I'm going to illustrate this confusion of mine in a thought experiment (poorly):

enter image description here

Here we have origins for $\vec r_2, \vec r_1,$ and $\vec r_0$.

What's undeniable is that this wheel is spinning in uniform circular motion on it, as I am defining it. There is no tangent acceleration on the particles on this wheel. Here we have $3$ different origins at different places, which have their own distance vectors $\vec r_0, \vec r_1, \vec r_2$ to that point. The particle is experiencing a centripetal force colored in yellow. From the origin as the center of mass of the wheel, the radius vector and the force vector would be anti-parallel and thus no torque would be active on the particle. However, according to these origins I've colored, there is a torque on the particle at each origin with the exception of $r_0$, although in the next instant of time there will be once the position vector changes as it follows the particle. This notion that it's feeling a torque with its only active force on that particle being the centripetal force bothers me. The particle on the wheel is not actually experiencing a torque! How could it? It doesn't actually, or at least relative to the center of the wheel, but it does relative to the radius vectors? I know torque is different relative to your origin, but how can we have this scenario I've underscored? It makes me feel like torque can give unreliable interpretations on the motion of this rotating particle.

And, in addition, like its linear analog, applying a torque on an object will give it energy, just like applying a force on an object will give it energy. But the torque here depends on the vantage point, so what it seems like to me is that the wheel is gaining different amounts of energy from different vantage points depending on the torque it's experiencing according to each radius vector (or vantage point/origin).

How do I make sense of all this? The kinetic energy on the wheel has to be uniform regardless of vantage point.

  • $\begingroup$ A wheel spinning in uniform circular motion has, by definition, no net torque on it. $\endgroup$ Nov 23, 2017 at 15:27
  • $\begingroup$ I agree emphatically. This highlights my question though. According to those vantage points, where the radius vectors and the force vector on that particle are not antiparallel or parallel, the particle appears to experience a torque. Yet it doesn't actually. So what's going on there? $\endgroup$
    – sangstar
    Nov 23, 2017 at 16:01
  • 3
    $\begingroup$ I don't see the connection between torque and kinetic energy. Anyway, kinetic energy does depend on the frame of reference. A ball being juggled on a train has different K.E. if viewed from the train or the ground. $\endgroup$ Nov 23, 2017 at 17:54
  • 1
    $\begingroup$ Torque is only an expression of a force at a distance and doesn't stand on its own. See physics.stackexchange.com/a/367347/392. $\endgroup$ Nov 23, 2017 at 17:58
  • $\begingroup$ Read also this answer to understand the fundamental geometry of mechanics. How positional information enters into the common concepts of velocity, momentum and torque. $\endgroup$ Nov 23, 2017 at 18:00

3 Answers 3


The answer is that rotational kinetic energy is not, in general, invariant under coordinate transforms. What is invariant is the rotational kinetic energy about the center of mass. To see why, look at the kinetic energy in polar coordinates: $$ K = \frac{1}{2} m \dot{r}_{\mathrm{CM}}^2 + \frac{1}{2} m r_{\mathrm{CM}}^2 \dot{\theta}_{\mathrm{CM}}^2 + \frac{1}{2} I_{0} \omega^2 . $$ The third term is the moment of inertia around the center of mass ($I_0$) and $\omega$ is the rotational speed around the center of mass in the center of mass frame.

Both the second and third terms are considered rotational kinetic energy, but only the third is invariant under changes of reference frame. Consider, for example, if the center of mass is moving along the radial direction (or instantaneously at the origin), then the second term is $0$. Just shift the origin to a different position, though, and the second term has rotational kinetic energy.

This is actually closely related to the parallel axis theorem, where $$I_{\mathrm{tot}} = m r_{\mathrm{CM}}^2 + I_0.$$

Another way to see this is to write the kinetic energy in terms of angular momentum, which is different depending on which origin you choose, but is conserved by inertial motion: $$K = \frac{1}{2}m\dot{r}_{\mathrm{CM}}^2 + \frac{L_{\mathrm{CM}}^2}{2 m r_{\mathrm{CM}}^2} + \frac{1}{2} I_0 \omega^2.$$ For inertial motion the path followed is given by: \begin{align} \vec{r} & = (x_0 + v_x t)\hat{i} + (y_0 + v_y t)\hat{j} \Rightarrow \\ r & = \sqrt{\left(x_0 + v_x t\right)^2 + \left(y_0 + v_y t\right)^2} \\ \theta & = \operatorname{atan2}\left(y_0 + v_y t, x_0 + v_x t\right). \end{align} Since $L_{\mathrm{CM}}$ is a constant, the second term goes to zero like $t^{-2}$ for long times - so rotational kinetic energy is not even a constant for the inertial motion of an individual particle, let alone invariant under changes of coordinate system origin.

  • $\begingroup$ If I understand you correctly, you're saying the rotational kinetic energy is indeed actually variant under coordinate transforms. So, if we start at one origin and have the center of mass of an object move out radially from it, but then establish a new origin, say $+y$ from the original, whilst the object still moves the same way, it now has a rate of change of theta relative to this new point, and thus has a greater rotational kinetic energy from this new vantage point? $\endgroup$
    – sangstar
    Nov 24, 2017 at 0:57
  • $\begingroup$ And mustn't the total kinetic energy be invariant? And if so, how can it be when the second term can be zero or non-zero given a coordinate transform yet everything else stays the same I'd assume? $\endgroup$
    – sangstar
    Nov 24, 2017 at 0:59
  • $\begingroup$ Yes, total kinetic energy is invariant to shifts of origin and rotations, and how it is divided between "rotation" about the origin is not. It is also not invariant to what are called "boosts" (i.e. using moving coordinates). That rotational kinetic energy about the center of mass (third term) is invariant to all inertial coordinate systems, though you can make it vanish using a rotating coordinate system (non-inertial). $\endgroup$ Nov 24, 2017 at 3:27

I think I understand your confusion, but you've got the answer in your own question, just read calm and try to understand.

