Question

A solid sphere of mass \( M \) and radius \( R \) is free to rotate about an axis passing through its centre. A solid cylinder of mass \( \frac{M}{2} \) and radius \( R \) is free to rotate about its standard axis of symmetry. Initially both the sphere and the cylinder are at rest. If torques of equal magnitude are applied to the sphere and the cylinder, the ratio of the angular speeds acquired by the sphere and the cylinder after a given time is

• A. 2:5
• B. 5:8
• C. 1:2
• D. 3:4

Hint:

To find the ratio of angular speeds, use \( \omega = \frac{\tau}{I} t \). The ratio depends inversely on the moments of inertia \( I \). Calculate \( I \) for each object using standard formulas.

Answer 1

The Correct Option is B

The torque \( \tau \) applied to a rotating object is related to its angular acceleration \( \alpha \) by: \[ \tau = I \alpha \quad \text{or} \quad \alpha = \frac{\tau}{I} \] where \( I \) is the moment of inertia. The angular speed \( \omega \) after time \( t \) is: \[ \omega = \alpha t = \left( \frac{\tau}{I} \right) t \] Since the torques are equal and applied for the same time \( t \), the ratio of angular speeds \( \omega_{\text{sphere}} : \omega_{\text{cylinder}} \) is: \[ \frac{\omega_{\text{sphere}}}{\omega_{\text{cylinder}}} = \frac{\alpha_{\text{sphere}}}{\alpha_{\text{cylinder}}} = \frac{\tau / I_{\text{sphere}}}{\tau / I_{\text{cylinder}}} = \frac{I_{\text{cylinder}}}{I_{\text{sphere}}} \] - Moment of inertia of a solid sphere about its central axis: \[ I_{\text{sphere}} = \frac{2}{5} M R^2 \] - Moment of inertia of a solid cylinder about its axis of symmetry: \[ I_{\text{cylinder}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2 \] Now, the ratio of the moments of inertia: \[ \frac{I_{\text{cylinder}}}{I_{\text{sphere}}} = \frac{\frac{1}{4} M R^2}{\frac{2}{5} M R^2} = \frac{1}{4} \times \frac{5}{2} = \frac{5}{8} \] Thus, the ratio of the angular speeds \( \omega_{\text{sphere}} : \omega_{\text{cylinder}} \) is: \[ \frac{\omega_{\text{sphere}}}{\omega_{\text{cylinder}}} = \frac{5}{8} \quad \text{or} \quad 5:8 \] So, the ratio of the angular speeds is 5:8.

Answer 2

Step 1: Given Data
- Mass of sphere, \( M \)
- Radius of sphere, \( R \)
- Mass of cylinder, \( \frac{M}{2} \)
- Radius of cylinder, \( R \)
- Equal torque \( \tau \) applied to both
- Both initially at rest

Step 2: Moments of inertia
- For solid sphere about its center:
\[ I_{\text{sphere}} = \frac{2}{5} M R^2 \]
- For solid cylinder about its axis:
\[ I_{\text{cyl}} = \frac{1}{2} \times \frac{M}{2} \times R^2 = \frac{M R^2}{4} \]

Step 3: Angular acceleration
Using \(\tau = I \alpha \Rightarrow \alpha = \frac{\tau}{I}\)
- Sphere:
\[ \alpha_{\text{sphere}} = \frac{\tau}{(2/5) M R^2} = \frac{5 \tau}{2 M R^2} \]
- Cylinder:
\[ \alpha_{\text{cyl}} = \frac{\tau}{(M R^2 / 4)} = \frac{4 \tau}{M R^2} \]

Step 4: Ratio of angular speeds after time \( t \)
Since \(\omega = \alpha t\), and \(t\) is same for both,
\[ \frac{\omega_{\text{sphere}}}{\omega_{\text{cyl}}} = \frac{\alpha_{\text{sphere}}}{\alpha_{\text{cyl}}} = \frac{\frac{5 \tau}{2 M R^2}}{\frac{4 \tau}{M R^2}} = \frac{5/2}{4} = \frac{5}{8} \]

Final Answer:
\[ \boxed{5 : 8} \]