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Problem 81 Hard Difficulty

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.

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 11

Angular Momentum; General Rotation

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Video Transcript

for party 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 collapsed so we can say L initial equals our final. The initial angular momentum equals the final angular momentum. We used the equation for the angular momentum, and we find that here this would be to fifth times 8.0 times the mass of the Sun Times, the radius of the sun squared, multiplied by the initial angular velocity. This would be equaling to fifth time's 1/4 times, eight times the mass of the Sun Times the final radius squared, and then we're gonna multiply this by the final angular velocity. And so the final angular velocity would be equaling 2/5 times, 8.0 mass of the Sun Times the radius of the sun squared, divided by 2/5 times 1/4 times eight 0.0 times the mass of the sun times The radius for 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 weaken solve. So the final angular velocity would equaling four times the mass of the sun. 6.96 times tend to the eight meters quantity squared times of final radius squared, which was 12 times 10 to the third meters. Quantity squared, multiplied by the initial angular velocity of one revolution for every nine days. And so this is giving us 1.495 times 10 to the ninth revolutions her day and let's convert. So it should be again 1.495 times 10 to the ninth revolutions per day, and this is multiplied by one day for every 86,400 seconds. And this is giving us 1700 whether 1.730 times, 10 to the fourth revolutions per second. And we can then say that the final angular velocity is approximately equaling 17,000 300 revolutions For a second. This would be our final answer for Part a, and then for part B. We're now going to assume that the final, angular momentum of the neutron star is only 1/4 of the angular momentum before the collapse, since the rotation speed, of course, is directly proportional to the angular momentum again according to the definition of the angular momentum. We can then say that the final rotation speed for Part B will be 1/4 of that found in part A. And so the final angular velocity for Part B would be equaling 1/4 times 17,300 revolutions per second. This is giving us 4325 revolutions per second. This is a bit more, um, we really can't have this many significant figures. So we're gonna round 23 and the final angular velocity would be approximately equaling 4000 330 revolutions per second. Approximately this would be our final answer for Part B. That is the end of the solution. Thank you for watching

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Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Andy Chen

Farnaz Mohseni

Aspen Fenzl

Jared Enns

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Andy Chen

Farnaz Mohseni

Aspen Fenzl

Jared Enns

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Andy Chen

Farnaz Mohseni

Aspen Fenzl

Jared Enns

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics