Physics 2210 Fall smartphysics Rotational Statics 11/18/2015

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1 Physics 2210 Fall 2015 smartphysics Rotational Statics 11/18/2015

2 τ TT = L T 1 sin 150 = 1 T 2 1L Poll τ TT = L 2 T 2 sin 150 = 1 4 T 2L τ gg = L 2 MM sin +90 = 1 2 MMM +90 MM τ gg = L 2 MM sin +90 = 1 2 MMM +90 MM In both cases the beam has the same mass and length and is attached to the wall by a hinge. In which of the static cases shown is the tension in the supporting wire bigger? τ TT + τ gg = 0 T 1 = MM A. Case 1 B. Case 2 C. Same τ T2 + τ g2 = 0 T 2 = 2MM

3 Unit 17 3/3 Lever Arm r line-of-action This is exactly the same as τ = r F Some people have found the lever arm confusing here is my take on it The line-of-action of the force is a line parallel to the direction of the force and passing through the point of application of the force) The Lever arm, r, is the perpendicular distance (of closest approach) from the rotation axis to the line-of-action Sometimes makes calculating the cross product easier

4 Example 17-1 A traffic light hangs from a beam as shown in the figure. The uniform aluminum beam AB is 7.20 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD, and (b) the vertical and horizontal components of the force exerted by the pivot A on the aluminum beam. r = line-of-action T Lever Arm F pp F pp Solution This problem is all about balancing all forces and torques on the beam AB. We solve such problems by making a catalogue of all forces acting on the beam, and the torque each exerts. But first we must decide on a rotation axis about which to calculate torque. It is usually convenient to pick a pivot that exerts force in both x and y directions on the element in question: because then these forces exert NO torque!!! We pick A (1) Force of pivot: F pp in the x-direction, F pp in the y direction. Notice we are treating them like two separate forces for convenience. They exert no torque (they act at the chosen rotation axis). (2) Tension Force in the cable: T directed in the x direction, The line-of-action is horizontal, The lever arm, r, is given in the diagram to be 3.80m. The torque (it acts CCW so it is POSITIVE) is τ T = +r T = m T

5 Example 17-1 Uniform beam AB: l=7.20 m, m =12.0 kg. (gi09-019) Traffic light M=21.5 kg. Determine (a) tension T (b) components F pp, F pp of the force exerted by the pivot on the beam (3) Weight of the beam itself. mm in the y direction Torque: mm acts at radius r=l 2=3.60 m. Angle from radial vector to weight is 127 So the torque is τ m = l 2 mmsin 127 : Note the negative orientation of this torque (it wants to deflect beam CW around the pivot) is contained in sin 127 = sin 127 (4) Weight of traffic light: Mg in the y direction Acts at distance r=l=7.20 m, and angle from radial vector to weight is AGAIN 127 τ M = lmm sin 127 Force components in the x- and y-direction independently sum to zero: F x = F pp T = 0 (1) F y = F py mm MM = 0 2 (*** note you CANNOT add the magnitudes of the forces) Torques add to zero τ = τ T + τ m + τ M = 3.80 m T 3.60 m mm sin m Mg sin 127 = 0 (3) F pp F pp mm T MM 127

6 Example 17-1 Uniform beam AB: l=7.20 m, m =12.0 kg. (gi09-019) Traffic light M=21.5 kg. Determine (a) tension T (b) components F pp, F pp of the force exerted by the pivot on the beam From equation (3) we have 3.60 m mm sin m MM sin m m m M T = = g sin m 3.80 m = 9.8 m s m 12.0 kg + (7.20 m)(21.5 kg) m = m s kg m 3.80 m = 408 N From equation (1): F pp T = 0 F pp = T = 408 N From equation (2): F pp mm MM = 0 F pp = mm + MM = 12.0 kg kg 9.8 m s 2 = 328 N

7 Example 17-1 synopsis A traffic light hangs from a beam as shown in the figure. The uniform aluminum beam AB is 7.20 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD, and (b) the vertical and horizontal components of the force exerted by the pivot A on the aluminum beam. F pp F pp Choose pivot at A to be rotation axis. (1) Force components F pp in the +x direction, F pp in the +y direction. They exert no torque (2) Tension Force : T directed in the x direction τ T = +T (3.80m) (3) Weight of the beam: mm in the y direction, torque is τ m = l 2 mmsin 127 : (4) Weight of traffic light: Mg in the y direction τ M = lmm sin 127 F x = F pp T = 0 (1) F y = F pp mm MM = 0 2 τ = τ T + τ m + τ M = 3.80 m T 3.60 m mm sin m MM sin 127 = 0 3 T = 9.8 m s m 12.0 kg + (7.20 m)(21.5 kg) = 408 N 3.80 m F pp T = 0 F pp = T = 408 N F pp mm MM = 0 F pp = mm + MM = 12.0 kg kg 9.8 m s 2 = 328 N mm T MM 127

8 Unit 18 We already covered this in Unit 16 U g = MMY CC = MM l sin φ + b 2 φ b This statement is somewhat misleading: because du g dy CC = MM is never ZERO JUST IGNORE IT!!!! What we really want to say is du g dφ = 0 To minimize the potential energy

9 Unit 18

10 Poll y T A F x > 0 T B F x > 0 T B F x = 0!!!! x MM N A MM N A MM N A Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with a black spot. Which of the above configurations best represents the equilibrium condition of this setup? A. A B. B C. C

11 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

12 Unit 19

13 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.

14 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

15 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

16 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

17 Law of Conservation of Angular Momentum Definition (1) : Angular Momentum of a rotor (sign of L follows sign of ω) L II I P ω L II Definition (2) : Angular Momentum of a particle in linear motion L mmm Sign of L is positive if particle misses P to the right Negative if it misses P to the left 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.

18 Example 19-1 putty m=100 g, v 0 =25 m/s, sticks to a stick, M=1.4 kg, l=1.2 m. b=0.80 m Find ω of the stick + putty after the collision. Solution by Conservation of Angular Momentum The only external force acting on the system is that by the pivot P. But since this force acts at point P, it exerts no torque No net external torque in collision Angular Momentum is conserved Before: (stick at rest) L i = mm 0 b (note in this case b turns out to be the minimum approach distance) After: rod and putty become joined, with total moment-of-inertia: I (rrr+ppppp) = I rrr + mb 2 And the system now rotates at angular velocity ω L f = I (rrr+ppppp) ω = Ml mb 2 ω Total Angular Momentum is conserved: L f = L i, Ml mb 2 ω = mm 0 b ω = mm 0 b Ml mb = 0.10 kg 25 m s 0.80 m kg 1.2 m kg 0.80 m 2 = 2.72 rad/s

19 Unit 20

20 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

21 Example 20-1 (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

22 Example 20-1 (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

23 Precessing Gyropscope Demo Video from MIT

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