A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in…

Question

A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be 1/2√xgh m s^(-1). The value of x (rounded off to the nearest integer) is _____.

MARKS Content Team · Accepted Answer

Correct answer: 3. If we take the wheel as a ring then its moment of inertia will be, I=Mr^2. Using mechanical energy conservation, Mgh-mgh=1/2mv^2+1/2Mv^2+1/2Iω^2⇒12gh-3gh=1/23v^2+1/212v^2+1/212r^2v^2r^2 ⇒9gh=(15v^2+12v^2)/2 ⇒v=√23gh=1/2√83gh Therefore, x=8/3≃3

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