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 1014times 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 7.0×105km(comparable to sun) its final radius is 16km. If the original star rotated once in 30 days, find the angular speed of the neutron star.

Short Answer

Expert verified

Hence, the final angular speed is 4.6×103rad/s.

Step by step solution

01

Conservation of angular momentum.

If a system experiences no external torques from the environment, the total angular momentum is conserved.

ΔLt=0

Apply the law of conservation of angular momentum to a system whose moment of inertia changes gives:

Ijωj=lfωf=constant

02

The given data in the question

Given that, initial radius of the original star is Rj=7.0×105km, and the final radius of the neutron star is Rf=16km. The angular speed of the original star is

ωi=1rev30days2πrad1rev1day24h1h3600s=2.424×10-6rad/s

The star is modelled as a uniform, solid, rigid sphere both before and after the collapse.

03

The angular speed of the star.

Let the initial state of the system before the collapse and the final sate be after the collapse.

Consider there are no astronomical bodies are in the vicinity of the star, no forces or torques are exerted on the star. Thus, apply conservation of momentum principle from equation to the star between the initial and final states:

Ijωi=lfωf25MRi2ωi=25MRf2ωfRi2ωi=Rf2ωf

Solve for ωfas follows:

ωf=ωiRiRf2=2.424×10-67.0×10516=4.6×103rad/s

Hence, the final angular speed is 4.6×103rad/s.

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