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Suppose a star the size of our Sun, but of mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 12 days. If it were to undergo gravitational collapse to a neutron star of radius 12 km, losing 3/4 of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times.

(a) Assume that the thrown-off mass carries off no angular momentum, in rev/s

(b) Assume that the thrown-off mass carries off its proportional share (3/4) of the initial angular momentum, in rev/s

Respuesta :

Mindaka
You have two different phases of the star: 1 the Sun-like phase and 2 the neutron star phase. The given quantities are:
r₁ = r(Sun) = 695700km
m₁ = 8 Msun
f₁ = 1 rev / 12 days
m₂ = [tex] \frac{1}{4} [/tex]·m₁

First thing, you need to transform the frequency in units of revolution/seconds
f₁ = 1 rev / (12·24·60·60) = 1 rev / 1036800 s

and then into angular velocity through the formula
ω₁ = 2πf = 6.06E-6 rad/s

a) If the angular momentum stays the same: L₁ = L₂
where L = I·ω    
and the momentum of inertia I is given by I = m·r²

Therefore, substituting we have:
m₁·r₁²·ω₁ = m₂·r₂²·ω₂

And we can find:

ω₂ = [tex] \frac{r1²w1}{0.25r2²} [/tex]

(remember m₂=[tex] \frac{1}{4} [/tex]·m₁ so we can cancel out the two m₁)

We obtain:
ω₂ = (695700²·6.06E-6)/(0.25·12²) = 81473.1 rad/s

we can transform it back into frequency:
f₂ = ω₂/2π = 1 rev / 12967 s = 7.7e-5 rev/s

b) If L₂ = [tex] \frac{1}{4} [/tex]·L₁
we do expect an angular velocity 4 times smaller.
Using the same formulas as above:

[tex] \frac{1}{4} [/tex]·m₁·r₁²·ω₁ = [tex] \frac{1}{4} [/tex]·m₁·r₂²·ω₂

ω₂ = [tex] \frac{r1²w1}{r2²} [/tex] =
     = [(695700²·6.06E-6)(12²) = 20368 rad/s

and 

f₂ = 1 rev / 3242 s = 3.08e-4 rev/s



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