When a point mass slips down a smooth incline from top, it reaches the bottom with linear speed v. If same mass in the form of disc rolls down without slipping a rough incline of identical geometry through same distance, what will be its linear velocity at the bottom ?

Options

A. `vsqrt((2)/(3))`
B. `sqrt((v)/(3))`
C. v
D. `sqrt((2v)/(3))`

Answer

To solve the problem of finding the linear velocity of a disc rolling down a rough incline, we will use the principle of conservation of mechanical energy. We will compare the scenario of a point mass sliding down a smooth incline with that of a disc rolling down a rough incline.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A point mass slides down a smooth incline and reaches the bottom with a linear speed \( v \).
- A disc of the same mass rolls down a rough incline of identical geometry through the same distance.
- We need to find the linear velocity of the disc at the bottom.

2. **Energy Conservation for the Point Mass**:
- For the point mass, the initial potential energy (PE) at the top is converted entirely into kinetic energy (KE) at the bottom.
- Initial potential energy: \( PE_i = mgh \)
- Initial kinetic energy: \( KE_i = 0 \) (since it starts from rest)
- Final potential energy at the bottom: \( PE_f = 0 \)
- Final kinetic energy: \( KE_f = \frac{1}{2} mv^2 \)
- Applying conservation of energy:
\[ mgh = \frac{1}{2} mv^2 \]
- Canceling \( m \) from both sides:
\[ gh = \frac{1}{2} v^2 \]
- Rearranging gives:
\[ v^2 = 2gh \quad \Rightarrow \quad v = \sqrt{2gh} \]

3. **Energy Conservation for the Disc**:
- For the disc, the initial potential energy is the same:
\[ PE_i = mgh \]
- The disc rolls down, so it has both translational and rotational kinetic energy at the bottom:
- Final potential energy: \( PE_f = 0 \)
- Final kinetic energy:
\[ KE_f = \frac{1}{2} mv'^2 + \frac{1}{2} I \omega^2 \]
- For a disc, the moment of inertia \( I = \frac{1}{2} mr^2 \) and the relationship between linear velocity \( v' \) and angular velocity \( \omega \) is \( v' = r\omega \).
- Substituting \( \omega = \frac{v'}{r} \) into the kinetic energy expression:
\[ KE_f = \frac{1}{2} mv'^2 + \frac{1}{2} \left(\frac{1}{2} mr^2\right) \left(\frac{v'}{r}\right)^2 \]
- Simplifying gives:
\[ KE_f = \frac{1}{2} mv'^2 + \frac{1}{4} mv'^2 = \frac{3}{4} mv'^2 \]

4. **Applying Conservation of Energy for the Disc**:
- Setting initial potential energy equal to final kinetic energy:
\[ mgh = \frac{3}{4} mv'^2 \]
- Canceling \( m \) from both sides:
\[ gh = \frac{3}{4} v'^2 \]
- Rearranging gives:
\[ v'^2 = \frac{4}{3} gh \quad \Rightarrow \quad v' = \sqrt{\frac{4}{3} gh} \]

5. **Relating \( v' \) to \( v \)**:
- Since we already found \( v = \sqrt{2gh} \):
\[ v' = \sqrt{\frac{4}{3}} \cdot \sqrt{gh} = \sqrt{\frac{4}{3}} \cdot \sqrt{\frac{2}{2}} \cdot \sqrt{2gh} = \sqrt{\frac{8}{6}} \cdot v = v \cdot \sqrt{\frac{4}{3}} \]
- Therefore, the linear velocity of the disc at the bottom is:
\[ v' = v \cdot \sqrt{\frac{2}{3}} \]

### Final Answer:
The linear velocity of the disc at the bottom is:
\[ v' = v \cdot \sqrt{\frac{2}{3}} \]