Respuesta :
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 :
[tex]I_s\times \omega_s=I_d\times \omega_d[/tex]
Where
[tex]I_s\ and\ I_d[/tex] are moment of inertia of sphere and the disk respectively
Here, volume before = volume after
[tex]\dfrac{4}{3}\pi (D/2)^3=\pi r^2\times D/2[/tex]
[tex]r=\dfrac{D}{\sqrt{3} }=0.577\ D[/tex]
Initial angular momentum,
[tex]L_i=I_s\times \omega_s[/tex]
[tex]L_i=\dfrac{2mr^2}{5}\times \omega_s[/tex]
[tex]L_i=\dfrac{2m(D/2)^2}{5}\times \omega_s[/tex]
[tex]L_i=\dfrac{2mD^2}{20}\times \omega_s[/tex]..........(1)
Final angular momentum,
[tex]L_d=I_f\times \omega_d[/tex]
[tex]L_d=\dfrac{2mr^2}{5}\times \omega_d[/tex]
[tex]L_d=\dfrac{2m(0.577D)^2}{5}\times \omega_d[/tex]............(2)
From equation (1) and (2) :
[tex]\dfrac{2mD^2}{20}\times \omega_s=\dfrac{2m(0.577D)^2}{5}\times \omega_d[/tex]
[tex]\dfrac{\omega_d}{\omega_s}=0.75[/tex]
Hence, this is the required solution.
