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Question

# A massless cord is wound round the circumference of the wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about the centre is I. A weight mg is attached to the cord at the end. The weight falls from rest and performs pure rolling. After falling through distance h, angular velocity of wheel is

A
2ghr
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B
mghI+2mr2
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C
2mghI+mr2
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D
2mghI+2mr2
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Solution

## The correct option is C √2mghI+mr2 Loss of potential energy= gain in kinetic energy Loss of potential energy=mgh Gain of kinetic energy = translational kinetic energy + rotational kinetic energy Gain in Kinetic energy = 12mv2+12Iω2 ∴ mgh=12mv2+12Iω2 on solving above equation, we get ω2[mr2+I2]=mgh [2mghmr2+I]1/2=ω

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