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a neutron star with a moment of inertia of 18 x 10³⁷ kg m² is spinning with a period of 0.06 s. what is it's rotational kinetic energy (leave as x 10³⁷)?

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The rotational kinetic energy of the star is determined as 986.6 x 10³⁷ J.

How to calculate the rotational kinetic energy of the star?

The rotational kinetic energy of the star is calculated by applying the following formula as shown below.

K.E = ¹/₂Iω²

where;

  • I is the moment of inertia
  • ω is the angular speed

The given parameters include;

moment of inertia, I = 18 x 10³⁷ kg m²

angular speed, ω = 1 rev / 0.6 s = 2π / 0.6 s = 10.47 rad/s

The rotational kinetic energy of the star is calculated as follows;

K.E = ¹/₂Iω²

K.E = ¹/₂(18 x 10³⁷)(10.47)²

K.E = 986.6 x 10³⁷ J

Thus, the rotational kinetic energy of the star is 986.6 x 10³⁷ J.

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