JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Work, Power & Energy

A cord is wound round the circumference of wheel of radius

r

. The axis of the wheel is horizontal and the moment of inertia about it is I . A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance ' h ', the square of angular velocity of wheel will be :

(A) $\frac{2 m g h}{I + m r ^{2}}$ ✓ Correct
(B)2 gh
(C) $\frac{2 g h}{I + m r ^{2}}$
(D) $\frac{2 m g h}{I + 2 m r ^{2}}$

Answer: (A)

Solution

The cord does not slip, so the speed of the falling weight and the angular speed of the wheel are tied by

v = ω r

. The weight starts from rest and falls through

h

, and the only energy transfer is from gravity to the kinetic energy of the weight and of the wheel:

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} (m r^{2} + I) ω^{2}

Rearranging gives

ω^{2} = \frac{2 m g h}{I + m r ^{2}}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Work, Power & Energy

Solution

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