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I know that rotational kinetic energy is $\frac{1}{2}I\omega^2$. Therefore, the rotational kinetic energy will depend on the moment of inertia.

I came to the conclusion that since both have the same mass, both should have the same moment of inertia as
$$I = \sum_\text{over all mass} m r^2$$

But the answer says that since the hollow cylinder has greater moment of inertia, it has greater rotational kinetic energy.

How can the hollow cylinder have greater moment of inertia even though the masses of both the bodies is same?

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  • $\begingroup$ I think you mean to add the qualifier "at the same angular speed". $\endgroup$ – Selene Routley Dec 13 '13 at 11:44
  • $\begingroup$ Google "radius of gyration" and you will find that $I = m r_{gyr}^2$ $\endgroup$ – John Alexiou Dec 13 '13 at 13:32
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The answer to that is because the moment of inertia is not the same for the solid cylinder than for the hollow one.

As you write the formula for the moment of inertia, it depends on the distribution of the mass. The further away the mass is from the rotation axis, the more contributes to the moment of inertia (as in distance squared $r^2$).

So, since the hollow cylinder has all it's mass at the border in comparison with the solid one which distributes all it's mass from the center (with very small contribution) to the border, it has a higher moment of inertia and thus more rotational energy.

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You've failed to take into consideration that $r$ is the radius of a piece of mass $\delta m$ rotating about an axis. So that the product of $\delta m$ and $r$ must be done first before the sum, or more probably the integral.

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