The speed of a uniform solid cylinder after rolling down an inclined plane of vertical height H, from rest without sliding is :-

Options

A. `sqrt(gH)/(3)`
B. `sqrt((2gH)/(3))`
C. `sqrt(gH)`
D. `sqrt((4gH)/(3))`

Answer

To find the speed of a uniform solid cylinder after rolling down an inclined plane of vertical height \( H \) from rest without sliding, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Identify Initial and Final Energies
- **Initial Energy**: At the top of the incline, the cylinder has potential energy and no kinetic energy since it starts from rest.
\[ \text{Initial Potential Energy} = mgh \]
\[ \text{Initial Kinetic Energy} = 0 \]

- **Final Energy**: At the bottom of the incline, the potential energy is zero, and the cylinder has both translational and rotational kinetic energy.
\[ \text{Final Potential Energy} = 0 \]
\[ \text{Final Kinetic Energy} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

### Step 2: Write the Moment of Inertia and Angular Velocity
- For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]
- Since the cylinder rolls without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]

### Step 3: Substitute \( \omega \) in Kinetic Energy Equation
- Substitute \( \omega \) into the kinetic energy equation:
\[ \text{Final Kinetic Energy} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
- Simplifying gives:
\[ \text{Final Kinetic Energy} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 4: Apply Conservation of Energy
- According to the conservation of energy:
\[ \text{Initial Potential Energy} = \text{Final Kinetic Energy} \]
\[ mgh = \frac{3}{4} mv^2 \]
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{3}{4} v^2 \]

### Step 5: Solve for \( v \)
- Rearranging gives:
\[ v^2 = \frac{4}{3} gh \]
- Taking the square root:
\[ v = \sqrt{\frac{4}{3} gh} = \frac{2}{\sqrt{3}} \sqrt{gh} \]

### Final Answer
Thus, the speed of the uniform solid cylinder after rolling down the inclined plane is:
\[ v = \sqrt{\frac{4}{3} gh} \]