A solid cylinder of mass `M` and radius `R` rolls without slipping down an inclined plane of length `L` and height `h`. What is the speed of its center of mass when the cylinder reaches its bottom

Options

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

Answer

To find the speed of the center of mass of a solid cylinder rolling down an inclined plane, 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
Initially, when the cylinder is at the top of the incline, it has potential energy and no kinetic energy. At the bottom of the incline, all potential energy is converted into kinetic energy.

**Hint:** Remember that potential energy (PE) is given by \( PE = mgh \), where \( h \) is the height of the incline.

### Step 2: Write the expression for potential energy at the top
The potential energy (PE) at the top of the incline is given by:
\[ PE = mgh \]

**Hint:** The height \( h \) can be related to the length \( L \) of the incline using trigonometry if needed, but in this case, we will use \( h \) directly.

### Step 3: Write the expression for kinetic energy at the bottom
At the bottom of the incline, the cylinder has both translational kinetic energy and rotational kinetic energy. The total kinetic energy (KE) is given by:
\[ KE = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

For a solid cylinder, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} m R^2 \]

And since the cylinder rolls without slipping, we have the relation:
\[ \omega = \frac{v}{R} \]

**Hint:** Substitute \( I \) and \( \omega \) into the kinetic energy equation.

### Step 4: Substitute the values into the kinetic energy equation
Substituting \( I \) and \( \omega \) into the kinetic energy equation gives:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m R^2\right) \left(\frac{v}{R}\right)^2 \]
\[ = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

**Hint:** Combine the terms carefully to ensure you account for both translational and rotational kinetic energy.

### Step 5: Apply conservation of energy
By the conservation of energy, we set the potential energy equal to the total kinetic energy:
\[ mgh = \frac{3}{4} mv^2 \]

**Hint:** You can cancel \( m \) from both sides since it is non-zero.

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

**Hint:** Take the square root to find \( v \).

### Step 7: Find \( v \)
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{4gh}{3}} \]

**Final Answer:** The speed of the center of mass when the cylinder reaches the bottom is:
\[ v = \frac{2\sqrt{gh}}{\sqrt{3}} \]