Question

A solid sphere with uniform density and radius $ R $ is rotating initially with constant angular velocity ($ \omega_1 $) about its diameter. After some time during the rotation, it starts losing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius becomes $ \frac{R}{2} $ is $ \omega_2 $. The value of $ x $ is ________.

Hint:

Conservation of angular momentum is key when the object loses mass uniformly but retains its shape and rotation.

Answer 1

Correct Answer: 32

To resolve the problem, the principle of conservation of angular momentum is applied. The sphere initially possesses mass $M$, radius $R$, and angular velocity $\omega_1$. The moment of inertia for a solid sphere is calculated as $I = \frac{2}{5} MR^2$.

The initial angular momentum ($L_1$) is given by:

$L_1 = I \cdot \omega_1 = \frac{2}{5} MR^2 \omega_1$.

The sphere loses mass uniformly while maintaining its shape, indicating constant density. When the radius is reduced to $\frac{R}{2}$, the new mass ($M_2$) is determined by comparing densities, where $density=\frac{Mass}{Volume}$:

The original density is $\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{3M}{4\pi R^3}$.

The new density, with a radius of $\frac{R}{2}$, is also $\rho = \frac{M_2}{\frac{4}{3}\pi \left(\frac{R}{2}\right)^3} = \frac{3M_2}{\pi R^3}$.

Equating the original and new densities to solve for $M_2$:

$\frac{3M}{4\pi R^3} = \frac{3M_2}{\pi R^3}$.

This yields $M_2 = \frac{M}{8}$.

The moment of inertia at the new state is:

$I_2 = \frac{2}{5} M_2 \left(\frac{R}{2}\right)^2 = \frac{2}{5} \left(\frac{M}{8}\right) \frac{R^2}{4}$.

$I_2 = \frac{2}{5} \cdot \frac{M}{8} \cdot \frac{R^2}{4} = \frac{MR^2}{80}$.

By the conservation of angular momentum, $L_1 = L_2$:

$\frac{2}{5} MR^2 \omega_1 = \frac{MR^2}{80}\omega_2$.

$16 \cdot \frac{2}{5} \omega_1 = \omega_2$.

Simplifying this equation results in $\omega_2 = 32 \omega_1$.

Therefore, the calculated value of $x$ is 32, which is within the specified range of (32,32).