Question

A star rotates with a period of 37 days about an axis through its center. The period is the time interval required for a point on the star's equator to make one complete revolution around the axis of rotation. After the star undergoes a supernova explosion, the stellar core, which had a radius of 2.0 ✕ 104 km, collapses into a neutron star of radius 6.1 km. Determine the period of rotation of the neutron star. SOLUTION Conceptualize The change in the neutron star's motion is similar to that of the skater described earlier in the textbook, but in the reverse direction. As the mass of the star moves closer to the rotation axis, we expect the star to spin Correct: Your answer is correct. . Categorize Let us assume that during the collapse of the stellar core, (1) no external torque acts on it, (2) it remains spherical with the same relative mass d

Answer 1

T = 1.1285 10⁻² day

Step-by-step explanation:

For this exercise the forces in the premiere are internal, so the angular momentum is conserved

L₀ = I₀ w₀

L = I w

L₀ = L

I₀ w₀ = I w

Angular velocity and period are related

w₀ = 2π / T₀

w = 2π / T

The moment of inertia of a sphere is

I₀ = 2/5 M R²

I = 2/5 m r²

If we assume that the mass of the star does not change in the transformation

We substitute

2/5 M R² 2π/T₀ = 2/5 M r² 2π/ T

R² /T₀ = r² / T

T = (r / R)² T₀

T = (6.1 / 2.0 104) 37

T = 1.1285 10⁻² day