Conservation of angular momentum (Unit 20) requires that the...

Conservation of angular momentum (Unit 20) requires that the product of the speed at which an object is orbiting times the distance from the point around which it is orbiting remain constant. If the Sun is currently orbiting at $220 \mathrm{km} / \mathrm{sec}, 8 \mathrm{kpc}$ from the center of the Milky Way, how fast did the material that formed the Sun rotate if the gas began 400 kpe distant (about halfway to the next large galaxy)?

Key Concepts

Angular Momentum

Angular momentum is a physical quantity that represents the amount of rotational motion an object possesses. It is calculated as the product of the object's mass, its velocity, and the perpendicular distance from the axis of rotation. This concept helps in analyzing systems where rotation is involved and is crucial in understanding how motion responds to changes in distribution of mass or position relative to a fixed point.

Conservation of Angular Momentum

This principle states that in the absence of an external torque, the total angular momentum of a system remains constant. It is a fundamental law in physics that applies to a variety of systems, from spinning objects on Earth to large astronomical bodies in orbit, and it explains how changes in the radius of orbit affect orbital speed.

Orbital Mechanics

Orbital mechanics involves the study of the motions of objects under the influence of gravitational forces, particularly the relationships between orbital speed and distance from the center of mass. In contexts such as planetary systems or galaxies, understanding orbital mechanics is essential since it allows us to predict how changes in distance affect speed in order to conserve angular momentum.

Transcript

00:01 Question number 36 asks us to find the angular velocity of a neutron star whose current diameter is 20 kilometers.

00:10 We're given that the star's original radius was equal to that of the suns.

00:17 Our initial is 7 times 10 to the 8 meters.

00:23 We know its final diameter, but its final radius would be equal to that 20 kilometers.

00:30 Divided by 2 or 10 kilometers.

00:35 Keeping our units consistent, let's convert that to a distance of 1 times 10 to the 4 meters in radius.

00:44 We also know that its initial mass was equal to 1 .5 times that of the sun.

00:51 One solar mass is equal to 1 .99 times 10 to the 30 kilograms, 1 .5 times that much.

01:01 Mass is going to be 2 .99 times 10 to the 30 kilograms.

01:09 And we're also given that its initial rotation rate was equal to that of the sun, being about one revolution per month.

01:20 That's going to be our frequency.

01:22 What we want to find in this problem is its final angular velocity, omega sub -f.

01:29 We're given the initial frequency but to find its initial angular velocity we take two times pi times that initial frequency there are two pi radians in one revolution we can convert this one revolution per month into an angular velocity involving radiance so canceling out the revolutions that gives us an initial angular velocity of 2 pi radians per month.

02:06 Now, we have our known values.

02:09 Let's see if we can solve for final angular velocity.

02:14 By the conservation of angular momentum, our initial angular momentum, l sub i, will be equal to the final angular momentum, l sub f.

02:25 Angular momentum is equal to an object's moment of inertia times its angular velocity.

02:34 So, since our initial and final angular moment are equal to each other, our initial moment of inertia and initial angular velocity multiplied together should be the same as the product of the final moment of inertia and the final angular velocity.

02:54 What we want to find in this problem is our final angular velocity...

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