I am reposting this question because the previous solution was incorrect. See be
ID: 1482602 • Letter: I
Question
I am reposting this question because the previous solution was incorrect. See below:
Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 9.0×105 km (comparable to our sun); its final radius is 16 km .
If the original star rotated once in 35 days, find the angular speed of the neutron star.
Express your answer using two significant figures.
Expert Answer
pf = final density = 1014 pi
ri =initial radius = 9 x 108 m
rf =final radius = 16 x 103 m
initial mass = mi = pi Vi = (3.1 x 1027 ) pi
final mass = mf = pf Vf = (1.7 x 1013 ) pf = (1.7 x 1013 ) 1014 pi = (1.7 x 1027 ) pi
initial moment of inertia = Ii = (2/5) mi ri2 = (2/5) ((3.1 x 1027 ) pi ) (9 x 108 )2
final moment of inertia = If = (2/5) mf rf2 = (2/5) ((1.7 x 1027 ) pi ) (16 x 103 )2
Ti = initial time = 35 days
Tf = final time
Using conservation of angular momentum
Ii Wi = If Wf
Ii /Ti = If /Tf
(2/5) ((3.1 x 1027 ) pi ) (9 x 108 )2 (1/35) = (2/5) ((1.7 x 1027 ) pi ) (16 x 103 )2 (1/Tf)
(2/5) ((3.1 x 1027 )) (9 x 108 )2 (1/35) = (2/5) ((1.7 x 1027 )) (16 x 103 )2 (1/Tf)
Tf = 6.1 x 10-9 days
I need the answer in rads/sec. I tried converting and got 1.9*10^8, but this answer is incorrect. Help?
Explanation / Answer
Conserve angular momentum:
initial I = final I
For a solid sphere, I = (2/5)mr²
initial = 2 rads / T =
2 / (35days * 24hr/day * 3600s/hr) = 2.076×10^-6 rad/s
(2/5)*m*(9×10^5 km)² * 2.076×10^-6rad/s = (2/5)*m*(16 km)² *
(2/5)m cancels;
= 6568.59 rad/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.