Animated Solution for Physics - Rotational Motion: A sphere of mass $2\text{ kg}$ and radius $0.5\text{ m}$ is rolling with an initial speed of $1{\text{ ms}}^{-1}$ goes up an inclined plane which makes an angle of $30^{\circ }$ with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A ?

Rolling motion on an inclined plane is one of the most elegant and frequently tested concepts in rotational mechanics. It beautifully marries Newton's laws of translation with the dynamics of rotation. Let's embark on a detailed journey to solve this problem, uncovering a few conceptual traps and numerical subtleties along the way.

We are given a solid sphere of mass $m=2\text{ kg}$ and radius $R=0.5\text{ m}$. It is projected up an inclined plane of angle $\theta =30^{\circ }$ with an initial velocity $u=1\text{ m/s}$. The sphere rolls without slipping. Our goal is to find the total time it takes to travel up the incline, momentarily stop, and roll back down to its starting point.

To find the time, we first need to understand the kinematics of the sphere. Since the forces acting on it are constant, it will experience a uniform deceleration as it moves up. If we can find this deceleration $a$, we can easily use the first equation of motion, $v=u+at$, to find the time of ascent.

A Conceptual Trap

Before jumping into formulas, let's visualize the forces. Gravity acts downwards, and the normal force acts perpendicular to the surface. But what about static friction?

When the sphere rolls up the incline, its translational velocity $v$ is directed upwards. Because it rolls without slipping, its angular velocity $\omega =v/R$ must be clockwise. As the sphere moves up, gravity slows down its translational velocity. Consequently, its angular velocity must also decrease to maintain the rolling condition ($v=\omega R$).

To decrease a clockwise angular velocity, we need a counter-clockwise torque. Gravity acts through the center of mass, so it provides zero torque. The normal force also passes through the center. The only force capable of providing this torque is static friction acting at the point of contact. For the torque to be counter-clockwise, the friction force must point UP the incline! This is a highly counter-intuitive but fascinating reality of rolling motion.

Instead of deriving the equations of motion from scratch every time, we can rely on the master formula for the acceleration of a body rolling on an inclined plane:

This formula is incredibly powerful. Notice how the acceleration depends on the ratio $\frac{I}{m{R}^2}$, which is a purely geometric factor. For a solid sphere, the moment of inertia is $I=\frac{2}{5}m{R}^2$. Therefore, the ratio $\frac{I}{m{R}^2}$ is simply $\frac{2}{5}$.

This means the actual mass ($2\text{ kg}$) and radius ($0.5\text{ m}$) given in the problem are completely irrelevant! Any solid sphere, whether it's a marble or a bowling ball, will experience the exact same deceleration.

The $g$ Value Catch

Let's substitute our values into the acceleration formula:

Here lies a classic numerical trap. Which value of $g$ should we use?

If we use $g=10{\text{ m/s}}^2$:

If we use $g=9.8{\text{ m/s}}^2$:

Let's see which one aligns with the options.

Using the first equation of motion for the ascent, where the final velocity $v=0$:

If we used $g=10$, $t=\frac{7}{25}=0.28\text{ s}$. The total time for the round trip (ascent + descent) is $T=2t=0.56\text{ s}$. This is close, but not exactly in the options.

If we used $g=9.8$, $t=\frac{1}{3.5}=\frac{2}{7}\text{ s}$. The total time is:

This perfectly matches option (c)! The examiner specifically designed the problem expecting you to use $g=9.8{\text{ m/s}}^2$ to arrive at the clean, exact answer of $0.57\text{ s}$.

Always let the options guide your numerical assumptions in competitive exams. The sphere takes exactly $0.57\text{ s}$ to complete its elegant dance up and down the incline.