# Rotational Energy, Moment of Inertia Problem

1. Apr 2, 2014

### chongkuan123

1. The problem statement, all variables and given/known data

Energy is released by the Crab Nebula at a rate of about 5×10^31W, about 105 times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. This object rotates once every 0.0331 s, and this period is increasing by 4.22×10^−13s for each second of time that elapses.

a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star.

2. Relevant equations

KE = Iw^2 , w = 2pi/ T , P = dE/dT

3. The attempt at a solution

KE = Iw^2 , w = 2pi/ T , so KE = 1/2 I 4 pi^2 T^-2
take the derivative of that i get P = -4 I pi^2 T^-3
set it equal to -5×10^31
solve for I and i got 4.59*10^25 which is wrong :(

2. Apr 2, 2014

### slider142

Hmm, as far as I know, power is dE/dt, not dE/dT. The period T is itself a function of time in this problem, T(t), so when you take the derivative of K(t), you must use the chain rule (the derivative in your answer attempt is missing the required extra factor of T'(t)).

3. Apr 2, 2014

### MortalWombat

I actually solved a quite similar problem last semester in my astrophysics course, maybe that will help you:

My problem was like this: given that $\tau = 33$ ms and $\Delta \tau = 1.3 \cdot 10^{-5} s/year$, and given the mass $M = 1.44 M_{sun}$ and the radius $R = 12 km$, we calculate

$$E_{rot} = \frac{1}{2} I_{sphere} \omega^2 = \frac{1}{2} \frac{2}{5} M R^2 \omega^2 = \frac{1}{5} M R^2 \frac{4 \pi^2}{\tau^2}, \quad using \quad \omega = 2\pi f = 2 \pi \frac{1}{\tau}$$
$$E_{rot} = 2.9888 \cdot 10^{42} J$$
$$\Rightarrow \frac{\Delta E_{rot}}{s} = \frac{4}{5} M R^2 \pi^2 \left( \frac{1}{\tau_1} - \frac{1}{\tau_2} \right), \quad where \quad \tau_2 = \tau + \Delta \tau = \tau_1 + \Delta \tau$$
Now we use that $E_{rot}$ must be equal to $E_{pot}$ (I think that was Virial's theorem)
$$E_{rot}=\frac{\omega^2}{2} \frac{2}{5} M R^2 \overset{!}{=} \frac{M^2 G}{R} \frac{3}{5} = E_{pot}$$
$$\rightarrow \omega_{max} = \sqrt{\frac{3 M G}{R^3}} \approx 15180 s^{-1}$$

4. Apr 2, 2014

### chongkuan123

So T'(t) is the increasing period 4.22×10^−13s?

5. Apr 2, 2014

### slider142

Right. :) I assume the problem intends T(0) = 0.0331 s and T'(0) = 4.22x10^-13.

Last edited: Apr 2, 2014
6. Apr 2, 2014

### haruspex

... except that T' is dimensionless.

7. Apr 2, 2014

### slider142

Good point, that!