Crab Nebula Pulsar: Angular Acceleration Explained

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SUMMARY

The Crab Nebula pulsar is a rapidly rotating neutron star with a rotation period of T = 0.033 seconds, which is increasing at a rate of 1.26 x 10^-5 seconds per year. The discussion centers on calculating the pulsar's angular acceleration, utilizing the relationship T = 2π/ω, where ω represents angular velocity. The participants confirm that while the period is changing, the angular speed remains relatively constant during a single revolution due to the slow rate of change in the period.

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A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. The pulsar in the Crab nebula has a period of rotation of T=.033s that is increasing at the rate of 1.26 x 10^-5 s/y

a)What is the pulsar's angular acceleration?
I know that T=2pi/w when omega is constant. Does it make sense to that that T(t)=2pi/w(t) ? If this is correct then I can get the answer, but even if it is correct I'm not sure why it works.
 
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I can't do the math. Just remember that the beams are emitted from the magnetic poles of the star, so you can expect that once in a while the orientation will be such that we receive two pulses per revolution. I don't know if any currently known ones are like that, though.
 
Judging by the context of the problem this is irrelevant.
 
Skomatth said:
I know that T=2pi/w when omega is constant. Does it make sense to that that T(t)=2pi/w(t) ?
Makes sense to me. Note that the rate of change of the period is so slow that for all practical purposes the angular speed hardly changes during one revolution.
 

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