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 forces 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
Both the disc and the sphere experience gravitational force (mg), normal force (N), and frictional force (F). The gravitational force can be resolved into two components: one parallel to the incline (mg sin θ) and one perpendicular to the incline (mg cos θ).

### Step 2: Write the Equation of Motion for the Disc and Sphere
For both objects, we can write the equation of motion along the incline:
\[ mg \sin \theta - F = ma \]
where \( F \) is the frictional force, \( m \) is the mass, and \( a \) is the linear acceleration of the center of mass.

### Step 3: Write the Torque Equation
The frictional force also causes a torque about the center of mass. The torque (τ) can be expressed as:
\[ \tau = F \cdot R \]
where \( R \) is the radius of the object. This torque is also related to the moment of inertia (I) and angular acceleration (α) by:
\[ \tau = I \cdot \alpha \]
Since \( \alpha = \frac{a}{R} \), we can write:
\[ F \cdot R = I \cdot \frac{a}{R} \]
From this, we can express the frictional force as:
\[ F = \frac{I \cdot a}{R^2} \]

### Step 4: Substitute into the Equation of Motion
Substituting the expression for \( F \) into the equation of motion gives:
\[ mg \sin \theta - \frac{I \cdot a}{R^2} = ma \]
Rearranging this, we can express \( a \) as:
\[ mg \sin \theta = ma + \frac{I \cdot a}{R^2} \]
Factoring out \( a \):
\[ mg \sin \theta = a \left( m + \frac{I}{R^2} \right) \]
Thus, we can solve for \( a \):
\[ a = \frac{mg \sin \theta}{m + \frac{I}{R^2}} \]

### Step 5: Calculate the Moment of Inertia for Both Objects
- For the disc, the moment of inertia \( I_D = \frac{1}{2} m R^2 \).
- For the sphere, the moment of inertia \( I_S = \frac{2}{5} m R^2 \).

### Step 6: Substitute the Moments of Inertia into the Acceleration Equation
For the disc:
\[ a_D = \frac{mg \sin \theta}{m + \frac{1/2 m R^2}{R^2}} = \frac{mg \sin \theta}{m + \frac{1}{2}m} = \frac{mg \sin \theta}{\frac{3}{2}m} = \frac{2g \sin \theta}{3} \]

For the sphere:
\[ a_S = \frac{mg \sin \theta}{m + \frac{2/5 m R^2}{R^2}} = \frac{mg \sin \theta}{m + \frac{2}{5}m} = \frac{mg \sin \theta}{\frac{7}{5}m} = \frac{5g \sin \theta}{7} \]

### Step 7: Calculate the Friction Forces
Using the expression for friction:
For the disc:
\[ F_D = \frac{I_D \cdot a_D}{R^2} = \frac{\frac{1}{2} m R^2 \cdot \frac{2g \sin \theta}{3}}{R^2} = \frac{mg \sin \theta}{3} \]

For the sphere:
\[ F_S = \frac{I_S \cdot a_S}{R^2} = \frac{\frac{2}{5} m R^2 \cdot \frac{5g \sin \theta}{7}}{R^2} = \frac{2mg \sin \theta}{7} \]

### Step 8: Find the Ratio of the Friction Forces
Now, we can find the ratio of the friction forces:
\[ \frac{F_D}{F_S} = \frac{\frac{mg \sin \theta}{3}}{\frac{2mg \sin \theta}{7}} = \frac{7}{6} \]

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

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