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

Options

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

Explanation

To find the speed of the center of mass of a solid cylinder rolling down an inclined plane without slipping, we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Identify the initial and final energy states
- **Initial State**: At the top of the inclined plane, the cylinder has potential energy and no kinetic energy since it starts from rest.
- **Final State**: At the bottom of the inclined plane, the cylinder has both translational kinetic energy and rotational kinetic energy.

### Step 2: Write the expression for potential energy at the top
The potential energy (PE) at the top of the incline can be expressed as:
\[ PE = mgh \]
where \( m \) is the mass of the cylinder, \( g \) is the acceleration due to gravity, and \( h \) is the height of the inclined plane.

### Step 3: Write the expression for kinetic energy at the bottom
At the bottom of the incline, the total kinetic energy (KE) consists of translational kinetic energy and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]
Since the cylinder rolls without slipping, we have the relationship:
\[ v = r \omega \quad \Rightarrow \quad \omega = \frac{v}{r} \]
Substituting \( \omega \) into the kinetic energy expression:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 4: Apply the conservation of energy principle
According to the conservation of energy, the potential energy at the top equals the total 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 5: Solve for \( v \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{4gh}{3} \]
Taking the square root gives:
\[ v = \sqrt{\frac{4gh}{3}} \]

### Final Result
Thus, the speed of the center of mass of the solid cylinder when it reaches the bottom of the inclined plane is:
\[ v = \frac{2}{\sqrt{3}} \sqrt{gh} \]