A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period.

Explanation

To find the time period of a spherical ball of mass \( m \) and radius \( r \) rolling without slipping on a rough concave surface of large radius \( R \), we can follow these steps:

### Step 1: Understand the System
The spherical ball is rolling without slipping on a concave surface. The motion of the ball can be approximated as simple harmonic motion (SHM) when it makes small oscillations about the lowest point.

### Step 2: Identify Relevant Parameters
- Mass of the ball: \( m \)
- Radius of the ball: \( r \)
- Radius of the concave surface: \( R \)
- Angular displacement: \( \theta \)
- Gravitational acceleration: \( g \)

### Step 3: Relate Angular Displacement and Acceleration
For small oscillations, we can approximate the motion as that of a body on an inclined plane. The acceleration of the center of mass of the rolling ball can be expressed as:
\[ a_{cm} = \frac{g \sin \theta}{1 + \frac{r^2}{k^2}} \]
where \( k \) is the radius of gyration.

### Step 4: Moment of Inertia and Radius of Gyration
For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
The radius of gyration \( k \) is related to the moment of inertia by:
\[ k^2 = \frac{I}{m} = \frac{2}{5} r^2 \]

### Step 5: Substitute for Radius of Gyration
Substituting \( k^2 \) into the acceleration equation gives:
\[ a_{cm} = \frac{g \sin \theta}{1 + \frac{r^2}{\frac{2}{5} r^2}} = \frac{g \sin \theta}{1 + \frac{5}{2}} = \frac{5g \sin \theta}{7} \]

### Step 6: Relate Linear and Angular Acceleration
The linear acceleration \( a_{cm} \) is related to the angular acceleration \( \alpha \) by:
\[ a_{cm} = \alpha (R - r) \]
Thus, we have:
\[ \alpha = \frac{a_{cm}}{R - r} = \frac{5g \sin \theta}{7(R - r)} \]

### Step 7: Approximate for Small Angles
For small angles, \( \sin \theta \approx \theta \). Therefore:
\[ \alpha = \frac{5g \theta}{7(R - r)} \]

### Step 8: Use the Formula for Time Period
The time period \( T \) for small oscillations is given by:
\[ T = 2\pi \sqrt{\frac{\theta}{\alpha}} \]
Substituting for \( \alpha \):
\[ T = 2\pi \sqrt{\frac{7(R - r)}{5g}} \]

### Final Answer
Thus, the time period of the oscillations of the spherical ball is:
\[ T = 2\pi \sqrt{\frac{7(R - r)}{5g}} \]

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