A bob of mass ' m ' is tied by a massless string whose other end is wound on a flywheel (disc) of radius ' R ' and mass ' m '. When released from the rest, the bob starts

Question

A bob of mass ' m ' is tied by a massless string whose other end is wound on a flywheel (disc) of radius ' R ' and mass ' m '. When released from the rest, the bob starts falling vertically downwards. If the bob has covered a vertical distance ' h ', then angular speed of wheel will be (There is no slipping between string and wheel, g - acceleration due to gravity)

TestHub · Accepted Answer

By conservation of energy, the loss in potential energy of the bob is equal to the gain in translational kinetic energy of the bob and rotational kinetic energy of the flywheel.

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

For a disc, the moment of inertia is $I = \frac{1}{2} m R^{2}$ .

Since there is no slipping, the linear velocity of the bob is $v = ω R$ .

Substituting $I$ and $v$ into the energy equation:

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

$m g h = \frac{1}{2} m ω^{2} R^{2} + \frac{1}{4} m ω^{2} R^{2}$

$m g h = \frac{3}{4} m ω^{2} R^{2}$

$ω^{2} = \frac{4 g h}{3 R ^{2}}$

$ω = \frac{2}{R} \frac{g h}{3}$

Answer: $\frac{2}{R} \frac{gh}{3}$

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