Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius 1.0% of its existing radius. Assuming the lost mass carries away no angular momentum, what would the Sun's new rotation rate be? (Take the Sun's current period to be about 30 days.) What would be its final KE in terms of its initial KE of today?

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## Answers ( 2 )

Answer question

Now, the formula for angularmomentum is:

L= Iw

So L before = L after:

I

_{1}w_{1}= I_{2}w_{2 }The trick to solving these isto figure out what the change in moment of inertia is, and thenapply the concept of conservation of angular momentum toit.The sun and the white dwarf will both be spherical

A uniform sphere (p. 223) is

^{2}/_{5}mr^{2}:I

_{1}=^{2}/_{5}mr^{2}=^{2}/_{5}MR^{2}Where M and R are the current massand radius

And

w

_{1}=1 rev/30daysAfter, the sun has half the mass,and 1/100 the radius:

I

_{2}=^{2}/_{5}(^{M}/_{2})(^{R}/_{100})^{2}= (^{2}/_{5}MR^{2})/(20,000)Plugging into conservation ofangular momentum:

I

_{1}w_{1}= I_{2}w_{2 }(^{2}/_{5}MR^{2})(1 rev/30 days) =(^{2}/_{5}MR^{2})/(20,000)w_{2}(1 rev/30 days) = w

_{2}/_{(20,000)}w

_{2}=(20,000)(1 rev/30 days)w

_{2}=(20,000)(1 rev/30 days)(2p radians/revolution)(1day/243600 sec) =0.048481368 rad/sec = .048 rad/sec^{.}As for the ratio of kineticenergies,

Before:

E

_{k rot}=^{1}/_{2}I_{o}w_{o}^{2}=^{1}/_{2}(^{2}/_{5}MR^{2})(1 rev/30 days)^{2}After:

E

_{k rot}=^{1}/_{2}I_{o}w_{o}^{2}=^{1}/_{2}(^{2}/_{5}(^{M}/_{2})(^{R}/_{100})^{2})((20,000)(1rev/30 days))^{2}E

_{krot}=^{1}/_{2}I_{o}w_{o}^{2}= (20,000)^{2}/(20,000){^{1}/_{2}(^{2}/_{5}MR^{2})(1rev/30 days)^{2}}Which is just 20,000 times what it was before