A pulley of radius 2 m is rotated about its axis by a force =20t-5t^2 newton (where t is measured in seconds) applied tangentially. If the moment of…

Question

A pulley of radius 2 m is rotated about its axis by a force =20t-5t^2 newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m^2, the number of rotation made by the pulley before its direction of motion is reversed is

MARKS Content Team · Accepted Answer

Correct answer: (A) more than 3 but less than 6. We have, Torque =r×F=lα or 220t-5t^2=10α α=4t-t^2 But α=(dω)/dt, so (dω)/dt=4t-t^2 dω=4t-t^2dt Integrating ω=2t^2-t^33 ω will be zero at t=6 s So, ω=(dθ)/dt=2t^2-t^33 or dθ=2t^2-t^33dt Again, integrating both sides θ=(2t^3)/3-t^412 since. t=6 s so, θ=(2×6^3)/3-6^412=36 Number of turns, n=(θ)/(2π)=36/(2π)=5.73

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A pulley of radius 2 m is rotated about its axis by a force =20t-5t2 newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2, the number of rotation made by the pulley before its direction of motion is reversed is

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