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Let's assume a point mass is moving with constant speed on a linear trajectory. Its angular kinetic energy is given by the formula: $$E=\frac{1}{2}I\omega^2$$ Does this energy correspond to the linear kinetic energy of the point $$E=\frac{1}{2}mv^2$$ or to the energy of just one component (the perpendicular component to the radius) of the speed?

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  • $\begingroup$ What angular motion is there to speak of if the particle is moving in a straight line? $\endgroup$ Commented May 2 at 7:52
  • $\begingroup$ A particle moving in a straight line is actually rotating around any point not lying on the line, because the angle formed by the particle with the point changes over time. $\endgroup$
    – Antozoa
    Commented May 2 at 11:22
  • $\begingroup$ The two equations describe the same motion. just substitute $mR^2$ for $I$ and $v^{2}/R^2$ for $\omega ^2$. $\endgroup$
    – Bob D
    Commented May 2 at 13:21
  • $\begingroup$ Bob, your comment is an answer, please repeat it as answer. $\endgroup$
    – trula
    Commented May 2 at 14:07

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Let's assume the point mass is moving at constant velocity $v$ and does not pass through the origin. Let $b$ be its closest distance to the origin and $\theta$ be the angle between its velocity and position vectors. This gives $$\frac{1}{2}I\omega^2 = \frac{1}{2}mr^2\omega^2 = \frac{1}{2}m\left(\frac{b}{\sin\theta}\right)^2\left(\frac{v\sin\theta}{r}\right)^2 = \frac{1}{2}m\left(\frac{b}{r}\right)^2 v^2 \neq \frac{1}{2}mv^2.$$ Therefore, it underestimates the kinetic energy as it does not account for the radial component of the velocity. So it is incorrect and $mv^2/2$ is the correct formula. As an extreme example, consider the case where $b=0$, i.e. the trajectory passes through the origin. $I\omega^2/2$ will then give zero even though the mass is moving. The reason the formula $I\omega^2/2$ does not apply here is because it requires that $\mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}$. This condition is not satisfied here as the point mass is not moving in a pure rotation about the origin.

If the point mass were instead moving in a circular trajectory, then $\mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}$ holds and both formulae will be valid and equivalent.

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