The acceleration down the plane of spherical body of mass m radius R and moment of inertia I having inclination `theta` to the horizontal is

Options

A. `(g sin theta)/(1+1^2//R^2)`
B. `(g sin theta)/(1+I//R^2)`
C. `(g sin theta)/(1+I//MR^2)`
D. `(g sin theta)/( MR^2 +1)`

Answer

To solve the problem of finding the acceleration down the plane of a spherical body with mass \( m \), radius \( R \), and moment of inertia \( I \) inclined at an angle \( \theta \) to the horizontal, we can follow these steps:

### Step 1: Identify the Forces Acting on the Body
The forces acting on the spherical body include:
- The gravitational force \( mg \) acting downward.
- The component of gravitational force acting down the incline, which is \( mg \sin(\theta) \).
- The frictional force \( f_s \) acting up the incline.

### Step 2: Apply Newton's Second Law for Linear Motion
According to Newton's second law, the net force acting on the body along the incline can be expressed as:
\[ mg \sin(\theta) - f_s = m a \]
where \( a \) is the acceleration of the center of mass of the spherical body.

### Step 3: Apply the Torque Equation
Next, we consider the rotational motion about the center of mass. The torque \( \tau \) due to the frictional force is given by:
\[ \tau = f_s \cdot R \]
This torque is also related to the angular acceleration \( \alpha \) of the body:
\[ \tau = I \alpha \]
Since the linear acceleration \( a \) is related to angular acceleration by the equation \( a = R \alpha \), we can substitute \( \alpha \) as \( \frac{a}{R} \):
\[ f_s \cdot R = I \left(\frac{a}{R}\right) \]
From this, we can express the frictional force:
\[ f_s = \frac{I a}{R^2} \]

### Step 4: Substitute the Frictional Force Back into the Linear Motion Equation
Now, we substitute the expression for \( f_s \) back into the linear motion equation:
\[ mg \sin(\theta) - \frac{I a}{R^2} = m a \]

### Step 5: Rearrange the Equation to Solve for Acceleration \( a \)
Rearranging the equation gives:
\[ mg \sin(\theta) = m a + \frac{I a}{R^2} \]
Factoring out \( a \) from the right side:
\[ mg \sin(\theta) = a \left(m + \frac{I}{R^2}\right) \]
Now, we can solve for \( a \):
\[ a = \frac{mg \sin(\theta)}{m + \frac{I}{R^2}} \]

### Final Result
Thus, the acceleration down the plane of the spherical body is:
\[ a = \frac{g \sin(\theta)}{1 + \frac{I}{mR^2}} \]