A solid cylinder, a circular disc, a solid sphere and a hollow cylinder of the same radius are placed on an inclined plane. Which of the following will have maximum acceleration at the bottom of the plane?

Options

A. Circular disc.
B. Solid cylinder.
C. Solid sphere.
D. Hollow cylinder.

Answer

To determine which object will have the maximum acceleration at the bottom of an inclined plane, we need to analyze the motion of each object (solid cylinder, circular disc, solid sphere, and hollow cylinder) in terms of their moment of inertia and how they roll down the incline.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Objects**:
- Each object experiences gravitational force (mg) acting downwards.
- The component of gravitational force acting down the incline is \( mg \sin \theta \).
- There is also a frictional force (F) acting up the incline.

2. **Write the Equation of Motion**:
- For an object rolling down the incline, the net force can be expressed as:
\[ mg \sin \theta - F = ma \]
- Here, \( a \) is the linear acceleration of the object.

3. **Relate Linear Acceleration to Angular Acceleration**:
- Since the object is rolling without slipping, the relationship between linear acceleration (a) and angular acceleration (α) is given by:
\[ a = \alpha r \]
- The torque (τ) caused by the frictional force is:
\[ \tau = F \cdot r = I \alpha \]
- Substituting \( \alpha \) in terms of \( a \):
\[ F \cdot r = I \left(\frac{a}{r}\right) \]
- Rearranging gives:
\[ F = \frac{I a}{r^2} \]

4. **Substitute F into the Equation of Motion**:
- Substitute \( F \) back into the equation of motion:
\[ mg \sin \theta - \frac{I a}{r^2} = ma \]
- Rearranging gives:
\[ mg \sin \theta = ma + \frac{I a}{r^2} \]
- Factor out \( a \):
\[ mg \sin \theta = a \left(m + \frac{I}{r^2}\right) \]

5. **Solve for Acceleration (a)**:
- Rearranging for \( a \):
\[ a = \frac{mg \sin \theta}{m + \frac{I}{r^2}} \]

6. **Calculate the Moment of Inertia for Each Object**:
- **Solid Cylinder**: \( I = \frac{1}{2} m r^2 \) → \( \frac{I}{mr^2} = \frac{1}{2} \)
- **Circular Disc**: \( I = \frac{1}{2} m r^2 \) → \( \frac{I}{mr^2} = \frac{1}{2} \)
- **Solid Sphere**: \( I = \frac{2}{5} m r^2 \) → \( \frac{I}{mr^2} = \frac{2}{5} \)
- **Hollow Cylinder**: \( I = m r^2 \) → \( \frac{I}{mr^2} = 1 \)

7. **Determine the Object with Minimum \( \frac{I}{mr^2} \)**:
- The object with the smallest \( \frac{I}{mr^2} \) will have the maximum acceleration.
- Comparing the values:
- Solid Cylinder: \( \frac{1}{2} \)
- Circular Disc: \( \frac{1}{2} \)
- Solid Sphere: \( \frac{2}{5} \) (smallest)
- Hollow Cylinder: \( 1 \)

8. **Conclusion**:
- The solid sphere has the minimum \( \frac{I}{mr^2} \) value, which means it will have the maximum acceleration at the bottom of the incline.

### Final Answer:
The solid sphere will have the maximum acceleration at the bottom of the inclined plane.