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 principles of energy conservation. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
When the cylinder rolls down the incline, its potential energy (PE) at the top is converted into kinetic energy (KE) at the bottom. The total mechanical energy is conserved.

### Step 2: Write the Potential Energy
The potential energy of the cylinder 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.

### Step 3: Write the Kinetic Energy
The kinetic energy of a rolling object consists of two parts: translational kinetic energy and rotational kinetic energy. For a solid cylinder, the total kinetic energy (KE) is:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

### Step 4: Substitute the Moment of Inertia
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 relation:
\[ v = r\omega \implies \omega = \frac{v}{r} \]
Now substituting \( \omega \) in the kinetic energy equation:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 \]
This simplifies to:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 5: Set Potential Energy Equal to Kinetic Energy
According to the conservation of energy:
\[ mgh = \frac{3}{4} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{3}{4} v^2 \]

### Step 6: Solve for \( v^2 \)
Rearranging the equation gives:
\[ v^2 = \frac{4gh}{3} \]

### Step 7: Take the Square Root
Taking the square root of both sides to find \( v \):
\[ v = \sqrt{\frac{4gh}{3}} \]

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