A solid cylinder and a solid sphere, both having the same mass and radius, are released from a rough inclined plane of inclination `theta` one by one. They roll on the inclined plane without slipping. The force of friction that acts

Options

A. Same for both the bodies
B. More for cylinder
C. More for sphere
D. Double for sphere than for cylinder

Answer

To solve the problem of comparing the force of friction acting on a solid cylinder and a solid sphere rolling down a rough inclined plane, we can follow these steps:

### Step 1: Understand the System
We have a solid cylinder and a solid sphere, both with the same mass \( m \) and radius \( r \). They roll down an inclined plane with an angle \( \theta \) without slipping.

### Step 2: Apply Conservation of Energy
The initial potential energy when the objects are at height \( h \) is converted into kinetic energy as they roll down.

The initial potential energy (PE) is given by:
\[ PE = mgh \]
The final kinetic energy (KE) when they reach the bottom consists of translational and rotational kinetic energy:
\[ KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]
For rolling without slipping, we have the relation \( v = r\omega \).

### Step 3: Moment of Inertia
The moment of inertia \( I \) for the solid cylinder is:
\[ I_{\text{cylinder}} = \frac{1}{2} mr^2 \]
For the solid sphere, it is:
\[ I_{\text{sphere}} = \frac{2}{5} mr^2 \]

### Step 4: Set Up the Energy Equation
Using conservation of energy:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \left(\frac{v}{r}\right)^2 \]
Substituting the moment of inertia for both objects, we can express the equation for each.

#### For the Cylinder:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} mr^2\right) \left(\frac{v^2}{r^2}\right) \]
This simplifies to:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]
Thus,
\[ v^2 = \frac{4gh}{3} \]

#### For the Sphere:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5} mr^2\right) \left(\frac{v^2}{r^2}\right) \]
This simplifies to:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 = \frac{7}{10} mv^2 \]
Thus,
\[ v^2 = \frac{10gh}{7} \]

### Step 5: Find the Acceleration
Using \( v^2 = u^2 + 2as \) where \( u = 0 \):
\[ v^2 = 2as \Rightarrow a = \frac{v^2}{2s} \]
Substituting \( s = \frac{h}{\sin \theta} \):
- For the cylinder:
\[ a_{\text{cylinder}} = \frac{4gh/3}{2(h/\sin \theta)} = \frac{2g\sin \theta}{3} \]
- For the sphere:
\[ a_{\text{sphere}} = \frac{10gh/7}{2(h/\sin \theta)} = \frac{5g\sin \theta}{7} \]

### Step 6: Calculate the Friction Force
Using the equation of motion and torque:
\[ F_{\text{friction}} \cdot r = I \cdot \alpha \]
Where \( \alpha = \frac{a}{r} \):
- For the cylinder:
\[ F_{\text{friction}} = \frac{I_{\text{cylinder}} \cdot a}{r^2} = \frac{\frac{1}{2} mr^2 \cdot \frac{2g\sin \theta}{3}}{r^2} = \frac{mg\sin \theta}{3} \]
- For the sphere:
\[ F_{\text{friction}} = \frac{I_{\text{sphere}} \cdot a}{r^2} = \frac{\frac{2}{5} mr^2 \cdot \frac{5g\sin \theta}{7}}{r^2} = \frac{2mg\sin \theta}{7} \]

### Step 7: Compare the Friction Forces
- Friction force for the cylinder: \( F_{\text{friction, cylinder}} = \frac{mg\sin \theta}{3} \)
- Friction force for the sphere: \( F_{\text{friction, sphere}} = \frac{2mg\sin \theta}{7} \)

Since \( \frac{mg\sin \theta}{3} > \frac{2mg\sin \theta}{7} \), the force of friction is greater for the solid cylinder.

### Conclusion
The force of friction that acts more for the solid cylinder than for the solid sphere.