Moment of inertia of a neutron star

  • #1
kent davidge
933
56

Homework Statement



kJFqrs7.jpg


Homework Equations



rotational kinetic energy = 0.5 I ⋅ M ⋅ ω², where I is the moment of intertia, M the mass and ω the angular speed

The Attempt at a Solution



T = period of revolution
K = kinetic energy associated with rotational moviment
Since T increases with time according the text,

T = ƒ(t) = 3.31(10^-2) + 1.40(10^-14)t
and I = 2K / ω²M = 2 ⋅ 5(10^31)t / [2π / [3.31(10^-2) + 1.40(10^-14)t]² ⋅ 1.4(1.99)(10^30)]

I'm not sure about this answer. Could someone review it for me?
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
45,410
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rotational kinetic energy = 0.5 I ⋅ M ⋅ ω², where I is the moment of intertia, M the mass and ω the angular speed
Quick comment: Get the mass out of that formula (the mass is already accounted for in the moment of inertia): rot KE = 0.5 I ⋅ ω²
 
  • #3
gneill
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Hint: The power will be equal to the rate of change of the rotational energy with respect to time. So think about differentiation of the rotational energy equation, taking into account that ω is a function of time.
 
  • #4
kent davidge
933
56
Doc Al Ok.
gneill Would it becomes dk/dt = - 5 × (10^31) = - 4.22(10^-13)4π²⋅I / (4.22(10^-13)t + 0.0331)³ and for t = 0, I ≅ 1.09 × 10^25 ?
 
  • #5
gneill
Mentor
20,945
2,886
Let's use symbols and avoid plugging in any numbers until the end.

The power: ##p = 5 ⋅ 10^{31}~W ##
The current period: ##\tau = 0.0331~s ##
The rate of change of the period: ##\frac{Δ\tau}{Δt} = -4.22⋅10^{-13} ##

Write the simple relationship for the angular speed ω given the period ##\tau##. Assume that ω and ##\tau## are functions of time, ##ω(t)## and ##\tau(t)##, and differentiate (hint: chain rule). You'll end up with ##\frac{d ω}{dt}## in terms of ##\tau## and ##\frac{d \tau}{dt}##. Note that a useful approximation is ##\frac{d \tau}{dt} ≈ \frac{Δ\tau}{Δt}##

I suspect that that should be sufficient hints...
 
  • #6
kent davidge
933
56
Ok. I will do that.
 

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