Question

An isolated star with the radius of the Sun (6.95 × 108 m) rotates on an axis through its centre once every 2.16 × 106 s. It collapses under gravity to form a much denser star with the radius of the Earth (6.38 × 106 m). For the purposes of this question, you may assume that no matter is lost from the star during its collapse. You may also treat the star as a homogeneous sphere before and after the collapse.

Given that the moment of inertia of a homogeneous sphere is (2/5) MR^2, where M is the mass of the sphere and R is its radius, calculate the period of rotation of the star after its collapse. Give your answer by entering a number, specified to an appropriate number of significant figures, into the empty box below.

Answer 1

Here, we shall apply the conservation of angular momentum L to calculate the time period as follows

Li=L

(1)........122 = 1لI

Here, I is the moment of inertia and omega is the angular speed. We are using subscripts 1 (2) before(after) the collapse.

57

11 Τη 12 Το

MR MR Τι

The mass of the star will remain the same M1=M2.Putting all the values and solving for T2, time period after collapse.

(6.5 x (6.38 x 10) 2.16 x 10

(6.38 x 1062 T2 = 2.16 x 106 6.95 x 108)2