A solid cylinder rolls down from an inclined plane of height h. What is the velocity of the cylinder when it reaches at the bottom of the plane ?

Options

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

Answer

To find the velocity of a solid cylinder when it reaches the bottom of an inclined plane of height \( h \), we can use the principle of conservation of energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
When the solid cylinder rolls down the inclined plane, its potential energy at the top is converted into kinetic energy (both translational and rotational) at the bottom.

### Step 2: Write the Potential Energy at the Top
The potential energy (PE) 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 of the inclined plane.

### Step 3: Write the Total Kinetic Energy at the Bottom
When the cylinder reaches the bottom, its kinetic energy (KE) consists of translational kinetic energy and rotational kinetic energy. The total kinetic energy is given by:
\[ KE_{total} = KE_{translational} + KE_{rotational} \]
The translational kinetic energy is:
\[ KE_{translational} = \frac{1}{2} mv^2 \]
The rotational kinetic energy for a solid cylinder is:
\[ KE_{rotational} = \frac{1}{2} I \omega^2 \]
For a solid cylinder, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} m r^2 \]
where \( r \) is the radius of the cylinder. The angular velocity \( \omega \) can be related to the linear velocity \( v \) by:
\[ \omega = \frac{v}{r} \]
Thus, substituting \( \omega \) into the rotational kinetic energy formula gives:
\[ KE_{rotational} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{4} mv^2 \]

### Step 4: Combine the Kinetic Energies
Now, we can express the total kinetic energy as:
\[ KE_{total} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

### Step 5: Set Potential Energy Equal to Total Kinetic Energy
According to the conservation of energy:
\[ mgh = KE_{total} \]
Substituting in the expression for total kinetic energy:
\[ mgh = \frac{3}{4} mv^2 \]

### Step 6: Simplify and Solve for \( v \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{3}{4} v^2 \]
Now, solving for \( v^2 \):
\[ v^2 = \frac{4gh}{3} \]
Taking the square root gives:
\[ v = \sqrt{\frac{4gh}{3}} \]

### Final Answer
Thus, the velocity of the cylinder when it reaches the bottom of the inclined plane is:
\[ v = \sqrt{\frac{4gh}{3}} \]