Animated Solution for Physics - Rotational Motion: A circular disc reaches from top to bottom of an inclined plane of length $L$. When it slips down the plane, it takes time $t_1$. When it rolls down the plane, it takes time $t_2$. The value of $\frac{t_2}{t_1}$ is $\sqrt{\frac{3}{x}}$. The value of $x$ will be …… .

This problem is a classic exploration of the difference between pure translation (slipping) and a combination of translation and rotation (pure rolling). Let's break down the physics of both scenarios to understand why a rolling object takes longer to reach the bottom of an incline.

Imagine a circular disc placed at the top of an inclined plane of length $L$ and angle $\theta$. We are given two distinct cases: 1. Slipping: The plane is perfectly smooth. There is no friction, so the disc simply slides down without rotating. 2. Rolling: The plane is rough enough to provide the necessary static friction for pure rolling. The disc translates and rotates simultaneously.

Our goal is to find the ratio of the time taken in the rolling case ($t_2$) to the time taken in the slipping case ($t_1$).

Pure Slipping

When the disc slips down a smooth incline, the only force acting along the plane is the component of gravity, $mg\sin \theta$. There is no friction to oppose this motion or cause rotation.

Therefore, the linear acceleration $a_1$ is simply:

Since the disc starts from rest ($u=0$), we can use the second equation of motion to find the time $t_1$ it takes to cover the distance $L$:

Pure Rolling

When the disc rolls, static friction acts up the incline. This friction does two things: it reduces the net downward force (lowering the linear acceleration), and it provides the torque necessary for the disc to rotate.

The linear acceleration $a_2$ for a body rolling down an incline is given by the standard formula: where $K$ is the radius of gyration of the body.

For a uniform circular disc, the moment of inertia about its central axis is $I=\frac{1}{2}M{R}^2$. Since $I=M{K}^2$, we have:

Substituting this into our acceleration formula:

Notice that $a_2<a_1$. Because some of the gravitational potential energy is converted into rotational kinetic energy, less energy is available for translational motion, resulting in a lower linear acceleration.

Now, let's find the time $t_2$ for rolling:

We are asked to find the ratio $\frac{t_2}{t_1}$:

The terms $L$ and $g\sin \theta$ cancel out beautifully, leaving us with:

The problem states that this ratio is equal to $\sqrt{\frac{3}{x}}$. Comparing our result with the given expression, it is clear that:

$$x=2$$

This elegant result shows that the ratio of times depends only on the mass distribution (shape) of the object, not on its mass, radius, or the specifics of the incline!