A hollow cylinder and a solid cylinder are rolling without slipping down an inclined plane, then which of these reaches earlier ?

Options

A. Solid cylinder
B. Hollow cylinder
C. Both simultaneously
D. Can't say anything

Answer

To solve the problem of which cylinder reaches the bottom of the inclined plane first, we will analyze the motion of both the hollow cylinder and the solid cylinder as they roll down the incline.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Cylinders**:
- Both cylinders experience gravitational force (mg) acting downwards.
- The component of gravitational force acting down the incline is \( mg \sin \theta \).
- There is a frictional force acting up the incline, which prevents slipping.

2. **Set Up the Equations of Motion**:
- For both cylinders, we can write the equation of motion along the incline:
\[ mg \sin \theta - f = ma \]
where \( f \) is the frictional force and \( a \) is the linear acceleration of the center of mass.

3. **Torque Equation**:
- The frictional force also creates a torque about the center of mass:
\[ f \cdot r = I \alpha \]
- For rolling without slipping, the relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is given by:
\[ a = \alpha r \]

4. **Substituting for Angular Acceleration**:
- From the torque equation, we can express \( f \):
\[ f = \frac{I \alpha}{r} = \frac{I a}{r^2} \]

5. **Combine the Equations**:
- Substitute \( f \) back into the equation of motion:
\[ mg \sin \theta - \frac{I a}{r^2} = ma \]
- Rearranging gives:
\[ mg \sin \theta = ma + \frac{I a}{r^2} \]
- Factor out \( a \):
\[ mg \sin \theta = a \left( m + \frac{I}{r^2} \right) \]
- Thus, the acceleration \( a \) can be expressed as:
\[ a = \frac{mg \sin \theta}{m + \frac{I}{r^2}} \]

6. **Moment of Inertia for Each Cylinder**:
- For a hollow cylinder, \( I = mr^2 \).
- For a solid cylinder, \( I = \frac{1}{2}mr^2 \).

7. **Calculate the Accelerations**:
- For the hollow cylinder:
\[ a_{\text{hollow}} = \frac{mg \sin \theta}{m + \frac{mr^2}{r^2}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} \]
- For the solid cylinder:
\[ a_{\text{solid}} = \frac{mg \sin \theta}{m + \frac{1}{2}mr^2/r^2} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

8. **Compare Accelerations**:
- Since \( \frac{g \sin \theta}{2} < \frac{2g \sin \theta}{3} \), the solid cylinder has a greater acceleration than the hollow cylinder.

9. **Conclusion**:
- The solid cylinder will reach the bottom of the incline first because it has a greater acceleration compared to the hollow cylinder.

### Final Answer:
The solid cylinder reaches the bottom of the inclined plane earlier than the hollow cylinder.