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Does an object that only spins have translational kinetic energy?

Since $v=\omega r$ ($\omega$ is angular velocity), if substituting $v$ with $\omega r$, won't that allow an object that only spins to have translational kinetic energy?

Thinking about the other way, won't an object that only slides would have rotational kinetic energy?

Therefore, for an object that only spins, the total energy would be... $$E = \frac{1}{2}m(\omega r)^2 + \frac{1}{2}I\omega^2$$

And for an object that only slides, the total energy would be... $$E = \frac{1}{2}mv^2 + \frac{1}{2}m(v/r)^2$$

But, the second one doesn't make sense, because $E = \frac{1}{2}mv^2$ for object that only slides

I'm confused. Can someone explain about this please?

Andrea Di Pinto
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Anna
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3 Answers3

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For a rigid body, the total kinetic energy is the translational energy of the center of mass (CM) plus the rotational energy about the CM: ${1 \over 2} mv^2 + {1 \over 2} I \omega^2$ for rotation in a plane. (For general motion $T = {1 \over 2} mv^2 +{1 \over 2} \vec \omega \cdot \tilde I \cdot \vec \omega$ where $\tilde I $ is the inertia tensor.) Here, $v$ is the velocity of the CM, not the velocity of the rotating particles. See Goldstein, Classical Mechanics. If the CM is stationary there is no translational velocity and no translational kinetic energy, but there can be rotational energy about the CM. If the CM is moving but there is no rotation about the CM, there is translational kinetic energy but no rotational kinetic energy.

John Darby
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You are confusing different types of motion. If a body only spins but stays at the same point, say $r=0$, then it has no translational kinetic energy for $v = w r = 0$.

If a point mass moves on an orbit, say circular, then one can speak of both, linear and angular energies, which however must be identical $E = mv^2/2 = m(w r)^2/2$.

Nikodem
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An object spinning about a point not the center of mass will have translational kinetic energy since it has momentum.

To find the proportion of kinetic energy due to translation and the proportion due to rotation you must express the kinetic energy at the center of mass, and then the $\tfrac{1}{2} m v_{\rm CM}^2$ part is the translating energy and the $\tfrac{1}{2} I_{\rm CM} \omega^2$ is the rotational energy.

Consider a rod pinned on one end. It has at the center of mass $I_{\rm CM} = \tfrac{m}{12} \ell^2$ and $v_{\rm CM} = \omega \tfrac{\ell}{2}$

$$ \begin{aligned} K & = \tfrac{1}{2} m v_{\rm CM}^2 + \tfrac{1}{2} I_{\rm CM} \omega \\ & = \underbrace{ \tfrac{1}{8} m \ell^2 \omega^2 }_\text{trans} + \underbrace{ \tfrac{1}{24} m \ell^2 \omega^2 }_\text{rot} \\ & = \underbrace{ \tfrac{1}{6} m \ell^2 \omega^2 }_\text{total} \end{aligned}$$

But take the same situation and calculate the kinetic energy at the pivot A with $I_{\rm A} = \tfrac{m}{3} \ell^2$ and $v_{\rm A} = 0$

$$ \begin{aligned} K & = \tfrac{1}{2} m v_{\rm A}^2 + \tfrac{1}{2} I_{\rm A} \omega \\ & = 0 + \tfrac{1}{2} ( \tfrac{1}{3} m \ell^2) \omega^2 \\ & = \underbrace{ \tfrac{1}{6} m \ell^2 \omega^2 }_\text{total} \end{aligned}$$

Same total kinetic energy, but the proportion of translating and rotating isn't evident at all.

In the case where the body is purely translating with $v_{\rm CM} \neq 0$ and $\omega =0$ then the kinetic energy calculation is

$$ \require{cancel} K = \frac{1}{2} m v_{\rm CM} + \cancel{ \frac{1}{2} I_{\rm CM} \omega} $$

John Alexiou
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