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
