Solid cylinders of radii `r_1,r_2 and r_3` roll down an inclined plane from the same place simultaneously. If r1 > r2 > r3 , which one would reach the bottom first?

Options

A. Cylinder of radius `r_1 `
B. Cylinder of radius `r_2`
C. Cylinder of radius `r_3`
D. All the three cylinders simultaneously

Answer

To determine which solid cylinder reaches the bottom of the inclined plane first, we need to analyze the motion of the cylinders as they roll down the incline.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have three solid cylinders with different radii \( r_1, r_2, \) and \( r_3 \) such that \( r_1 > r_2 > r_3 \). All cylinders are released from the same height on an inclined plane.

2. **Identifying the Forces**:
When a solid cylinder rolls down an incline, it experiences gravitational force acting downwards and a normal force acting perpendicular to the surface of the incline. The component of gravitational force acting down the incline is \( mg \sin \theta \), where \( m \) is the mass of the cylinder and \( \theta \) is the angle of the incline.

3. **Acceleration of the Rolling Cylinder**:
The acceleration \( a \) of a rolling cylinder can be expressed as:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
where \( I \) is the moment of inertia of the cylinder. For a solid cylinder, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]

4. **Substituting Moment of Inertia**:
Substituting the moment of inertia into the acceleration formula, we get:
\[ a = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

5. **Conclusion on Acceleration**:
The derived expression for acceleration \( a \) shows that it is independent of the radius \( r \) of the cylinder. Therefore, all three cylinders will have the same acceleration when rolling down the incline.

6. **Time to Reach the Bottom**:
Since all cylinders start from rest and travel the same distance \( L \) down the incline, the time taken \( t \) to reach the bottom can be calculated using the equation of motion:
\[ L = \frac{1}{2} a t^2 \]
Rearranging gives:
\[ t = \sqrt{\frac{2L}{a}} = \sqrt{\frac{2L}{\frac{2g \sin \theta}{3}}} = \sqrt{\frac{3L}{g \sin \theta}} \]
This shows that the time taken is also independent of the radius.

7. **Final Answer**:
Since all cylinders have the same acceleration and travel the same distance, they will all reach the bottom of the incline simultaneously.