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, we can follow these steps:

### Step 1: Identify the Forces Acting on the Bodies
For both the sphere and the disc, the gravitational force \( mg \) acts downwards, which can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

### Step 2: Apply Newton's Second Law for Linear Motion
For the sphere and the disc, we can apply Newton's second law along the incline. The net force acting along the incline is given by:
\[ F_{\text{net}} = mg \sin \theta - F_s \]
where \( F_s \) is the frictional force.

Using Newton's second law:
\[ F_{\text{net}} = ma \]
we can write:
\[ mg \sin \theta - F_s = ma \]
(Equation 1)

### Step 3: Apply Torque Equation for Rotational Motion
The frictional force \( F_s \) also provides torque about the center of mass. The torque \( \tau \) is given by:
\[ \tau = F_s \cdot r \]
where \( r \) is the radius of the sphere or disc.

Using the relation between torque and angular acceleration:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration. For rolling without slipping, we have:
\[ \alpha = \frac{a}{r} \]
Thus, we can write:
\[ F_s \cdot r = I \cdot \frac{a}{r} \]
Rearranging gives:
\[ F_s = \frac{I a}{r^2} \]
(Equation 2)

### Step 4: Substitute Equation 2 into Equation 1
Now, we substitute \( F_s \) from Equation 2 into Equation 1:
\[ mg \sin \theta - \frac{I a}{r^2} = ma \]
Rearranging gives:
\[ mg \sin \theta = ma + \frac{I a}{r^2} \]
Factoring out \( a \):
\[ mg \sin \theta = a \left( m + \frac{I}{r^2} \right) \]
Thus, we can express the acceleration \( a \) as:
\[ a = \frac{mg \sin \theta}{m + \frac{I}{r^2}} \]
(Equation 3)

### Step 5: Calculate the Moment of Inertia for Sphere and Disc
For a solid sphere:
\[ I_s = \frac{2}{5} m r^2 \]
For a disc:
\[ I_d = \frac{1}{2} m r^2 \]

### Step 6: Substitute the Moments of Inertia into Equation 3
For the sphere:
\[ a_s = \frac{mg \sin \theta}{m + \frac{2}{5} m} = \frac{mg \sin \theta}{m \left( 1 + \frac{2}{5} \right)} = \frac{mg \sin \theta}{\frac{7}{5} m} = \frac{5}{7} g \sin \theta \]

For the disc:
\[ a_d = \frac{mg \sin \theta}{m + \frac{1}{2} m} = \frac{mg \sin \theta}{m \left( 1 + \frac{1}{2} \right)} = \frac{mg \sin \theta}{\frac{3}{2} m} = \frac{2}{3} g \sin \theta \]

### Step 7: Compare the Accelerations
Now we compare the two accelerations:
- For the sphere: \( a_s = \frac{5}{7} g \sin \theta \)
- For the disc: \( a_d = \frac{2}{3} g \sin \theta \)

To compare these values, we can convert \( \frac{5}{7} \) and \( \frac{2}{3} \) to decimal:
- \( \frac{5}{7} \approx 0.714 \)
- \( \frac{2}{3} \approx 0.667 \)

Thus, we find:
\[ a_s > a_d \]

### Conclusion
The correct statement is:
\[ g > a_s > a_d \]