Suppose there is a particle with mass $m$ moving with constant velocity $v$ towards the right on the $x$-axis. The position of the particle at one instance of time is shown below:

enter image description here

Here, $r = \sqrt{x^2+d^2}$, and $\theta$ is the angle formed between the y-axis and the line connecting the particle's current position and the point $(0, d)$ ($\theta = 0$ when $x = 0$ and $\theta = \pi/2$ when $x = \infty$).

I want to find the rotational kinetic energy of the particle about the point $(0,d)$.

First, the particle's moment of inertia is:

$$I = mr^2$$

To find the particle's angular velocity about the point $(0,d)$, I do the following:

$$x = d\tan(\theta)$$ $$ v = \frac{dx}{dt} = d\sec^2(\theta)\frac{d\theta}{dt} = d\frac{r^2}{d^2}\frac{d\theta}{dt} = \frac{r^2}{d}\frac{d\theta}{dt}$$

Since the angular velocity $\omega$ is $\frac{d\theta}{dt}$,

$$\omega = \frac{d\theta}{dt} = \frac{d}{r^2}v$$

The rotational kinetic energy $K_r$ is:

$$K_r = \frac{1}{2}I\omega^2 = \frac{1}{2}(mr^2)(\frac{d}{r^2}v)^2 = \frac{1}{2}m\frac{d^2}{r^2}v^2$$

This says that the rotational kinetic energy decreases as the particle goes to infinity. However, this should be equivalent to the linear kinetic energy, which is $K = \frac{1}{2}mv^2$ (which is constant). Where am I going wrong?

  • $\begingroup$ The definition of the rotational kinetic energy that you are using is missing a piece coming from the longitudinal velocity $\endgroup$
    – nwolijin
    Dec 23, 2020 at 20:37

3 Answers 3


There is nothing wrong with your calculation. At large distance there is not much rotation around that point. However this is not the total kinetic energy of the particle. What you call rotational KE is the KE associated with the component of velocity perpendicular to $ \vec{r}$. This is simply $1/2 mv^2 _{perp}$. You can call it rotational KE if you wish. But there is also $1/2 mv^2 _{parallel}$, associated with the component of velocity along $ \vec{r}$. The total KE is the sum $$ KE =1/2 m(v^2 _{perp} +v^2 _{parallel})$$ of the two and is constant (equal to $1/2 mv^2$). As r increases the perpendicular contribution decreases and the parallel increases, as expected.


In this case of a particle with a constant velocity, the angular momentum is also constant: $\mathbf L = \mathbf r \times \mathbf p \implies |\mathbf L| = dmv$. While the expression $L = I\omega$ works also here, the split in the product of that 2 components is artificial.

Moment of inertia is related to how much an object opposes to a change of its angular velocity. And here the angular velocity is decreasing spontaneously!

The same for rotational kinetic energy, that is meaningful when an object has a translational kinetic energy of its center of mass, and also rotates around it.


The rotational kinetic energy should go to zero. This also makes sense intuitively, since for very large distances $\theta$ is essentially zero and constant. Only the total kinetic energy, which consists of a rotational and a radial component, converges to the usual notion of kinetic energy (cartesian).

Let's have a look at the total kinetic energy:

$E_{tot}= E_{rot} + E_{rad} = \frac{1}{2}I \omega^2 + \frac{1}{2}m \dot{r}^2$

You already computed the rotational part:


To compute the radial kinetic energy we have to derive the time derivative of $r=\sqrt{d^2+x^2}$:

$\dot r = \frac{x \dot x }{\sqrt{d^2+x^2}}= \frac{\sqrt{r^2-d^2}v}{r} \quad \to \quad E_{rad} = \frac{1}{2}m \dot r ^2 = \frac{r^2-d^2}{2r^2}v^2 $

$ \quad \xrightarrow[r\to \infty]{} \frac{1}{2}mv^2$

We also see that as r goes to infinity the radial kinetic energy converges to the cartesian kinetic energy.

To sum up, the total kinetic energy is given by:

$E_{tot}=E_{rot} + E_{rad}= \frac{1}{2}I \omega^2 + \frac{1}{2}m \dot{r}^2= E_{rot}=\frac{md^2}{2r^2}v^2 + \frac{r^2-d^2}{2r^2}v^2 $

And the limiting behavior is:

$\lim_{r\to\infty} E_{tot} = \frac{1}{2}m v^2$


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