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Q.

A cord is wound around the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be

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a

$\sqrt{\frac{2 m g h}{I + 2 m r^{2}}}$

b

$\sqrt{\frac{2 m g h}{I + m r^{2}}}$

c

$\sqrt{2 g h}$

d

$\sqrt{\frac{2 g h}{I + m r^{2}}}$

answer is B.

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Detailed Solution

Velocity, $v = r ω$

Decrease in gravitational potential energy = Increase in kinetic energy

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$∴ m g h = \frac{1}{2} l ω^{2} + \frac{1}{2} m v^{2}$

$= \frac{1}{2} I ω^{2} + \frac{1}{2} m (r ω)^{2}$

$∴$ Angular velocity, $ω = \sqrt{\frac{2 m g h}{I + m r^{2}}}$

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A cord is wound around the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be