Question

nder 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 6.0×105 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.

Answer 1

Answer:

2.92 * 10³ rad/s

Explanation:

Given:

Initial Radius of Original Star (Ri) = 6.0 * 10^5 km

Final Radius of Neutron Star (Rf) = 16km

Angular Speed = 1 revolution in 35 days

We need to convert this to rad/s

To do that, we first convert to rad/day

i.e (1 rev/35 days) * (2π rad/ 1 rev)

We then convert the days to hour

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours)

Finally, we convert the hour to seconds (3600 seconds makes an hour)

i.e. (1 rev / 35 days) * (2π rad/ 1 rev) * (1 day / 24 hours) * (1 hour/ 3600 sec)

Angular Speed = 2π rad/ 3024000 secs

Angular Speed (wi) = 2.079 * 10^-6rad/s

From the question, we're asked to calculate the angular speed of the neutron star (wf)

Applying law of conservation of angular momentum to a system whose moment of Inertia changes, we have

Ii*wi = If * wf ----------------- Formula

Where Ii and If are the initial and final Inertia of the star

The relationship between Inertia and Radius of each object is I = 2/5MR²

So, Ii = 2/5(MRi²) and If = 2/5(MRf²)

Substitute the above in the formula quoted

We have 2/5(MRi²)wi = 2/5(MRf²)wf ---------------- Divide through by 2M/5

We are left with, Ri²wi = Rf²wf

Make wf the subject of the formula

wf = wi * (Ri/Rf)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km)²

wf = 2.079 * 10^-6rad/s * (6.0 * 10^5 km/16km) * (6.0 * 10^5 km/16km)

wf = 2.92 * 10³ rad/s