Question

Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 8.0 days. If it were to undergo gravitational collapse to a neutron star of radius 11 km , losing three-quarters of its mass in the process, what would its rotation speed be? Assume also that the thrown- off mass carries off either

a) no angular momentum

b)its proportional share three-quarters of the initial angular momentum

Express your answer using two significant figures.

Answer 1

Answer :-

Apply the conservation of angular momentum;

angular momentum at the beginning = angular momentum at the end

angular momentum = I w

I=moment of inertia for a sphere= 2/5 MR^2

where , M is the mass and R is the radius of the sun

According to the given situation ,

M1=8M

R1=R

w1=2 pi rad/ 8 days = 2*3.14 / 8 days =0.785 rad/days

After the star undergoes the gravitational collapse to the neutron star

then ,

M2=M/4 ( Three quarter mass has been lost , only one fourth quarter is left )

R2= 11km = 11000 m    (radius of the sun is 6.96x10^8 m )

I1 w1 = I2 w2

2/5 M1R1^(2) *w1 = 2/5M2*R2^(2) *w2

8M *(6.96*10^(8))^(2) * ( 0.785 rad / days ) = M/4 * (11000)^(2) *w2

w2 =1.0056 *10^(11) rad / days

Hence the rotation speed would be 1.0056*10^(11) rad / days.