A solid cylinder (i) rolls down (ii) slides down an inclined plane. The ratio of the accelerations in these conditions is

Options

A. `3:2`
B. `2:3`
C. `sqrt3:sqrt2`
D. `sqrt2:sqrt3`

Answer

To solve the problem of finding the ratio of accelerations of a solid cylinder rolling down an inclined plane versus sliding down the same plane, we can follow these steps:

### Step 1: Understand the two scenarios
- **Rolling**: The solid cylinder rolls down the incline without slipping. In this case, both translational and rotational motion are involved.
- **Sliding**: The solid cylinder slides down the incline without rolling. Here, only translational motion is considered.

### Step 2: Analyze the rolling motion
For a solid cylinder rolling down an inclined plane, the acceleration \( a_R \) can be derived from the forces acting on it. The component of gravitational force acting down the incline is \( mg \sin \theta \), and the moment of inertia \( I \) for a solid cylinder about its center is given by \( I = \frac{1}{2} m r^2 \).

Using the formula for acceleration in rolling motion:
\[ a_R = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
Substituting \( I = \frac{1}{2} m r^2 \):
\[ a_R = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

### Step 3: Analyze the sliding motion
For the sliding motion, the only force acting down the incline is the gravitational component:
\[ a_S = g \sin \theta \]

### Step 4: Calculate the ratio of accelerations
Now, we can find the ratio of the accelerations for rolling and sliding:
\[ \frac{a_R}{a_S} = \frac{\frac{2g \sin \theta}{3}}{g \sin \theta} \]
Cancelling \( g \sin \theta \) from the numerator and denominator:
\[ \frac{a_R}{a_S} = \frac{2}{3} \]

### Conclusion
Thus, the ratio of the accelerations when the solid cylinder rolls down versus when it slides down the inclined plane is:
\[ \frac{a_R}{a_S} = \frac{2}{3} \]