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 1.5 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 3.7R. Assume that all of the star's original mass is contained in the shell.

Question

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 1.5 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 3.7R. Assume that all of the star's original mass is contained in the shell.

Answer 1

conservation of momentum

Io wi = If wf

Io for a sphere...mass m, radius R
If for a sphericalshell, massm, radisu 3R
find wf.

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Answer 2

Why is R1 equal to 1?

Answer 3

To find the angular velocity of the expanding supernova shell, we can use the principle of conservation of angular momentum. According to this principle, the angular momentum of an object remains constant unless acted upon by an external torque.

Let's denote the initial angular velocity of the star as ω₁, and the final angular velocity of the supernova shell as ω₂. The angular momentum L₁ of the star before the explosion is given by:

L₁ = I₁ * ω₁

where I₁ is the moment of inertia of the star. Since the star is a solid sphere, its moment of inertia can be expressed as:

I₁ = (2/5) * M₁ * R₁²

where M₁ is the mass of the star and R₁ is its radius.

After the explosion, all of the star's mass is contained within the expanding supernova shell. Therefore, the final moment of inertia I₂ of the shell can be expressed as:

I₂ = (2/5) * M₁ * R₂²

where R₂ is the radius of the expanding shell.

From the conservation of angular momentum, we have:

L₁ = L₂

I₁ * ω₁ = I₂ * ω₂

Substituting the expressions for I₁ and I₂:

(2/5) * M₁ * R₁² * ω₁ = (2/5) * M₁ * R₂² * ω₂

Simplifying:

R₁² * ω₁ = R₂² * ω₂

Now we can solve for ω₂:

ω₂ = (R₁² * ω₁) / R₂²

Plugging in the known values, ω₁ = 1.5 revolutions per day and R₂ = 3.7R:

ω₂ = (R₁² * 1.5) / (3.7R)²

Simplifying further:

ω₂ = (R₁² * 1.5) / (3.7² * R²)

Finally, to find the angular velocity in revolutions per day, we can divide ω₂ by 2π (since there are 2π radians in one revolution) and multiply by the number of days in a day:

Angular velocity (revolutions per day) = (R₁² * 1.5) / (3.7² * R² * 2π) * 1 day

Simplifying this expression will give you the final answer for the angular velocity of the expanding supernova shell in revolutions per day.

Answer 4

Angular momentum is conserved. That means, in this case,

I1*w1 = I2*w2
where w is the angular velocity and i is the moment of inertia.

The moment of inertia of a solid sphere of uniform density with mass M and radius R1 is
I1 = (2/5)M*R1^2

The moment of inertia of a thin spherical shell of radius R2 is
I2 = (2/3)M*R2^2

The angular velocity w decreases by the ration factor
w2/w1 = I1/I2 = [(2/5)/(2/3)](R1/R2)^2
= (3/5)*(1/3.7)^2 = 0.044

The supernova shell will require 1.5/0.044 = 34 days for a single rotation.

reference:
http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html