Animated Solution for Physics - Rotational Motion: A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same, which one of the following will not be affected?

The beauty of physics often lies in what doesn't change when everything else seems to be in flux. In this problem, we are presented with a fascinating thought experiment: a solid sphere rotating in the absolute void of free space, which suddenly begins to expand.

Imagine a solid sphere of mass $m$ and initial radius $R_1$ floating in free space. It is spinning smoothly around its axis with an initial angular velocity ${\omega }_1$. The term "free space" is the crucial key here. It means there is no gravity pulling on the sphere, no air resistance slowing it down, and absolutely no external forces applying a twist or torque to it.

Mathematically, we state this as:

Newton's second law for rotation tells us that the net external torque acting on a system is equal to the rate of change of its angular momentum ($L$).

Since ${\tau }_{ext}=0$, it immediately follows that $\frac{dL}{dt}=0$. This means the angular momentum $L$ is a constant of motion. It is strictly conserved, no matter what internal changes the sphere undergoes.

Now, the problem states that the radius of the sphere increases to a new value $R_2$ (where $R_2>R_1$), while its mass $m$ remains exactly the same. Let's see how this geometric change affects the other rotational properties.

First, consider the moment of inertia ($I$). For a solid sphere, the moment of inertia about its central axis is given by:

Since the mass $m$ is constant and the radius $R$ increases, the moment of inertia $I$ must increase. The mass is now distributed further away from the axis of rotation, making it harder to change the sphere's rotational state.

We established that the angular momentum $L$ is conserved. We also know that angular momentum is the product of moment of inertia and angular velocity:

If $I$ increases, then the angular velocity $\omega$ must decrease proportionally to keep the product $L$ constant. This is the classic "ice skater effect"—as the skater extends their arms (increasing $I$), their spin slows down (decreasing $\omega$).

Finally, let's look at the rotational kinetic energy ($K$). The formula for rotational kinetic energy is $K=\frac{1}{2}I{\omega }^2$. However, when dealing with conserved angular momentum, it is often much more insightful to rewrite this formula in terms of $L$ and $I$:

Since $L$ is a constant, the numerator remains unchanged. But as we found earlier, the denominator $I$ has increased. Dividing a constant by a larger number results in a smaller value. Therefore, the rotational kinetic energy $K$ must decrease.

Through our analysis, we have seen that the moment of inertia increases, the angular velocity decreases, and the rotational kinetic energy decreases. The only quantity that stands unwavering, completely unaffected by the sphere's expansion, is the angular momentum.