A disc of radius `R` and mass `M` is pivoted at the rim and it set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

Options

A. `5/4R`
B. `2/3R`
C. `3/4R`
D. `3/2R`

Answer

To solve the problem, we need to find the length of a simple pendulum that has the same period as a disc of radius \( R \) and mass \( M \) pivoted at the rim. We will follow these steps:

### Step 1: Calculate the Moment of Inertia of the Disc
The moment of inertia \( I \) of a disc about its center is given by:
\[ I_c = \frac{1}{2} M R^2 \]
Since the disc is pivoted at the rim, we will use the parallel axis theorem to find the moment of inertia about the pivot point (the rim). The parallel axis theorem states:
\[ I = I_c + M d^2 \]
where \( d \) is the distance from the center of mass to the pivot point. For our disc, \( d = R \):
\[ I = \frac{1}{2} M R^2 + M R^2 = \frac{1}{2} M R^2 + \frac{2}{2} M R^2 = \frac{3}{2} M R^2 \]

### Step 2: Calculate the Time Period of the Disc
The time period \( T \) of a rigid body undergoing small oscillations is given by:
\[ T = 2\pi \sqrt{\frac{I}{M g d}} \]
Here, \( d \) is the distance from the pivot point to the center of mass, which is \( R \). Substituting \( I \) and \( d \):
\[ T = 2\pi \sqrt{\frac{\frac{3}{2} M R^2}{M g R}} \]
Now, simplify the equation:
\[ T = 2\pi \sqrt{\frac{3 R}{2g}} \]

### Step 3: Calculate the Time Period of the Simple Pendulum
The time period \( T_p \) of a simple pendulum of length \( l \) is given by:
\[ T_p = 2\pi \sqrt{\frac{l}{g}} \]

### Step 4: Set the Time Periods Equal
To find the length \( l \) of the simple pendulum that has the same period as the disc, we set the two time periods equal:
\[ 2\pi \sqrt{\frac{3 R}{2g}} = 2\pi \sqrt{\frac{l}{g}} \]
We can cancel \( 2\pi \) and \( g \) from both sides:
\[ \sqrt{\frac{3 R}{2}} = \sqrt{l} \]

### Step 5: Solve for \( l \)
Squaring both sides gives:
\[ \frac{3 R}{2} = l \]
Thus, the length of the simple pendulum is:
\[ l = \frac{3 R}{2} \]

### Final Answer
The length of the simple pendulum should be \( \frac{3R}{2} \).

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