Question

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ω. The sphere is slowly heated until it reaches its melting temperature, at which point it flattens into a uniform disk of thickness D/2. By what factor is the angular velocity changed?

Answer 1

Explanation:

It is given that, a solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ω.The sphere is slowly heated until it reaches its melting temperature, at which point it flattens into a uniform disk of thickness D/2.

The angular momentum remains conserved in this case. The relation between the angular momentum and the angular velocity is given by :

$I_s\times \omega_s=I_d\times \omega_d$

Where

$I_s\ and\ I_d$ are moment of inertia of sphere and the disk respectively

Here, volume before = volume after

$\dfrac{4}{3}\pi (D/2)^3=\pi r^2\times D/2$

$r=\dfrac{D}{\sqrt{3} }=0.577\ D$

Initial angular momentum,

$L_i=I_s\times \omega_s$

$L_i=\dfrac{2mr^2}{5}\times \omega_s$

$L_i=\dfrac{2m(D/2)^2}{5}\times \omega_s$

$L_i=\dfrac{2mD^2}{20}\times \omega_s$..........(1)

Final angular momentum,

$L_d=I_f\times \omega_d$

$L_d=\dfrac{2mr^2}{5}\times \omega_d$

$L_d=\dfrac{2m(0.577D)^2}{5}\times \omega_d$............(2)

From equation (1) and (2) :

$\dfrac{2mD^2}{20}\times \omega_s=\dfrac{2m(0.577D)^2}{5}\times \omega_d$

$\dfrac{\omega_d}{\omega_s}=0.75$

Hence, this is the required solution.