A hollow cylinder and a solid cylinder are rolling without slipping down an inclined plane, then which of these reaches earlier ?

Options

A. Solid cylinder
B. Hollow cylinder
C. Both simultaneously
D. Can't say anything

Answer

To solve the problem of determining which cylinder reaches the bottom of an inclined plane first, we need to analyze the motion of both the hollow cylinder and the solid cylinder as they roll down the incline. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the motion of rolling cylinders
Both cylinders are rolling without slipping down an inclined plane. The acceleration of a rolling object can be expressed in terms of gravitational acceleration and its moment of inertia.

### Step 2: Write the formula for acceleration
The acceleration \( a \) of a rolling object down an incline is given by:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}} \]
where:
- \( g \) is the acceleration due to gravity,
- \( \theta \) is the angle of inclination,
- \( I \) is the moment of inertia,
- \( m \) is the mass of the cylinder,
- \( r \) is the radius of the cylinder.

### Step 3: Calculate the moment of inertia for both cylinders
1. **Hollow Cylinder (or Ring)**:
\[ I = mr^2 \]
2. **Solid Cylinder (or Disc)**:
\[ I = \frac{1}{2}mr^2 \]

### Step 4: Substitute the moment of inertia into the acceleration formula
1. **For the Hollow Cylinder**:
\[ a_{\text{hollow}} = \frac{g \sin \theta}{1 + \frac{mr^2}{mr^2}} = \frac{g \sin \theta}{2} \]

2. **For the Solid Cylinder**:
\[ a_{\text{solid}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

### Step 5: Compare the accelerations
- The acceleration of the hollow cylinder is \( \frac{g \sin \theta}{2} \).
- The acceleration of the solid cylinder is \( \frac{2g \sin \theta}{3} \).

### Step 6: Determine which acceleration is greater
To compare:
- \( \frac{g \sin \theta}{2} \) (hollow cylinder) is approximately \( 0.5g \sin \theta \).
- \( \frac{2g \sin \theta}{3} \) (solid cylinder) is approximately \( 0.67g \sin \theta \).

Since \( 0.67g \sin \theta > 0.5g \sin \theta \), the solid cylinder has a greater acceleration.

### Step 7: Relate acceleration to time
The time \( t \) taken to travel a distance \( s \) is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
Since both cylinders start from rest (\( u = 0 \)):
\[ s = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2s}{a}} \]
Thus, the time taken is inversely proportional to the square root of the acceleration.

### Step 8: Conclusion
Since the solid cylinder has a greater acceleration, it will take less time to reach the bottom of the incline compared to the hollow cylinder. Therefore, the solid cylinder reaches the bottom first.

### Final Answer
The solid cylinder reaches the bottom of the inclined plane first.

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