A hollow cylinder of mass M and radius R rolls down an inclined plane. A block of mass M slides down an identical inclined plane. If both objects are released at the same time, then

Options

A. the cylinder reaches the bottom first.
B. ) the block reaches the bottom first.
C. the block reaches the bottom with the greater kinetic energy.
D. the cylinder reaches the bottom with the greater kinetic energy

Answer

To solve the problem, we need to analyze the motion of both the hollow cylinder and the block as they roll and slide down the inclined plane. We will determine which one reaches the bottom first and compare their kinetic energies at the bottom.

### Step-by-Step Solution:

1. **Identify the Objects and Their Motion**:
- We have a hollow cylinder of mass \( M \) and radius \( R \) that rolls down the inclined plane.
- We have a block of mass \( M \) that slides down an identical inclined plane.

2. **Understand the Forces Acting on Each Object**:
- The hollow cylinder experiences both translational and rotational motion.
- The block only experiences translational motion.

3. **Determine the Time Taken for Each Object to Reach the Bottom**:
- For the hollow cylinder, the time \( t \) to reach the bottom can be derived from the equation of motion considering both translational and rotational dynamics:
\[ t = \sqrt{\frac{2h}{g} \cdot \frac{1}{\sin \theta} \cdot \sqrt{1 + \frac{I}{MR^2}}} \]
where \( I \) is the moment of inertia of the hollow cylinder, which is \( I = MR^2 \). Thus, \( \frac{I}{MR^2} = 1 \), leading to:
\[ t = \sqrt{\frac{2h}{g} \cdot \frac{2}{\sin \theta}} = \sqrt{\frac{4h}{g \sin^2 \theta}} \]

- For the block, the time \( t' \) to reach the bottom can be derived from the motion equations:
\[ t' = \sqrt{\frac{2h}{g \sin \theta}} \]

4. **Compare the Times**:
- We can see that:
\[ t = \sqrt{\frac{4h}{g \sin^2 \theta}} \quad \text{and} \quad t' = \sqrt{\frac{2h}{g \sin \theta}} \]
- Since \( t > t' \), this means the block reaches the bottom first.

5. **Calculate the Kinetic Energies at the Bottom**:
- For the block, using conservation of energy:
\[ KE_{\text{block}} = \frac{1}{2} M v^2 = Mgh \]
where \( v = \sqrt{2gh} \).

- For the hollow cylinder, the kinetic energy is:
\[ KE_{\text{cylinder}} = \frac{1}{2} M v^2 = \frac{1}{2} M \left(\sqrt{\frac{2gh}{1 + \frac{I}{MR^2}}}\right)^2 \]
Since \( \frac{I}{MR^2} = 1 \):
\[ KE_{\text{cylinder}} = \frac{1}{2} M \left(\frac{2gh}{2}\right) = \frac{Mgh}{2} \]

6. **Compare the Kinetic Energies**:
- The kinetic energy of the block is \( Mgh \) and the kinetic energy of the hollow cylinder is \( \frac{Mgh}{2} \).
- Therefore, the block has greater kinetic energy when it reaches the bottom.

### Conclusion:
- The block reaches the bottom first (Option 2).
- The block also has greater kinetic energy than the hollow cylinder when they reach the bottom (Option 3).