The torque denotes the "capability" of a force to make the body rotate around the point you chose.

So there is no contradiction. If you choose the one of $r_0$, the force cannot make the ball rotate around that point. However, the force actually CAN make the ball rotate around other point like the one in $r_1$, or $r_2$.

By the way, you can set the reference of torques outside your origin of coordinates. The torque will just be $\vec{r}_{ref-to-object} \times \vec{F}=(\vec{r}_{obj}-\vec{r}_{ref})\times \vec{F}$.

Finally, kinetic energy is not directly related to torque, but $E_k$ does also depend on the reference frame, and that's not a problem.


You can actually prove that the kinetic energy of any system is generally invariant under changing the reference point of your torque and angular momentum. An easy way to see this is by using the relation for kinetic energy in terms of the angular momentum and angular velocity (with some arbitrary reference point): $$E_K = \frac12 \mathbf L . \boldsymbol \omega$$ Now from the definition of the torque with respect to our arbitrary reference point: $$E_K = \frac12 \mathbf L . \boldsymbol \omega =\frac 12 (\sum_i m_i \ \mathbf r_i \times \mathbf v_i ).\boldsymbol \omega =\frac 12 \sum_i m_i \ \boldsymbol \omega.(\mathbf r_i \times \mathbf v_i)=\frac 12 \sum_i m_i \ \mathbf v_i.(\boldsymbol \omega \times\mathbf r_i ) $$ Now since the velocity of the i$^{th}$ particle is $\mathbf v_i = \boldsymbol \omega\times \mathbf r_i$ , we get: $$E_K=\frac 12 \sum_i m_i \ \mathbf v_i.\mathbf v_i = \frac 12 \sum_i m_i |\mathbf v_i|^2 $$ But because the velocities are intrinsically independent of the reference point, $E_K$ must also be independent of the reference point. You can easily extend this argument to continuous systems by changing the sums to integrals.

As you can see the above relation was expected because it's simply the sum of the individual kinetic energies of the constituent particles; which, again, is intrinsically independent of any reference point.


As Sean E. Lake points out in his answer, how you divide the total kinetic energy into rotational and non-rotational kinetic energies does depend on the reference point. However, this is not important because as shown in my answer, the total kinetic energy is invariant. Also, the kinetic energy does depend on the reference frame because the two reference frames can have a relative velocity with respect to each other. I have just shown the invariance of the kinetic energy with respect to the reference point, i.e. the two reference frames are stationary with respect to eachother, with just different origins, in my derivation.

  • $\begingroup$ I see, this helps me understand Sean's answer a little better too. But isn't that definition a definition for rotational kinetic energy only? It's from $E_{rot}=1/2 I\omega ^2$ is it not? So how can it be shown to be the general kinetic energy? $\endgroup$
    – sangstar
    Nov 24, 2017 at 1:04
  • $\begingroup$ Whether $E= \frac 12 I \omega ^2$ is the total or the rotational kinetic energy depends on your reference frame. If you work in the center of mass frame , $E= \frac 12 I \omega ^2$ (with I and $\omega$ calculated in this frame) is going to be just the rotational kinetic energy, and the total kinetic energy seen in the lab frame is going to be that plus $\frac 12 M v_{cm}^2$. But if you work in the lab frame from the start, $E=\frac 12 L. \omega$ is going to be the total kinetic energy because as shown above, it's just equal to the sum of the kinetic energies of the particles in the system. $\endgroup$ Nov 24, 2017 at 1:13
  • $\begingroup$ Why does working in the lab frame from start change your equation from $\frac{1}{2}I\omega ^2 + \frac{1}{2}Mv^2_{cm}$ to just $E = \frac{1}{2}L\omega$? I don't see where this comes from. Also, if $\frac{1}{2}I\omega ^2 = \frac{1}{2}L\omega$, how can we say that $E = \frac{1}{2}L\omega$ is the total energy when you just said $\frac{1}{2}I\omega ^2 $ is the rotational kinetic energy? $\endgroup$
    – sangstar
    Nov 24, 2017 at 2:16
  • $\begingroup$ The I and $\omega$ in the equation $E= \frac 12 I \omega^2 + \frac 12 M v_{cm}^2$ are measured in the center of mass frame. For example, consider a spinning top which also has translational motion in the lab frame. To use that equation you plug in the moment of inertia and angular velocity with respect to the axis of the top (which goes through its center of mass!), i.e. I and $\omega$ in that relation are defined in the cm frame. If you use I and $\omega$ (and L) as defined in the lab frame itself, the equation for the total kinetic energy just becomes $E= \frac 12 L. \omega$ $\endgroup$ Nov 24, 2017 at 2:24
  • $\begingroup$ There is a misunderstanding between us because of bad notation. The equation for the total kinetic energy (with better notation) is: $$E=\frac 12 \mathbf L_{lab}. \boldsymbol\omega_{lab}=\frac 12 \mathbf L_{cm}. \boldsymbol \omega_{cm}+\frac 12 M v_{cm}^2$$ Also, the reason that $E=\frac 12 \mathbf L_{lab}.\boldsymbol\omega_{lab}$ is the total kinetic energy is because as I have shown in my answer, it's equal to the sum of kinetic energies of the constituent particles (which is the definition of total kinetic energy!) $\endgroup$ Nov 24, 2017 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.