A long thin rod of mass M = 2:00 kg and length L = 75:0 cm is free to rotate abo

Question

A long thin rod of mass M = 2:00 kg and length L = 75:0 cm is free to rotate about its center as shown. Two identical masses (each of mass m = .461 kg) slide without friction along the rod. The two masses begin at the rod's point of rotation when the rod is rotating at 10.0 rad/s. (a) When they have moved halfway to the end of the rod, how fast (rad/s) is the rod rotating? (b) When the masses are halfway to the end of the rod, what is the ratio of the nal kinetic energy to the initial kinetic energy (Kf=Ki)? (c) When they reach the end, how fast is the rod rotating (rad/s)?

Explanation / Answer

Use conservation of angular momentum. I*w = constant I is the moment of inertia, which is (1/12)Mrod*L^2 + 2m*R^2 R is the distance of the masses from the center of the rod. w = 10.0 when R = 0 For (a) and (b), R = L/4 For (c), R = L/2 Let's do (a) Angular momentum with massesw m at R=0: = (1/12)*2.00*(0.75)^2*10 = 9.38*10^-1 kg m^2/s (This remains constant). When R = L/4 = 0.1875 m, I*w = (2/12)*(0.75)^2*w + 2*(0.403)*(0.1875)^2*w = (9.38*10^-2 + 2.83*10^-2)w = 1.221*10^-1*w = 9.38*10^-1 w = 7.68 m/s For (b), compare initial and final values of (1/2) I w^2 For (c), repeat the process of (a), but use R = L/2

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