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 Question:
A mass attached to a string falls under gravity, causing a solid disc (flywheel) to unspool and spin. We must find the angular velocity ($\omega$) of the disc after the mass has fallen a vertical distance $h$.

Step 2: Key Formula or Approach:
Use the Conservation of Mechanical Energy.
The loss in gravitational potential energy of the falling bob exactly equals the sum of the translational kinetic energy gained by the bob AND the rotational kinetic energy gained by the unwinding disc.
$$\Delta PE_{bob} = \Delta KE_{trans(bob)} + \Delta KE_{rot(disc)}$$
$$mgh = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$

Step 3: Detailed Explanation:
1. Identify the moment of inertia ($I$) for the solid disc:
$$I = \frac{1}{2} m_{disc} R^2$$
Since the mass of the disc is given as '$m$' (identical to the bob), $I = \frac{1}{2} m R^2$.
2. Relate linear velocity ($v$) of the falling bob to the angular velocity ($\omega$) of the edge of the disc:
The string does not slip, so $v = R\omega$.
3. Substitute these relations into the energy conservation equation:
$$mgh = \frac{1}{2} m (R\omega)^2 + \frac{1}{2} \left( \frac{1}{2} m R^2 \right) \omega^2$$
Expand the terms:
$$mgh = \frac{1}{2} m R^2 \omega^2 + \frac{1}{4} m R^2 \omega^2$$
The mass '$m$' perfectly cancels out from all terms!
$$gh = \frac{1}{2} R^2 \omega^2 + \frac{1}{4} R^2 \omega^2$$
Find a common denominator to add the fractions ($2/4 + 1/4$):
$$gh = \frac{3}{4} R^2 \omega^2$$
Rearrange to solve for $\omega^2$:
$$\omega^2 = \frac{4gh}{3R^2}$$
Take the square root of both sides:
$$\omega = \sqrt{\frac{4gh}{3R^2}} = \frac{2}{R} \sqrt{\frac{gh}{3}}$$

Step 4: Final Answer:
The angular speed is $\frac{2}{R} \sqrt{\frac{gh}{3}}$, matching option (a).