A solid cylinder of mass M and radius R rolls from rest down a plane inclined at an angle `theta` to the horizontal. The velocity of the centre of mass of the cylinder after it has rolled down a distance d is :

Options

A. `sqrt(2/3 gd tan theta)`
B. `sqrt(gd tan theta)`
C. `sqrt(3/4 gd sin theta)`
D. `sqrt(4/3 gd sin theta)`

Answer

To find the velocity of the center of mass of a solid cylinder rolling down an inclined plane, we can use the principles of energy conservation and the relationship between translational and rotational motion.

### Step-by-Step Solution:

1. **Identify the Parameters:**
- Mass of the cylinder: \( M \)
- Radius of the cylinder: \( R \)
- Incline angle: \( \theta \)
- Distance rolled down the incline: \( d \)

2. **Determine the Height:**
- The height \( h \) through which the cylinder descends can be calculated using trigonometry:
\[ h = d \sin \theta \]

3. **Apply the Work-Energy Theorem:**
- The work done by gravity as the cylinder rolls down the incline is equal to the change in kinetic energy of the cylinder.
- The work done by gravity is:
\[ W = Mgh = Mg(d \sin \theta) \]
- The kinetic energy of the cylinder consists of translational and rotational components:
\[ KE = KE_{translational} + KE_{rotational} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 \]

4. **Moment of Inertia for a Solid Cylinder:**
- The moment of inertia \( I \) for a solid cylinder about its axis is:
\[ I = \frac{1}{2}MR^2 \]

5. **Relate Linear and Angular Velocity:**
- Since the cylinder rolls without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = R\omega \quad \Rightarrow \quad \omega = \frac{v}{R} \]

6. **Substitute \( \omega \) into the Kinetic Energy Equation:**
- Substitute \( \omega \) in the kinetic energy equation:
\[ KE = \frac{1}{2}Mv^2 + \frac{1}{2} \left(\frac{1}{2}MR^2\right) \left(\frac{v}{R}\right)^2 \]
- Simplifying gives:
\[ KE = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2 = \frac{3}{4}Mv^2 \]

7. **Set Up the Energy Conservation Equation:**
- Equating the work done by gravity to the kinetic energy:
\[ Mg(d \sin \theta) = \frac{3}{4}Mv^2 \]

8. **Solve for \( v^2 \):**
- Cancel \( M \) from both sides:
\[ g(d \sin \theta) = \frac{3}{4}v^2 \]
- Rearranging gives:
\[ v^2 = \frac{4}{3}g(d \sin \theta) \]

9. **Take the Square Root to Find \( v \):**
- Finally, taking the square root gives:
\[ v = \sqrt{\frac{4}{3}g(d \sin \theta)} \]

### Final Answer:
The velocity of the center of mass of the cylinder after rolling down a distance \( d \) is:
\[ v = \sqrt{\frac{4}{3}g(d \sin \theta)} \]