A solid cylinder of mass `M` and radius `R` rolls down an inclined plane of height `h` without slipping. The speed of its centre when it reaches the bottom is.

Options

A. `sqrt(2gh)`
B. `sqrt((4gh)/3)`
C. `sqrt((3gh)/4)`
D. `sqrt((4g)/h)`

Answer

To solve the problem of a solid cylinder rolling down an inclined plane without slipping, we can use the principles of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the Initial and Final States
- **Initial State**: The cylinder is at rest at a height \( h \) on the inclined plane. Its potential energy is maximum, and kinetic energy is zero.
- **Final State**: The cylinder reaches the bottom of the incline. Its potential energy is zero, and it has both translational and rotational kinetic energy.

### Step 2: Write the Expression for Potential Energy
The potential energy (PE) at the height \( h \) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the cylinder, \( g \) is the acceleration due to gravity, and \( h \) is the height of the incline.

### Step 3: Write the Expression for Kinetic Energy
The total kinetic energy (KE) at the bottom consists of translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{\text{translational}} + KE_{\text{rotational}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m R^2 \]
where \( R \) is the radius of the cylinder.

### Step 4: Relate Angular Velocity and Linear Velocity
Since the cylinder rolls without slipping, we have the relationship:
\[ v = R \omega \quad \Rightarrow \quad \omega = \frac{v}{R} \]

### Step 5: Substitute \( \omega \) in the Kinetic Energy Expression
Substituting \( \omega \) into the kinetic energy equation gives:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m R^2\right) \left(\frac{v}{R}\right)^2 \]
Simplifying this, we get:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 6: Apply Conservation of Energy
According to the conservation of energy, the potential energy at the top is equal to the kinetic energy at the bottom:
\[ mgh = \frac{3}{4} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{3}{4} v^2 \]

### Step 7: Solve for \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{4}{3} gh \]
Taking the square root gives:
\[ v = \sqrt{\frac{4}{3} gh} \]

### Final Result
Thus, the speed of the center of the cylinder when it reaches the bottom of the incline is:
\[ v = \sqrt{\frac{4}{3} gh} \]