When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of a star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius R that is initially rotating at 2.8 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 4.5R. Assume that all of the star's original mass is contained in the shell.

To solve this problem, we need to apply the principle of conservation of angular momentum. The angular momentum of the rotating star before the explosion is equal to the angular momentum of the expanding shell after the explosion. The formula for angular momentum is given by:

L = Iω

Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

The moment of inertia for a solid sphere rotating about its axis is given by:

I = (2/5)MR^2

Where M is the mass of the sphere and R is the radius of the sphere.

Given that the star's mass is initially contained in the shell, we can assume that the mass of the shell after the explosion remains the same. Let's denote the mass of the shell as M.

The initial angular momentum of the star is equal to:

L1 = (2/5)MR^2 * ω1

Where ω1 is the initial angular velocity of the star.

Now, let's find the angular velocity of the expanding supernova shell when its radius is 4.5R. We'll denote this angular velocity as ω2.

The final angular momentum of the shell is given by:

L2 = (2/5)MR^2 * ω2

Since the angular momentum is conserved, we can set L1 equal to L2:

L1 = L2

(2/5)MR^2 * ω1 = (2/5)MR^2 * ω2

Canceling out the common terms, we get:

ω1 = ω2

This means that the angular velocity of the expanding supernova shell is equal to the initial angular velocity of the star.

Therefore, the angular velocity of the expanding supernova shell when its radius is 4.5R is 2.8 revolutions per day.

To find the angular velocity of the expanding supernova shell, we need to apply the law of conservation of angular momentum.

The law of conservation of angular momentum states that the angular momentum of a system remains constant if no external torque acts upon it. In this case, we can consider the star and the expanding supernova shell as a closed system.

The angular momentum of an object is given by the equation:

Angular momentum (L) = moment of inertia (I) * angular velocity (ω)

Since the shell expands while conserving its mass, we can assume that its moment of inertia remains constant.

Given the initial conditions:
Radius of the star (R) = R
Initial angular velocity (ω_initial) = 2.8 revolutions per day

And the final condition:
Radius of the supernova shell (4.5R) = 4.5R
Unknown final angular velocity (ω_final)

Using the conservation of angular momentum, we can write:

I_initial * ω_initial = I_final * ω_final

Since the star is a solid sphere, its moment of inertia (I_initial) can be calculated as:
I_initial = (2/5) * M * R^2, where M is the mass of the star

The mass of the star (M) is assumed to be contained in the supernova shell, so we can consider it constant throughout the explosion.

The moment of inertia of the supernova shell (I_final) can be calculated as:
I_final = (2/3) * M * (4.5R)^2

Now we can plug in the values:

(2/5) * M * R^2 * 2.8 = (2/3) * M * (4.5R)^2 * ω_final

Simplifying the equation:

(2/5) * 2.8 = (2/3) * (4.5^2) * ω_final

1.12 = 20.25 * ω_final

Dividing both sides by 20.25:

ω_final = 0.0554 revolutions per day

Therefore, the angular velocity of the expanding supernova shell when its radius is 4.5R is approximately 0.0554 revolutions per day.