A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be :

Options

A. `(1)/(2R)sqrt(gh)`
B. `(2)/(R )sqrt(gh)`
C. `(2)/(R )sqrt((gh)/(3))`
D. `(2)/(R )sqrt((gh)/(2))`

Answer

To find the angular velocity of a solid cylinder of mass \( M \) and radius \( R \) when it rolls down an inclined plane of height \( h \), we can use the principle of conservation of energy.

### Step-by-step Solution:

1. **Identify the Energy Changes**:
- When the cylinder rolls down the incline, it loses potential energy and gains kinetic energy. The potential energy lost is given by:
\[ \text{Potential Energy Lost} = Mgh \]

2. **Kinetic Energy Components**:
- The kinetic energy at the bottom consists of two parts: translational kinetic energy and rotational kinetic energy.
- The translational kinetic energy (\( KE_{\text{trans}} \)) is:
\[ KE_{\text{trans}} = \frac{1}{2} M v^2 \]
- The rotational kinetic energy (\( KE_{\text{rot}} \)) is:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
- For a solid cylinder, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} M R^2 \]
- The angular velocity \( \omega \) is related to the linear velocity \( v \) by:
\[ \omega = \frac{v}{R} \]

3. **Substituting for Rotational Kinetic Energy**:
- Substitute \( I \) and \( \omega \) into the rotational kinetic energy formula:
\[ KE_{\text{rot}} = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{4} M v^2 \]

4. **Total Kinetic Energy**:
- The total kinetic energy at the bottom of the incline is:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2 \]

5. **Apply Conservation of Energy**:
- Setting the potential energy lost equal to the total kinetic energy gained:
\[ Mgh = \frac{3}{4} M v^2 \]
- Cancel \( M \) from both sides:
\[ gh = \frac{3}{4} v^2 \]

6. **Solving for \( v^2 \)**:
- Rearranging gives:
\[ v^2 = \frac{4gh}{3} \]

7. **Finding Angular Velocity \( \omega \)**:
- Substitute \( v \) into the equation for \( \omega \):
\[ \omega = \frac{v}{R} = \frac{\sqrt{\frac{4gh}{3}}}{R} = \frac{2\sqrt{gh}}{R\sqrt{3}} \]

### Final Answer:
The angular velocity of the cylinder when it reaches the bottom of the plane is:
\[ \omega = \frac{2\sqrt{gh}}{R\sqrt{3}} \]