Question

Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days. If it were to undergo gravitational collapse to a neutron star of radius 13 km , losing three-quarters of its mass in the process, what would its rotation speed be

Answer 1

Answer:

Its rotation will be 3.89x10⁴ rad/s.

Explanation:

We can find the rotation speed by conservation of the angular momentum:

$L_{i} = L_{f}$

$I_{i}\omega_{i} = I_{f}\omega_{f}$   (1)

The initial angular speed is:

$\omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d}$

The moment of inertia (I) of a sphere is:

$I = \frac{2}{5}mr^{2}$    (2)

Where m is 9 times the sun's mass and r is the sun's radius

By entering equation (2) into (1) we have:

$\frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f}$

$9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}$

$\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s$          

Hence, its rotation will be 3.89x10⁴ rad/s.

I hope it helps you!