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the crab pulsar spins with a period of 33 ms its radius is about 25 km its mass is roughly 4 x 10^30kg (around twice the mass of the sun how much rotational kinetic energy does it have?​

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Answer:

1.81×10⁴³J [ or 2×10⁴³J if you care about sigfig rules]

Explanation:

Rotational Kinetic Energy = (1/2)Iω²

  • I [solid sphere] = (2/5)(mass)(radius)²
  • ω = period of rotation [in rad/seconds]

Energy of the Crab Nebula = (1/2)(2/5)(4×10³⁰)(25x10³)²((2π)/(33×10⁻³))² = 1.81×10⁴³J

The formulas can basically all be looked up (or even derived) if you can't remember them. Note that ω is equal to 2π/(period [in seconds]) and not one of its other whacky forms.

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