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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 IC-W11D3-2 Group Problem 2 A drum A of mass m and radius R is suspended from a drum B also of mass m and radius R , which is free to rotate about its axis. The suspension is in the form of a massless metal tape wound around the outside of each drum, and free to unwind. Gravity is directed downwards. Both drums are initially at rest. Find the initial acceleration of drum A , assuming that it moves straight down. Solution: The key to solving this problem is to determine the relation between the three kinematic quantities ! A , ! B and a A , the angular accelerations of the two drums and the linear acceleration of drum A . One way to do this is to introduce the auxiliary variable z for d 2z the length of the tape that is unwound from the upper drum. Then, ! B R = 2 . The dt dz linear velocity v A may then be expressed as the sum of two terms, the rate at which dt the tape is unwinding from the upper drum and the rate ! A R at which the falling drum is moving relative to the lower end of the tape. Taking derivatives, we obtain aA = d 2z + ! AR = !B R + ! AR . dt 2 Denote the tension in the tape as (what else) T . The net torque on the upper drum about its center is then ! B = TR , directed clockwise in the figure, and the net torque on the falling drum about its center is also ! A = TR , also directed clockwise. Thus, ! B = TR / I = 2T / MR , ! A = TR / I = 2T / MR . Where we have assumed that the moment of inertia of the drum and unwinding tape is I = (1/ 2) MR 2 . Newton’s Second Law, applied to the falling drum, with the positive direction downward, is Mg ! T = Ma A . We now have five equations, d 2z d 2z 2T 2T ! B R = 2 , a A = 2 + ! A R, ! B = , !A = , Mg " T = Ma A , dt dt MR MR in the five unknowns ! A , ! B , a A , d 2z and T . dt 2 It’s easy to see that ! A = !B . Therefore a A = ! B R + ! A R = 2! A R . The tension in the tape is then T= ! A MR a A MR Ma A = = 2 4R 4 Newton’s Second Law then becomes Mg ! Ma A = Ma A . 4 Therefore solving for the acceleration yields aA = 4 g 5 This result is certainly plausible. We expect a A < g , and we also expect that with both drums free to rotate, the acceleration will be almost but not quite g .