A round unifrom body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle theta with the horizontal. Then its acceleration

Options

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

Explanation

To find the acceleration of a round uniform body of radius \( R \), mass \( M \), and moment of inertia \( I \) rolling down an inclined plane at an angle \( \theta \), we can follow these steps:

### Step 1: Identify the forces acting on the body
The forces acting on the body include:
- The gravitational force \( Mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( f \) acting up the incline.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- The component parallel to the incline: \( F_{\parallel} = Mg \sin \theta \)
- The component perpendicular to the incline: \( F_{\perpendicular} = Mg \cos \theta \)

### Step 3: Apply Newton's second law in the direction of motion
Using Newton's second law in the direction parallel to the incline, we have:
\[ Mg \sin \theta - f = Ma \]
where \( a \) is the linear acceleration of the body.

### Step 4: Write the torque equation
The frictional force also causes a torque about the center of mass of the body. The torque \( \tau \) is given by:
\[ \tau = f \cdot R = I \alpha \]
where \( \alpha \) is the angular acceleration. For rolling without slipping, we have the relationship:
\[ a = \alpha R \quad \Rightarrow \quad \alpha = \frac{a}{R} \]

### Step 5: Substitute for torque
Substituting \( \alpha \) into the torque equation gives:
\[ f \cdot R = I \left(\frac{a}{R}\right) \]
Rearranging this, we find:
\[ f = \frac{I a}{R^2} \]

### Step 6: Substitute friction into the force equation
Now, substitute the expression for \( f \) back into the force equation:
\[ Mg \sin \theta - \frac{I a}{R^2} = Ma \]

### Step 7: Rearrange to solve for acceleration \( a \)
Rearranging the equation gives:
\[ Mg \sin \theta = Ma + \frac{I a}{R^2} \]
Factoring out \( a \) from the right side:
\[ Mg \sin \theta = a \left(M + \frac{I}{R^2}\right) \]
Now, solving for \( a \):
\[ a = \frac{Mg \sin \theta}{M + \frac{I}{R^2}} \]

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