Question

A spinning giant star of initial radius R1 and angular speed ω1 suddenly collapses radially inward reaching a new radius R2 = 0.5R1 ; its mass remains the same. What is the relation between the kinetic energies of rotation (spin) before and after the collapse? In both cases, approximate the star as a uniform solid sphere.

Answer 1

Answer:

K1/K2=4

Step-by-step explanation:

The kinetic energy of a rotating sphere is given by:

$K=\frac{I*\omega^{2} }{2}$

The moment of inertia of a solid sphere is given by

$I=\frac{2MR^{2} }{5}$

The initial kinetic energy is therefore

$K_1=\frac{2MR^{2}*\omega^{2} }{10}$

$K_1=\frac{MR_1^{2}*\omega^{2} }{5}$

The final kinetic energy is given by

$K_2=\frac{MR_2^{2}*\omega^{2} }{5}$

Therefore the relation K1/K2 if R2 = 0.5R1

$\frac{K_1}{K_2} =\frac{5M(R_1)^{2}*\omega^{2} }{5*M(0.5R_1)^{2} \omega^{2}}$

The text says nothing about the final angular velocity just the collapse of the collapse of the radius

$\frac{K_1}{K_2} =\frac{1 }{(0.5)^{2} }=4$