A) Yes, you have done this correctly so far. To finish solving for the rotational inertia of the rod-and-block apparatus, you need to multiply the angular acceleration by the mass of the rod-and-block apparatus. This will give you the rotational inertia of the rod-and-block apparatus.
B) To determine the downward acceleration of the large block, you need to use Newton's Second Law of Motion. This states that the net force on an object is equal to its mass times its acceleration. Therefore, the downward acceleration of the large block can be calculated by dividing the net force on the block by its mass.
C) The instantaneous total kinetic energy of the three blocks will be greater than 4mgD. This is because the kinetic energy of the three blocks includes the kinetic energy of the large block, the kinetic energy of the two small blocks, and the kinetic energy of the rod-and-block apparatus.
A light string that is attached to a large block of mass 4m passes over a pulley with negligible rotational inertia and is wrapped around a vertical pole of radius r. The system is released from rest, and as the block descends the string unwinds and the vertical pole with its attached apparatus rotates. The apparatus consists of a horizontal rod of length 2L, with a small block of mass m attached at each end. The rotational inertia of the pole and the rod are negligible. A) Determine the rotational inertia of the rod-and-block apparatus attached to the top of the pole. B) Determine the downward acceleration of the large block. C) When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three blocks compare with the value 4mgD? (greater, equal, less)----For A, someone started me off with Torque=I*(ang.acc.)--I plugged in 4mgr=(1/12)m4t^2(ang.acc.), solved for ang.acc. and got 12gr/t^2---have I done this right so far?
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