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 after it has rolled down a distance \(d\) on an inclined plane at an angle \(\theta\), we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify Initial and Final Energy
Initially, the cylinder is at rest at a height \(h\) above the reference point. The initial energy is purely gravitational potential energy:
\[ E_{\text{initial}} = mgh \]
where \(h = d \sin \theta\).

### Step 2: Calculate Initial Potential Energy
Substituting for \(h\):
\[ E_{\text{initial}} = mg(d \sin \theta) \]

### Step 3: Determine Final Energy
When the cylinder rolls down the incline, it has both translational and rotational kinetic energy. The total final energy \(E_{\text{final}}\) is given by:
\[ E_{\text{final}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid cylinder, the moment of inertia \(I\) about its center is:
\[ I = \frac{1}{2} mR^2 \]
Also, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is:
\[ \omega = \frac{v}{R} \]
Thus, \(\omega^2 = \frac{v^2}{R^2}\).

### Step 4: Substitute for Rotational Kinetic Energy
Substituting \(I\) and \(\omega\) into the final energy equation:
\[ E_{\text{final}} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} mR^2\right) \left(\frac{v^2}{R^2}\right) \]
This simplifies to:
\[ E_{\text{final}} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 5: Apply Conservation of Energy
According to the conservation of energy:
\[ E_{\text{initial}} = E_{\text{final}} \]
Substituting the expressions for energy:
\[ mg(d \sin \theta) = \frac{3}{4} mv^2 \]

### Step 6: Solve for Velocity \(v\)
Cancel \(m\) from both sides:
\[ g(d \sin \theta) = \frac{3}{4} v^2 \]
Now, rearranging gives:
\[ v^2 = \frac{4}{3} g(d \sin \theta) \]
Taking the square root:
\[ v = \sqrt{\frac{4}{3} g(d \sin \theta)} = \frac{2}{\sqrt{3}} \sqrt{g d \sin \theta} \]

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