A disc and a solid sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?

Options

A. solid sphere
B. both reach at the same time
C. depends on their masses
D. disc

Explanation

To determine which object, a disc or a solid sphere, reaches the bottom of an inclined plane first, we can analyze the motion of both objects using the principles of rotational dynamics and linear motion.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Objects:**
- When the disc and the sphere are placed on the inclined plane, the gravitational force acting on each object can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

2. **Set Up the Equations of Motion:**
- For both objects, we will apply Newton's second law. The net force acting down the incline is given by:
\[ F_{\text{net}} = mg \sin \theta - F_{\text{friction}} \]
- This net force causes linear acceleration \( a \) of the object:
\[ mg \sin \theta - F_{\text{friction}} = ma \]

3. **Consider the Torque Due to Friction:**
- Since both objects are rolling without slipping, friction provides the torque necessary for rotation. The torque \( \tau \) due to friction is:
\[ \tau = R \cdot F_{\text{friction}} \]
- This torque is related to the angular acceleration \( \alpha \) by:
\[ \tau = I \alpha \]
- Here, \( I \) is the moment of inertia of the object.

4. **Calculate the Moment of Inertia for Each Object:**
- For the disc:
\[ I_{\text{disc}} = \frac{1}{2} m R^2 \]
- For the solid sphere:
\[ I_{\text{sphere}} = \frac{2}{5} m R^2 \]

5. **Relate Linear and Angular Acceleration:**
- For pure rolling, the relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is:
\[ a = R \alpha \]
- Therefore, we can express \( \alpha \) as:
\[ \alpha = \frac{a}{R} \]

6. **Substitute and Solve for Acceleration:**
- For the disc:
\[ F_{\text{friction}} = I_{\text{disc}} \cdot \frac{a}{R} = \frac{1}{2} m R^2 \cdot \frac{a}{R} = \frac{1}{2} m a R \]
- Substituting into the net force equation:
\[ mg \sin \theta - \frac{1}{2} m a = ma \]
\[ mg \sin \theta = ma + \frac{1}{2} ma = \frac{3}{2} ma \]
\[ a = \frac{2g \sin \theta}{3} \]

- For the sphere:
\[ F_{\text{friction}} = I_{\text{sphere}} \cdot \frac{a}{R} = \frac{2}{5} m R^2 \cdot \frac{a}{R} = \frac{2}{5} m a R \]
- Substituting into the net force equation:
\[ mg \sin \theta - \frac{2}{5} m a = ma \]
\[ mg \sin \theta = ma + \frac{2}{5} ma = \frac{7}{5} ma \]
\[ a = \frac{5g \sin \theta}{7} \]

7. **Compare the Accelerations:**
- The acceleration of the disc is:
\[ a_{\text{disc}} = \frac{2g \sin \theta}{3} \approx 0.67 g \sin \theta \]
- The acceleration of the sphere is:
\[ a_{\text{sphere}} = \frac{5g \sin \theta}{7} \approx 0.71 g \sin \theta \]
- Since \( a_{\text{sphere}} > a_{\text{disc}} \), the solid sphere accelerates faster.

8. **Conclusion:**
- The solid sphere will reach the bottom of the inclined plane first.