# Does the angular kinetic energy of a point mass always correspond to its linear kinetic energy? [duplicate]

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?

• What angular motion is there to speak of if the particle is moving in a straight line? Commented May 2 at 7:52
• 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. Commented May 2 at 11:22
• The two equations describe the same motion. just substitute $mR^2$ for $I$ and $v^{2}/R^2$ for $\omega ^2$. Commented May 2 at 13:21
• Bob, your comment is an answer, please repeat it as answer. Commented May 2 at 14:07

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.