A drum has a radius of 0.40 m and a moment of inertiaof 1.7 kg * m^2. The fricti

Question

A drum has a radius of 0.40 m and a moment of inertiaof 1.7 kg * m^2. The frictional torque of the drum bearings is 3.0 N * m. A ring at one end of a rope is slipped on a short peg on a rim of the drum, and a 16 m length rope is wound upon it. The drum is initially at rest. A constant force is applied to the free end of the rope until the rope is completely unwound and slips off the peg. At that instant, the angular velocity of the drum is 14 rad/s. The drum then decelerates and coes to halt. In this situation, the total amount of negative work done by friction is closest to:

A) 187 J B) 287 J C) 237 J D) 453 J E) 120 J

Explanation / Answer

The radius of the drum is r = 0.40 m The moment of inertia of the drum is I = 1.7 kg * m^2 The frictional torque of the drum bearings is t = 3.0 N * m The length of the rope wound on the rim of the drum is l = 16 m The drum is initially at rest that is u = 0 The angular velocity of the drum is wo = 14 rad/s We know from the relation t = I * a or a = (t/I) where a is angular acceleration of the drum The angular displacement of the drum is l = r * ? or ? = (l/r) Let t be the time taken by the drum to come at rest We know that ? = wot + (1/2)at^2 Solving the above equation for t we get the time The angular velocity of the drum just before it comes to rest is w = (?/t) The total amount of negative work done by friction is W = (1/2)Iw2

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