A solid cylinder of mass M and radius R rolls without slipping on a flat horizontal surface. Its moment of inertia about the line of contact is ?

Options

A. `(MR^(2))/(2)`
B. `MR^(2)`
C. `3/2 MR^(2)`
D. `2MR^(2)`

Answer

To find the moment of inertia of a solid cylinder about the line of contact when it rolls without slipping on a flat horizontal surface, we can follow these steps:

### Step 1: Understand the Moment of Inertia of the Cylinder
The moment of inertia \( I \) of a solid cylinder about its center of mass (CM) is given by the formula:
\[ I_{CM} = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the cylinder and \( R \) is its radius.

### Step 2: Identify the Point of Contact
When the cylinder rolls without slipping, the point of contact with the ground (let's call it point P) is the point about which we want to find the moment of inertia.

### Step 3: Use the Parallel Axis Theorem
To find the moment of inertia about the point of contact (P), we can use the Parallel Axis Theorem. This theorem states that:
\[ I_P = I_{CM} + Md^2 \]
where \( d \) is the distance from the center of mass to the new axis (the line of contact in this case). For a cylinder, this distance \( d \) is equal to the radius \( R \).

### Step 4: Substitute the Values
Now we can substitute the values into the equation:
\[ I_P = I_{CM} + M R^2 \]
Substituting \( I_{CM} \):
\[ I_P = \frac{1}{2} M R^2 + M R^2 \]

### Step 5: Simplify the Equation
Combine the terms:
\[ I_P = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \]

### Final Answer
Thus, the moment of inertia of the solid cylinder about the line of contact is:
\[ I_P = \frac{3}{2} M R^2 \]