## Related questions with answers

Question

A neutron star of mass

$2 × 10^{30} kg$

and radius 10 km rotates with a period of 0.02 seconds. What is its rotational kinetic energy?

Solution

VerifiedStep 1

1 of 2We write the equation for the rotational kinetic energy of a rigid body and then, we continue to substitute $I$ for the inertia of a disk, and $\omega$ for $2\pi / T$ (where $T$ is the given period).

$\begin{align*} K&=\dfrac{1}{2} I \omega^2 \\ I&=\dfrac{2}{5} MR^2 \\ \omega&=\dfrac{2\pi}{T} \\ \implies K&=\dfrac{1}{2} \left(\dfrac{2}{5} MR^2\right) \left(\dfrac{2\pi}{T}\right)^2 \\ &=\dfrac{4\pi^2}{5} \cdot \dfrac{MR^2}{T^2} \\ &=\dfrac{4\pi^2}{5} \cdot \dfrac{\left(2\cdot 10^{30}\right) \left(10 \cdot 10^3\right)^2}{\left(0.02\right)^2} \\ &=3.95 \cdot 10^{42} \text{ J} \end{align*}$

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