Question

spherical star spinning at angular velocity omega suddenly collapses to half of its original radius without any loss of mass. Assume the star has uniform density before and after the collapse. What will the angular velocity of the star be after the collapse

Answer 1

The angular velocity of the star after the collapse will be four times greater than its initial angular velocity.

Step-by-step explanation:

Given:

The initial angular velocity of a spherical star is
`\omega`.

The final radius of the star is half of its initial radius.

The conservation of the angular momentum is given by

`I \omega = constant`

where
`I` is the moment of inertial of the star.

Consider that the initial moment of inertia of the star is
`I_(1)`, the final moment of inertia is
`I_(2)` and the final angular velocity is
`\omega_(f)`.

The moment of inertia of a sphere is given by

`I = (2)/(5)MR^(2)`

where
`M` is the mass of the sphere and
`R` is the radius of the sphere.

The expression for the conservation of angular momentum for the star is given by

`~~~~~&& I_(1) \omega = I_(2) \omega_(f)\\&or,& ((2)/(5)MR^(2)) \omega = ((2)/(5)M((R)/(2))^(2)) \omega_(f)\\&or,& \omega_(f) = 4 \omega`

Thus, the angular velocity of the star after the collapse will be four times greater than its initial angular velocity.