A block (mass = 2.6 kg) is hanging from a massless cord that is wrapped around a

Question

A block (mass = 2.6 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.2 x 10-3 kg·m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.038 m during the block's descent. Find (a) the angular acceleration of the pulley and (b) the tension in the cord.

Explanation / Answer

given,

mass = 2.6 kg

moment of inertia = 1.2 * 10^-3 kg.m^2

radius = 0.038 m

equation of motion of block

mg - T = ma

2.6 * 9.8 - T = 2.6 * a

equation of motion of pulley

T * R = I * a / R

T * 0.038 = 1.2 * 10^-3 * a / 0.038

on solving we'll get

acceleration a = 7.42635 m/s^2

angular acceleration = a / R

angular acceleration = 7.42635 / 0.038

angular acceleration = 195.43 rad/sec^2

tension T = 6.17148 N

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