A uniform solid sphere of mass m and radius R is suspended in vertical plane from a point on its periphery. The time period of its oscillation is

Options

A. `2pi sqrt(r/g)`
B. `2pi sqrt((5 R)/(3 g))`
C. `2pi sqrt ((7 R)/(5 g))`
D. `2pi sqrt((2 R)/(5 g))`

Answer

To find the time period of oscillation of a uniform solid sphere of mass \( m \) and radius \( R \) suspended from a point on its periphery, we can follow these steps:

### Step 1: Understand the System
We have a uniform solid sphere suspended from a point on its surface. This setup acts like a physical pendulum. The time period of a physical pendulum is given by the formula:

\[ T = 2\pi \sqrt{\frac{I}{mgh}} \]

where:
- \( T \) is the time period,
- \( I \) is the moment of inertia about the pivot point,
- \( m \) is the mass of the sphere,
- \( g \) is the acceleration due to gravity,
- \( h \) is the distance from the pivot to the center of mass of the sphere.

### Step 2: Calculate the Moment of Inertia
The moment of inertia \( I \) of a solid sphere about its center of mass is given by:

\[ I_{cm} = \frac{2}{5} m R^2 \]

Since we need the moment of inertia about the point on its periphery, we can use the parallel axis theorem:

\[ I = I_{cm} + md^2 \]

where \( d \) is the distance from the center of mass to the pivot point. For a sphere, \( d = R \) (the radius of the sphere). Thus, we have:

\[ I = \frac{2}{5} m R^2 + m R^2 = \frac{2}{5} m R^2 + \frac{5}{5} m R^2 = \frac{7}{5} m R^2 \]

### Step 3: Determine the Distance \( h \)
The distance \( h \) from the pivot point to the center of mass of the sphere is equal to the radius \( R \):

\[ h = R \]

### Step 4: Substitute Values into the Time Period Formula
Now we can substitute \( I \) and \( h \) into the time period formula:

\[ T = 2\pi \sqrt{\frac{I}{mgh}} = 2\pi \sqrt{\frac{\frac{7}{5} m R^2}{mgR}} \]

### Step 5: Simplify the Expression
Now simplify the expression:

\[ T = 2\pi \sqrt{\frac{\frac{7}{5} R}{g}} = 2\pi \sqrt{\frac{7R}{5g}} \]

### Final Answer
Thus, the time period of oscillation of the uniform solid sphere is:

\[ T = 2\pi \sqrt{\frac{7R}{5g}} \]