A sphere and a disc of same radii and mass are rolling on an inclined plane without slipping. `a_(s)` & `a_(d)` are acceleration and g is acceleration due to gravity. Then which statement is correct?

Options

A. `a_(s) gt a_(d) gt g`
B. `g gt a_(s) gt a_(d)`
C. `a_(s) gt g gt a_(d)`
D. `a_(d) gt a_(s) gt g `

Answer

To solve the problem of comparing the accelerations of a sphere and a disc rolling down an inclined plane without slipping, we can follow these steps:

### Step 1: Understand the Moment of Inertia
- The moment of inertia (I) is a measure of how much an object resists rotational motion. For a solid sphere and a solid disc of the same mass (m) and radius (r), the moments of inertia are:
- For a sphere: \( I_s = \frac{2}{5} m r^2 \)
- For a disc: \( I_d = \frac{1}{2} m r^2 \)

### Step 2: Write the Equations of Motion
- When rolling down an inclined plane, the forces acting on the objects can be analyzed. The gravitational force component acting down the incline is \( mg \sin(\theta) \), where \( \theta \) is the angle of inclination.
- The net torque (\( \tau \)) about the center of mass due to this force can be expressed as:
- \( \tau = I \alpha \)
- For rolling without slipping, \( a = r \alpha \) (where \( a \) is the linear acceleration).

### Step 3: Relate Linear Acceleration to Angular Acceleration
- For the sphere:
\[ mg \sin(\theta) - f = ma_s \]
\[ f \cdot r = I_s \alpha \]
Substituting \( \alpha = \frac{a_s}{r} \):
\[ f \cdot r = I_s \frac{a_s}{r} \]
Therefore, the frictional force \( f \) can be expressed in terms of \( a_s \).

- For the disc:
\[ mg \sin(\theta) - f = ma_d \]
\[ f \cdot r = I_d \alpha \]
Similarly, substituting \( \alpha = \frac{a_d}{r} \):
\[ f \cdot r = I_d \frac{a_d}{r} \]

### Step 4: Solve for Accelerations
- For the sphere:
\[ mg \sin(\theta) - f = ma_s \]
\[ f = \frac{I_s a_s}{r^2} \]
Substituting \( f \) into the first equation gives:
\[ mg \sin(\theta) - \frac{I_s a_s}{r^2} = ma_s \]
Rearranging leads to:
\[ a_s \left( m + \frac{I_s}{r^2} \right) = mg \sin(\theta) \]
\[ a_s = \frac{mg \sin(\theta)}{m + \frac{I_s}{r^2}} \]

- For the disc:
\[ a_d = \frac{mg \sin(\theta)}{m + \frac{I_d}{r^2}} \]

### Step 5: Compare the Accelerations
- Now, substituting the moments of inertia:
- For the sphere: \( I_s = \frac{2}{5} m r^2 \)
- For the disc: \( I_d = \frac{1}{2} m r^2 \)

- The expressions for acceleration become:
\[ a_s = \frac{mg \sin(\theta)}{m + \frac{2}{5} m} = \frac{mg \sin(\theta)}{m \left(1 + \frac{2}{5}\right)} = \frac{mg \sin(\theta)}{\frac{7}{5} m} = \frac{5g \sin(\theta)}{7} \]
\[ a_d = \frac{mg \sin(\theta)}{m + \frac{1}{2} m} = \frac{mg \sin(\theta)}{m \left(1 + \frac{1}{2}\right)} = \frac{mg \sin(\theta)}{\frac{3}{2} m} = \frac{2g \sin(\theta)}{3} \]

### Conclusion
- Since \( \frac{5g \sin(\theta)}{7} > \frac{2g \sin(\theta)}{3} \), we conclude that:
\[ a_s > a_d \]
- Thus, the acceleration of the sphere is greater than that of the disc.

### Final Answer
The correct statement is that the acceleration of the sphere will be greater than the acceleration of the disc.