Moment of inertia of a hoop suspended from a peg about the peg is

Options

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

Explanation

To find the moment of inertia of a hoop suspended from a peg about the peg, we can follow these steps:

### Step 1: Understand the System
A hoop is a circular object with mass distributed uniformly along its circumference. When suspended from a peg, it can rotate about the peg.

### Step 2: Define the Parameters
Let:
- \( R \) = radius of the hoop
- \( m \) = mass of the hoop

### Step 3: Use the Formula for Moment of Inertia
The moment of inertia \( I \) of a hoop about its center is given by:
\[ I_{center} = mR^2 \]

### Step 4: Apply the Parallel Axis Theorem
Since the hoop is rotating about a peg (which is not at its center), we need to use the parallel axis theorem to find the moment of inertia about the peg. The parallel axis theorem states:
\[ I = I_{center} + md^2 \]
where \( d \) is the distance from the center of mass to the new axis (the peg).

### Step 5: Calculate the Distance \( d \)
In this case, the distance \( d \) from the center of the hoop to the peg is equal to the radius \( R \) of the hoop.

### Step 6: Substitute Values into the Formula
Now substituting the values into the parallel axis theorem:
\[ I = mR^2 + mR^2 = 2mR^2 \]

### Step 7: Conclusion
Thus, the moment of inertia of the hoop about the peg is:
\[ I = 2mR^2 \]

### Final Answer
The moment of inertia of a hoop suspended from a peg about the peg is \( 2mR^2 \).

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