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ps11sol

# ps11sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 2010 Problem Set 11: Two Dimensional Rotational and Translational Motion Solution Problem 1: 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 the length of the tape that is unwound from the upper drum. Then, 2 2 B d z R dt ! = . The linear velocity A v may then be expressed as the sum of two terms, the rate dz dt at which 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 2 2 A A B A d z a R R R dt ! ! ! = + = + . 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

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2 falling drum about its center is also A TR ! = , also directed clockwise. Thus, / 2 / B TR I T MR ! = = , / 2 / A TR I T MR ! = = . Where we have assumed that the moment of inertia of the drum and unwinding tape is 2 (1/ 2) I MR = . Newton’s Second Law, applied to the falling drum, with the positive direction downward, is A Mg T Ma ! = . We now have five equations, 2 2 2 2 2 2 , , , , , B A A B A A d z d z T T R a R Mg T Ma dt dt MR MR ! ! ! ! = = + = = " = in the five unknowns A ! , B ! , A a , 2 2 d z dt and T . It’s easy to see that A B ! ! = . Therefore 2 A B A A a R R R ! ! ! = + = . The tension in the tape is then 2 4 4 A A A MR a MR Ma T R ! = = = Newton’s Second Law then becomes 4 A A Ma Mg Ma ! = . Therefore solving for the acceleration yields a A = 4 5 g 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 .