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  • 4 years ago

A pulsar is a rapidly rotating neutron star from which we receive radio pulses with precise synchronization, there being one pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. At present, the pulsar in the central region of the Crab nebula has a period of rotation of T = 0.033s, and this is observed to be increasing at a rate of 1.26*10^-5 seconds per year. What is the value of the angular acceleration in rad/s^2? TL;DR: T= 0.033s, T is increasing at a rate of 0.0000126s/year. Angular acceleration = ?

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  1. TuringTest
    • 4 years ago
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    \[\large\omega=\frac{2\pi}T\]Where omega is the angular velocity and T is the period. The angular acceleration is linear, so it is given by\[\large\alpha=\frac{\Delta\omega}{\Delta t}=\frac{\frac{2\pi}{T+\Delta T}-\frac{2\pi}T}{t}\]where DeltaT is the change in the period, and t is the time it takes for that change to occur (in years)

  2. anonymous
    • 4 years ago
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    Aha! I totally forgot that omega = 2pi/T. Thanks! So angular velocity at t=1 year minus angular velocity at t=0 and divide by t and that's the answer, I got \[-2.2*10^{-9}rad/s^{2}\]

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