A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from res

Question

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it 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

TestHub · Accepted Answer

By conservation of mechanical energy, the decrease in potential energy of the falling weight equals the total increase in kinetic energy of the system.

Loss in potential energy of the weight = $m g h$

Gain in kinetic energy of the system = Kinetic energy of the weight + Rotational kinetic energy of the wheel

$K = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$

Since the cord does not slip, the linear velocity of the weight is related to the angular velocity of the wheel by $v = ω r$ .

Substituting $v = ω r$ into the kinetic energy equation:

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

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

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

$ω = [\frac{2 m g h}{I + m r ^{2}}]^{\frac{1}{2}}$

Answer: $[\frac{2 m g h}{I + m r ^{2}}]^{\frac{1}{2}}$

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