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 solve the problem, we need to apply the principle of conservation of angular momentum. Initially, the sphere has mass $M$, radius $R$, and is rotating with angular velocity $\omega_1$. The moment of inertia of a solid sphere is given by $I= \frac{2}{5} MR^2$.

The initial angular momentum $L_1$ is:

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

As the sphere loses mass uniformly with no change in shape, its density remains constant. Thus, when the radius is $\frac{R}{2}$, its new mass $M_2$ can be determined using the density ratio since $density=\frac{Mass}{Volume}$:

Original density $\rho=\frac{M}{\frac{4}{3}\pi R^3} = \frac{3M}{4\pi R^3}$.

New density when radius becomes $\frac{R}{2}$ is also $\rho=\frac{M_2}{\frac{4}{3}\pi \left(\frac{R}{2}\right)^3} = \frac{3M_2}{\pi R^3}$.

Setting the original and new densities equal, solve for $M_2$:

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

Therefore, $M_2 = \frac{M}{8}$.

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 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, $\omega_2 = 32 \omega_1$.

Thus, the computed value of $x$ is 32, which falls within the provided range of (32,32).

Answer 2

When the sphere is of radius \( R \), its mass is \( M \), and when the radius is reduced to \( \frac{R}{2} \), the mass will reduce to \( \frac{M}{8} \). 
This is due to the conservation of angular momentum. 
Using the conservation of angular momentum (\( \tau_{\text{ext}} = 0 \)): \[ I_1 \omega_1 = I_2 \omega_2 \] \[ \left( \frac{2}{5} M R^2 \right) \omega_1 = \left( \frac{2}{5} \times \frac{M}{8} \times \left( \frac{R}{2} \right)^2 \right) \omega_2 \] Simplifying this: \[ \omega_2 = 32 \omega_1 \] 
Thus, the value of \( x \) is 32.