The M.I. of a hollow cylinder of mass M, radius R and length L about a tangential axis parallel to the length will be:

Options

A. `2MR^(2)`
B. `MR^(2)//2`
C. `M(R^(2)//4+L^(2)//12)`
D. `ML^(2)//12`

Answer

To find the moment of inertia (M.I.) of a hollow cylinder of mass \( M \), radius \( R \), and length \( L \) about a tangential axis parallel to its length, we can follow these steps:

### Step 1: Identify the Moment of Inertia about the Center of Mass
For a hollow cylinder, the moment of inertia about its central axis (which passes through the center of mass) is given by the formula:
\[ I_{cm} = MR^2 \]

### Step 2: Apply the Parallel Axis Theorem
The Parallel Axis Theorem states that if you know the moment of inertia of a body about an axis through its center of mass, you can find the moment of inertia about any parallel axis by adding the product of the mass of the body and the square of the distance between the two axes. The formula is:
\[ I = I_{cm} + Md^2 \]
where \( d \) is the distance between the center of mass axis and the new axis.

### Step 3: Determine the Distance \( d \)
In this case, the tangential axis is at a distance equal to the radius \( R \) from the center of mass axis. Therefore, we have:
\[ d = R \]

### Step 4: Substitute Values into the Parallel Axis Theorem
Now, substituting the values into the Parallel Axis Theorem:
\[ I = I_{cm} + Md^2 = MR^2 + M(R^2) \]

### Step 5: Simplify the Expression
Combining the terms gives us:
\[ I = MR^2 + MR^2 = 2MR^2 \]

### Final Answer
Thus, the moment of inertia of the hollow cylinder about the tangential axis parallel to its length is:
\[ I = 2MR^2 \]