Suppose a star the size of our Sun, but with mass 8.0 times...

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 9.0 days. If it were to undergo gravitational collapse to a neutron star of radius $12 \mathrm{km},$ losing $\frac{3}{4}$ of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times. Assume also that the thrown-off mass carries off either $(a)$ no angular momentum, or $(b)$ its proportional share $\left(\frac{3}{4}\right)$ of the initial angular momentum.

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
9.0 days. If it were to undergo gravitational collapse to
a neutron star of radius $12 \mathrm{km},$ losing $\frac{3}{4}$ of its mass in
the process, what would its rotation speed be? Assume
the star is a uniform sphere at all times. Assume also
that the thrown-off mass carries off either $(a)$ no angular
momentum, or $(b)$ its proportional share $\left(\frac{3}{4}\right)$ of the initial
angular momentum.

Key Concepts

Gravitational Collapse

Gravitational collapse refers to the process by which a star contracts under its own gravity, often accompanied by a change in size and mass distribution. This collapse significantly alters the moment of inertia, and under angular momentum conservation, leads to an increase in the rotation speed.

Effects of Mass Loss on Angular Momentum

In processes where mass is lost, determining whether the lost mass carries away angular momentum or not is important. This affects the remaining system's angular momentum budget and alters the resulting rotation speed, necessitating different treatments depending on whether the lost mass takes a proportional share of the original angular momentum.

Conservation of Angular Momentum

This principle states that in an isolated system with no external torques, the total angular momentum remains constant over time. It is crucial when analyzing how changes in the distribution of mass, such as during collapse, affect the rotation rate of an object.

Moment of Inertia for a Uniform Sphere

The moment of inertia quantifies how mass is distributed relative to the axis of rotation. For a uniform sphere, it is given by a formula involving the mass and the square of the radius. Changes in the radius dramatically affect the moment of inertia, which in turn influences the angular velocity when angular momentum is conserved.

Transcript

00:01 For part 8, assume that there is no angular momentum in the thrown -off mass, and so the final angular momentum of the neutron star is equaling the angular momentum before it collapse.

00:11 So we can say l initial equals l final, the initial angular momentum equals the final angular momentum.

00:23 We use the equation for the angular momentum, and we find that here, this would be 2 5th times 8 .0 times the mass of the size.

00:41 Sun times the radius of the sun squared multiplied by the initial angular velocity.

00:49 This would be equalling two -fifths times one -fourth times eight times the mass of the sun times the final radius squared.

01:05 And then we're going to multiply this by the final angular velocity.

01:08 And so the final angular velocity would be equaling two -fifths times 8 .0 mass.

01:16 Of the sun times the radius of the sun squared divided by two -fifths times one -fourth times eight point zero times the mass of the sun times the radius or the final radius squared rather times the initial angular velocity and so this would equal four times the radius of the sun squared divided by the final angular the final radius squared multiplied by the initial angular velocity and we can solve so the final angular velocity would be equaling four times the mass of the sun 6 .96 times 10 to the 8 meters quantity squared times the final radius squared which was 12 times 10 to the third meters quantity squared.

02:15 Multiplied by the initial angular velocity of one revolution for every nine days.

02:24 And so this is giving us 1 .495 times 10 to the 9th revolutions per day.

02:33 And let's convert.

02:37 So this would be again 1 .495 times 10 to the 9th revolutions per day...

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