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Suppose a star the size of our Sun, but with mass 8.0 timesas great, were rotating at a speed of 1.0 revolution every9.0 days. If it were to undergo gravitational collapse toa neutron star of radius $12 \mathrm{km},$ losing $\frac{3}{4}$ of its mass inthe process, what would its rotation speed be? Assumethe star is a uniform sphere at all times. Assume alsothat the thrown-off mass carries off either $(a)$ no angularmomentum, or $(b)$ its proportional share $\left(\frac{3}{4}\right)$ of the initialangular momentum.

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Physics 101 Mechanics

Chapter 11

Angular Momentum; General Rotation

Moment, Impulse, and Collisions

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

University of Winnipeg

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

07:30

Suppose a star the size of…

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01:44

A typical neutron star may…

03:58

Consider a neutron star wh…

10:35

(a) Show that Eq. 10.57 fo…

01:27

(a) Show that Eq. 10.58 fo…

07:44

A star with mass $3.00 \ti…

09:57

A neutron star is an extre…

02:07

A rotating star collapses …

01:34

Neutron star An extremely …

04:18

Suppose a uniform spherica…

05:09

Under some circumstances, …

02:26

Use conservation of angula…

06:42

Motion in a galaxy. Consid…

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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