Calculating New Rotation Period of a Star After Shrinking Diameter

[Hollywood]

The mass of a star is 1.4 × 10^31 kg and it performs one rotation in 26.3 day. Find its new period if the diameter suddenly shrinks to 0.55 times its present size. Assume a uniform mass distribution before and after.

I don't know what I am doing wrong here:

Iw(intial)=Iw(final)

I(intial)=(2/5)(1.4 × 10^31 kg)(1)^2
= 5.6 x 10^30

I(final)=(2/5)(1.4 × 10^31 kg/.55)(.55)^2
=9.317 x 10^29

(5.6 x 10^30)(26.3 day) = (9.317 x 10^29)(w)

w(final)=158.1 day

This answer is wrong. Can someone tell me what I am doing wrong here?
Thank You!

 

[Vandrerol]

Hi Hollywood!

Now I'm not ever sure of anything I say or do, but I would try to equate the initial and final angular momentums L = m * v * r since they should be conserved.
Where the tangential velocity v = (2 * pi * r )/ TSo the initial Lo = (Mo * 2pi * Ro ^2)/To
And the final Lf= (Mf * 2pi * Rf ^2)/Tf

Since the mass of your star does not change (and hopefully, neither does '2*pi'), you can just write

Tf * Ro ^2 = To * Rf ^2

Since Ro = 1 and Rf = .55

Tf = 26.3 * (.55^2) = 7.9...days

Hope that's correct!

 

[Moetasim]

I would first like to commend you for attempting to solve this problem using the principles of conservation of angular momentum. However, there are a few errors in your approach that may have led to the incorrect answer.

Firstly, the equation Iw(initial) = Iw(final) is only applicable for a system where the moment of inertia (I) and angular velocity (w) remain constant. In this case, the moment of inertia changes as the diameter of the star shrinks, so we cannot use this equation.

Secondly, the moment of inertia for a uniform spherical object is given by I = (2/5)mr^2, where m is the mass and r is the radius. In your calculation of I(final), you have used the mass and diameter instead of the mass and radius. This may have led to a significantly different value for the moment of inertia and hence the incorrect answer.

To correctly solve this problem, we need to use the equation for moment of inertia as I = (2/5)mr^2. Using the given mass and initial radius, we can calculate the initial moment of inertia as:

I(initial) = (2/5)(1.4 x 10^31 kg)(1)^2 = 2.8 x 10^31 kg m^2

Next, we need to calculate the final radius of the star after it shrinks to 0.55 times its initial size. This can be done by multiplying the initial radius by 0.55:

r(final) = 0.55(1) = 0.55

Now, we can use this final radius to calculate the final moment of inertia as:

I(final) = (2/5)(1.4 x 10^31 kg)(0.55)^2 = 2.1785 x 10^31 kg m^2

Finally, we can use the equation w(initial)I(initial) = w(final)I(final) to find the new rotation period (w(final)):

w(final) = (w(initial)I(initial)) / I(final)

Plugging in the values, we get:

w(final) = (2.8 x 10^31 kg m^2)(26.3 day) / (2.1785 x 10^31 kg m^2) = 33.9 day

Therefore, the new rotation period of the star after shrinking its diameter to