Question #291247

A star having rotational inertia of 10 1049 kgm2 is rotating at an angular speed of 2.0 revolutions per month about its axis. The only force on it is the force of gravitation. When its nuclear fuel is exhausted, it shrinks to a neutron star having rotational inertia of 6.0 x 1048 kgm². Determine the angular speed of the neutron star in revolutions per month.

1
Expert's answer
2022-01-27T13:17:26-0500

Rotational inertia

I1=10×1049kgm2I_1=10\times10^{49}kgm^2

I2=6×1048kgm2I_2=6\times10^{48}kgm^2

w1=2π×230×24×60×60rad/secw_1=\frac{2\pi\times2}{30\times24\times60\times60}rad/sec

w1=4.84××106rad/secw_1=4.84\times\times10^{-6}rad/sec

We know that

I1w1=I2w2I_1w_1=I_2w_2

w2=I1w1I2w_2=\frac{I_1w_1}{I_2}


w2=10×1049×4.84×1066×1048=8.066×105rad/sec2w_2=\frac{10\times10^{49}\times4.84\times10^{-6}}{6\times10^{48}}=8.066\times10^{-5}rad/sec^2


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS