A disc and a solid sphere of same mass and radius roll down an inclined plane. The ratio of thhe friction force acting on the disc and sphere is

Options

A. `(7)/(6)`
B. `(5)/(4)`
C. `(3)/(2)`
D. depends on angle of inclination

Explanation

To find the ratio of the friction force acting on a disc and a solid sphere rolling down an inclined plane, we can follow these steps:

### Step 1: Identify the Forces Acting on the Disc and Sphere
When a disc and a solid sphere roll down an inclined plane, the forces acting on them include:
- Weight (mg) acting downwards.
- Normal force (N) acting perpendicular to the surface.
- Frictional force (F) acting opposite to the direction of motion.

### Step 2: Resolve the Weight into Components
The weight can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

### Step 3: Write the Equations of Motion
For both the disc and the sphere, the net force acting along the incline can be expressed as:
\[ mg \sin \theta - F = ma \]
Where \( a \) is the linear acceleration.

### Step 4: Write the Torque Equation
The frictional force is responsible for the rotational motion. The torque (\( \tau \)) about the center of mass is given by:
\[ \tau = F \cdot r = I \alpha \]
Where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.

### Step 5: Relate Linear and Angular Acceleration
The angular acceleration (\( \alpha \)) is related to the linear acceleration (\( a \)) by:
\[ \alpha = \frac{a}{r} \]

### Step 6: Substitute and Rearrange
Substituting \( \alpha \) into the torque equation gives:
\[ F \cdot r = I \left(\frac{a}{r}\right) \]
Rearranging gives:
\[ F = \frac{I a}{r^2} \]

### Step 7: Substitute F into the Equation of Motion
Substituting the expression for \( F \) into the equation of motion:
\[ mg \sin \theta - \frac{I a}{r^2} = ma \]
Rearranging gives:
\[ mg \sin \theta = ma + \frac{I a}{r^2} \]
Factoring out \( a \):
\[ mg \sin \theta = a \left(m + \frac{I}{r^2}\right) \]

### Step 8: Solve for Linear Acceleration (a)
Now we can solve for \( a \):
\[ a = \frac{mg \sin \theta}{m + \frac{I}{r^2}} \]

### Step 9: Calculate the Friction Force for Both Objects
For the disc (moment of inertia \( I_d = \frac{1}{2}mr^2 \)):
\[ F_d = \frac{I_d a}{r^2} = \frac{\frac{1}{2}mr^2 \cdot \frac{mg \sin \theta}{m + \frac{1}{2}m}}{r^2} = \frac{mg \sin \theta}{3} \]

For the solid sphere (moment of inertia \( I_s = \frac{2}{5}mr^2 \)):
\[ F_s = \frac{I_s a}{r^2} = \frac{\frac{2}{5}mr^2 \cdot \frac{mg \sin \theta}{m + \frac{2}{5}m}}{r^2} = \frac{mg \sin \theta}{\frac{7}{5}} = \frac{5mg \sin \theta}{7} \]

### Step 10: Find the Ratio of Friction Forces
Now, we can find the ratio of the friction forces:
\[ \frac{F_d}{F_s} = \frac{\frac{mg \sin \theta}{3}}{\frac{5mg \sin \theta}{7}} = \frac{7}{15} \]

### Final Answer
The ratio of the friction force acting on the disc to that acting on the sphere is:
\[ \frac{F_d}{F_s} = \frac{7}{6} \]