small puck of mass m=100 g travels with initial speed v 0 =25m/s to the right. It strikes and sticks to a rod (initially at rest) that is free both to translate and rotate in the plane of the page. The rod has mass M=1.4 kg and length l=1.2 m. The point of impact is at a distance of b=0.80 m from the pivot. Find the final linear and angular velocity. TOP VIEW OVER ICE m v 0 b M l In general, collisions are NOT elastic CANNOT assume energy is conserved NO external force (in the x and y directions) Total Momentum is conserved NO (EXTERNAL) TORQUE abut the center-of-mass (CM) of puck+rod assembly Total Angular Momentum of the system (puck+rod) about CM is CONSERVED How do you conserve both linear and angular momentum??? The linear momentum is that of a (fictional) point particle at the CM with total mass m + M Angular momentum is conserved about the CM of the system at moment of collision

16 Example 19-2 (2/4) Puck of mass m=100 g, v 0 =25m/s, strikes and sticks to a rod: M=1.4 kg, l=1.2 m, at b=0.80 m. Find v f and ω f b r cc l/2 y ω f v f m M l x The puck and rod forms a new system whose CM is offset from the center of the rod toward the poi t at which the puck is stuck r cc = 1 m + M l m b + M 2 = 0.1kg 0.8m + 1.4kg 0.6m 1.4kg kg The CM of the puck+rod assembly will travel in a straight line Before the collision:p ii = mv 0, P iy = 0. = 0.613m After the collision the CM of the assembly must be traveling in the +x direction at v f. And the assembly is rotating about the CM at angular velocity ω f

17 Example 19-2 (3/4) Puck of mass m=100 g, v 0 =25m/s, strikes and sticks to a rod: M=1.4 kg, l=1.2 m, at b=0.80 m. Find v f and ω f r cc = 0.613m. Before the collision: P ii = mv 0, P iy = 0. After the collision P fx = m + M v f, P fy = 0. Momentum is conserved m + M v f = mv 0 v f = mv 0 0.1kg 25 m = s m + M 0.1kg + 1.4kg = 1.67 m s Angular momentum about CM before collision: L i = +mv 0 r = +mv 0 b r cc = 0.1kg 25 m s 0.8m 0.613m = 0.466kg m 2 s After the collision: L f = Iω f Where I is the moment-of-inertia about the CM of the puck+rod assembly, I = I rrr + I pppp By Parallel Axes Theorem: I rrr = 1 12 Ml2 + M r cc l 2 = kg 1.2m kg 0.613m 0.600m 2 = kg m 2 2

18 Example 19-2 (4/4) Puck of mass m=100 g, v 0 =25m/s, strikes and sticks to a rod: M=1.4 kg, l=1.2 m, at b=0.80 m. Find v f and ω f r cc = 0.613m, L i = +mv 0 b r cc = 0.466kg m 2 s After the collision: L f = Iω f, I = I rrr + I pppp By Parallel Axes Theorem: I rrr = 1 12 Ml2 + M r cc l 2 2 = kg m 2 And the puck is a point mass at a distance of b r cc from the CM I pppp = m b r cc 2 = m b r cc 2 = 0.1kg 0.8m 0.613m 2 = kg m 2 So the total moment of inertia of the puck+rod assembly about its own CM is I = I rrr + I pppp = kg m kg m 2 = kg m 2 Angular momentum about the CM of the rod+puck assembly is conserved Iω f = L i ω f = L i 0.466kg m2 s = = 2.72 rad/s I kg m2 Total final kinetic energy: K f = 1 2 m + M v f Iω f 2 = 2.72 J

19 Example 20-1 (1/2) (gi08-69) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1360 kg m 2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)? Answer: (a) 0.43 rad/s; (b) 0.80 rad/s. Solution: (a) An external force acts at the center of the merry-go-round to keep the disk from moving away from the pivot. But this force exerts no external torque (to the disk + people system) So the total angular momentum is conserved. Before collision (jumping on) L i = I MMM ω i Where I MMM =1360 kg m 2 is the moment-of-inertia of the merry-go-round given, and ω i =0.80 rad/s is its initial angular velocity. The people were at rest on the ground and do not contribute to the angular momentum. After the four people jump on the rim of the disk, the total moment of inertia changes to I = I MMM + 4mR 2 Where m is the mass of each person, R=2.1m is the radius of the merry-go-round And so after the collision : L f = Iω f = I MMM + 4mR 2 ω f

20 Example 20-1 (2) (gi08-69) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1360 kg m 2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)? Answer: (a) 0.43 rad/s; (b) 0.80 rad/s. Solution: (a) continued: Total angular momentum is conserved: L i = L f I MMM ω i = I MMM + 4mR 2 ω f I MMM ω i ω f = I MMM + 4mR 2 = 1360 kg m rad s 1360kg m 2 = rad s kg 2.1 m 2 (b) Here we start with all 4 people on board at the rim, so wehad, before the collision: L i = Iω i = I MMM + 4mR 2 ω i When the people jump off, they pushed radially, so their tangential speeds are unchanged: v f = v i = Rω i And so after the collision: L f = I MMM ω f + 4mv f b = I MMM ω f + 4mv f R = I MMM ω f + 4mR 2 ω i L i = L f I MMM + 4mR 2 ω i = I MMM ω f + 4mR 2 ω i I MMM ω i + 4mR 2 ω i = I MMM ω f + 4mR 2 ω i I MMM ω i = I MMM ω f ω f = ω i = 0.80 rad s