A uniform circular disc of mass M and radius R is pivoted at distance x above the centre of mass of the disc, such that the time period of the disc in the vertical plane is infinite. What is the distance between the pivoted point and centre of mass of the disc ?

Answer

To solve the problem, we need to find the distance \( x \) between the pivot point and the center of mass of a uniform circular disc such that the time period of the disc in a vertical plane is infinite.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform circular disc of mass \( M \) and radius \( R \).
- The disc is pivoted at a point that is a distance \( x \) above its center of mass.

2. **Torque Calculation**:
- When the disc is displaced by an angle \( \theta \), the torque \( \tau \) about the pivot point due to the weight of the disc (acting downward) can be expressed as:
\[ \tau = M g x \sin(\theta) \]
- For small angles, \( \sin(\theta) \approx \theta \), so we can write:
\[ \tau \approx M g x \theta \]

3. **Moment of Inertia Calculation**:
- The moment of inertia \( I \) of the disc about the pivot point can be calculated using the parallel axis theorem:
\[ I = I_{cm} + M x^2 \]
- The moment of inertia about the center of mass \( I_{cm} \) for a disc is given by:
\[ I_{cm} = \frac{1}{2} M R^2 \]
- Therefore, the moment of inertia about the pivot point becomes:
\[ I = \frac{1}{2} M R^2 + M x^2 \]

4. **Finding the Time Period**:
- The time period \( T \) of oscillation can be expressed as:
\[ T = 2\pi \sqrt{\frac{I}{\tau}} \]
- Substituting the expressions for \( I \) and \( \tau \):
\[ T = 2\pi \sqrt{\frac{\frac{1}{2} M R^2 + M x^2}{M g x}} \]
- Simplifying this gives:
\[ T = 2\pi \sqrt{\frac{\frac{1}{2} R^2 + x^2}{g x}} \]

5. **Condition for Infinite Time Period**:
- The time period \( T \) approaches infinity when the denominator approaches zero. This occurs when \( x \) approaches zero:
\[ g x \to 0 \implies x \to 0 \]

6. **Conclusion**:
- Therefore, the distance \( x \) between the pivot point and the center of mass of the disc must be:
\[ x = 0 \]
- This means that the pivot point is at the center of mass of the disc.

### Final Answer:
The distance between the pivoted point and the center of mass of the disc is \( 0 \).