Question

A bob of mass 'm' is tied by a string wound on a flywheel (disc) of radius 'R' and mass 'm'. If the bob has covered a vertical distance 'h', then the angular speed of the wheel will be \dots

• A. $\frac{2}{R} \sqrt{\frac{gh}{3}}$
• B. $\frac{1}{R} \sqrt{\frac{2gh}{3}}$
• C. $R \sqrt{\frac{2gh}{3}}$
• D. $2R \sqrt{\frac{gh}{3}}$

Hint:

Always double-check the stated geometry of the spinning object. If it were a ring/hoop instead of a disc, $I$ would simply be $mR^2$, leading to $mgh = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$, making the final answer completely different!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the Law of Conservation of Energy. Potential energy lost by the bob = Kinetic energy gained by the bob + Rotational kinetic energy gained by the flywheel.

Step 2: Formula Application:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For a disc, $I = \frac{1}{2}mR^2$. Since there is no slipping, $v = R\omega$.

Step 3: Explanation:
$mgh = \frac{1}{2}m(R\omega)^2 + \frac{1}{2}(\frac{1}{2}mR^2)\omega^2$ $mgh = \frac{1}{2}mR^2\omega^2 + \frac{1}{4}mR^2\omega^2 = \frac{3}{4}mR^2\omega^2$ $gh = \frac{3}{4}R^2\omega^2 \implies \omega^2 = \frac{4gh}{3R^2}$ $\omega = \sqrt{\frac{4gh}{3R^2}} = \frac{2}{R}\sqrt{\frac{gh}{3}}$.

Step 4: Final Answer:
The angular speed is $\frac{2}{R}\sqrt{\frac{gh}{3}}$.