A circular hoop of radius R is hung over a knife edge. The period of oscillation is equal to that of a simple pendulum of length

Options

A. R
B. 2R
C. 3R
D. `(3R)/(2)`

Explanation

To solve the problem, we need to find the length of a simple pendulum whose period of oscillation is equal to that of a circular hoop of radius \( R \) oscillating about a knife edge.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia**:
The moment of inertia \( I \) of the circular hoop about an axis through its center is given by:
\[ I_c = mR^2 \]
where \( m \) is the mass of the hoop and \( R \) is its radius.

2. **Apply the Parallel Axis Theorem**:
Since the hoop is oscillating about a knife edge (which is at a distance \( R \) from the center of mass), we can use the parallel axis theorem to find the moment of inertia about the knife edge:
\[ I = I_c + m d^2 \]
where \( d \) is the distance from the center of mass to the knife edge, which is equal to \( R \). Therefore:
\[ I = mR^2 + mR^2 = 2mR^2 \]

3. **Calculate the Period of Oscillation**:
The period of oscillation \( T \) for a physical pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{mgh}} \]
Here, \( h \) is the distance from the knife edge to the center of mass, which is \( R \). Substituting \( I \) and \( h \):
\[ T = 2\pi \sqrt{\frac{2mR^2}{mgR}} = 2\pi \sqrt{\frac{2R}{g}} \]

4. **Compare with the Simple Pendulum**:
The period of a simple pendulum of length \( L \) is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
Setting the two expressions for \( T \) equal:
\[ 2\pi \sqrt{\frac{2R}{g}} = 2\pi \sqrt{\frac{L}{g}} \]

5. **Solve for Length \( L \)**:
Squaring both sides and simplifying:
\[ \frac{2R}{g} = \frac{L}{g} \]
Thus, we find:
\[ L = 2R \]

### Final Answer:
The length of the simple pendulum whose period of oscillation is equal to that of the circular hoop is:
\[ L = 2R \]