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)/(4)` R
B. `(2)/(3)` R
C. `(3)/(4)` R
D. `(3)/(2)` R

Explanation

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 its rim. We will follow these steps:

### Step 1: Understand the Time Period of the Disc
The time period \( T \) of a physical pendulum (in this case, the disc) is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgh}} \]
where:
- \( I \) is the moment of inertia about the pivot,
- \( m \) is the mass of the disc,
- \( g \) is the acceleration due to gravity,
- \( h \) is the distance from the pivot to the center of mass.

### Step 2: Calculate the Moment of Inertia
For a disc of mass \( M \) and radius \( R \), the moment of inertia about its center is:
\[ I_{center} = \frac{1}{2} MR^2 \]
Using the parallel axis theorem to find the moment of inertia about the pivot (which is at the rim), we add \( Mh^2 \) where \( h = R \):
\[ I = I_{center} + MR^2 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2 \]

### Step 3: Determine the Distance to the Center of Mass
The distance \( h \) from the pivot to the center of mass of the disc is equal to the radius \( R \):
\[ h = R \]

### Step 4: Substitute Values into the Time Period Formula
Substituting \( I \) and \( h \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{\frac{3}{2} MR^2}{MgR}} = 2\pi \sqrt{\frac{3R}{2g}} \]

### Step 5: Time Period of the Simple Pendulum
The time period \( T' \) of a simple pendulum of length \( L \) is given by:
\[ T' = 2\pi \sqrt{\frac{L}{g}} \]

### Step 6: Set the Time Periods Equal
To find the length \( L \) of the simple pendulum that has the same period as the disc, we set \( T = T' \):
\[ 2\pi \sqrt{\frac{3R}{2g}} = 2\pi \sqrt{\frac{L}{g}} \]

### Step 7: Simplify the Equation
Dividing both sides by \( 2\pi \) and squaring both sides gives:
\[ \frac{3R}{2g} = \frac{L}{g} \]
Canceling \( g \) from both sides:
\[ \frac{3R}{2} = L \]

### Step 8: Conclusion
Thus, the length of the simple pendulum should be:
\[ L = \frac{3R}{2} \]

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

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