A sphere and a hollow cylinder roll without slipping down two separate inclined planes and travel the same distance in the same time. If the angle of the plane down which the sphere rolls is `30^(@)` , the angle of the other pane is

Options

A. `60^@`
B. `53^@`
C. `37^@`
D. `45^@`

Answer

To solve the problem, we need to analyze the motion of both the sphere and the hollow cylinder as they roll down their respective inclined planes. We will use the equations of motion and the properties of rolling objects.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The angle of the incline for the sphere, \( \theta_1 = 30^\circ \).
- Both objects (sphere and hollow cylinder) travel the same distance in the same time.

2. **Understand the Acceleration of Rolling Objects:**
- The acceleration \( a \) of a rolling object on an incline can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
- Here, \( g \) is the acceleration due to gravity, \( \theta \) is the angle of the incline, \( I \) is the moment of inertia, \( m \) is the mass, and \( r \) is the radius of the object.

3. **Determine the Moment of Inertia:**
- For the sphere:
\[ I_{\text{sphere}} = \frac{2}{5} m r^2 \implies \frac{I}{m r^2} = \frac{2}{5} \]
- For the hollow cylinder:
\[ I_{\text{cylinder}} = \frac{1}{2} m r^2 \implies \frac{I}{m r^2} = \frac{1}{2} \]

4. **Set Up the Acceleration Equations:**
- For the sphere:
\[ a_1 = \frac{g \sin \theta_1}{1 + \frac{2}{5}} = \frac{g \sin 30^\circ}{1 + \frac{2}{5}} = \frac{g \cdot \frac{1}{2}}{\frac{7}{5}} = \frac{5g}{14} \]
- For the hollow cylinder (with unknown angle \( \theta_2 \)):
\[ a_2 = \frac{g \sin \theta_2}{1 + \frac{1}{2}} = \frac{g \sin \theta_2}{\frac{3}{2}} = \frac{2g \sin \theta_2}{3} \]

5. **Equate the Accelerations:**
- Since both objects travel the same distance in the same time, their accelerations must be equal:
\[ a_1 = a_2 \implies \frac{5g}{14} = \frac{2g \sin \theta_2}{3} \]

6. **Cancel \( g \) and Solve for \( \sin \theta_2 \):**
- Cancel \( g \) from both sides:
\[ \frac{5}{14} = \frac{2 \sin \theta_2}{3} \]
- Cross-multiply to solve for \( \sin \theta_2 \):
\[ 5 \cdot 3 = 2 \cdot 14 \sin \theta_2 \implies 15 = 28 \sin \theta_2 \implies \sin \theta_2 = \frac{15}{28} \]

7. **Calculate \( \theta_2 \):**
- Use the inverse sine function to find \( \theta_2 \):
\[ \theta_2 = \sin^{-1}\left(\frac{15}{28}\right) \]
- Approximating \( \sin^{-1}(0.5357) \) gives approximately \( 32.5^\circ \).

8. **Final Result:**
- The angle of the other plane down which the hollow cylinder rolls is approximately \( 32.5^\circ \).