Why does the rod rotate only about its center of mass after being hit anywhere except its center of mass? Can it be proved mathematically? In case I'm wrong then show about which point the rod will rotate after being hit at any point other than the center of mass (mathematically). I don't need any physical interpretation cause its been discussed with logical support previously, please try to show it mathematically.[The rod is kept unhinged on a smooth horizontal plane]


4 Answers 4


The rod doesn't "only" rotate about its center of mass. The rod doesn't "only" rotate about any one axis, in fact, because the choice of axis of rotation is arbitrary. That being said, there is a reason why we often choose to measure rotation from the center of mass.

In order to properly study the rotation of an object, we must isolate its rotational motion, eliminating its translational motion. In other words, we must be in a frame in which the axis of rotation is stationary. It is also a basic fact of rotational dynamics that, in the absence of external forces, the center of mass of an object does not accelerate.

If you choose an axis of rotation that is not at the center of mass, then when you're in the frame in which the axis of rotation is stationary, the center of mass will rotate (i.e. accelerate). This means the reference frame in which you isolate rotation is non-inertial, which means there will be all sorts of fictitious forces (inertial forces, Coriolis forces, centrifugal forces) that you would have to add in to get the right physics. But if your axis of rotation is chosen to be the center of mass, then the center of mass is, by definition, stationary in this frame, so no fictitious forces will need to be applied.

What this means is that defining the axis of rotation to be through the center of mass makes it much easier to study rotation than putting it anywhere else, for a free object. If your rod is, for example, pinned at one end, then there will be external forces on the object (for example, the force that the pin exerts on the rod) that will make the center of mass accelerate. As such, if you were to choose the axis of rotation to be through the center of mass now, you would be in a non-inertial frame with fictitious forces. In this case, the frame that would make the study of rotation easy is one in which the axis of rotation is through the pin. Since the pin is stationary by definition, we know that this frame is inertial.

Note that a "natural" inertial frame that makes it easy to study rotation does not always exist, and in general requires either a complete absence of external forces or very special configurations of external forces (e.g. external forces that keep one point on the object fixed). But, for the case of an object rotating without any external forces applied, choosing the axis of rotation to be through the center of mass gives this "natural" inertial frame.

  • $\begingroup$ Your answer was fabulous, but if u calculate torque at the center of mass it won't come to happen to be zero, so how can it be the center of rotation? [Far I know, center of rotation must have total torque = 0] $\endgroup$ Apr 18, 2018 at 4:11
  • $\begingroup$ @BandhabKumarDas The situation I was assuming you were referring to was this: you hit an object (i.e. apply a force for a short period of time), and then you watch it as it rotates in the absence of further external forces. If there are no external forces after the rod is hit, then there are no external torques, either. $\endgroup$ Apr 18, 2018 at 4:12
  • $\begingroup$ But if u consider the situation at the moment the rod gets hit, and calculate torque; cause thats the moment its rotational motion is introduced. $\endgroup$ Apr 18, 2018 at 4:14
  • $\begingroup$ Your question asked about the rod "after being hit", which implies no external force. $\endgroup$ Apr 18, 2018 at 4:19
  • $\begingroup$ But u gotta consider that particular moment its getting hit, in order to calculate torque, or predict everything about the introduction of rotational motion in the rod, right? So can u answer how the net torque at center of mass is being 'zero' at the moment of collision? $\endgroup$ Apr 18, 2018 at 4:23

Let's assume for the sake of simplicity you hit a rod with an elastic ball, the energy of collusion which is the vector impulse difference of the ball before and after collision divides between:

the ball ricochet velocity and direction, the rod-spinning around it's CG and also moving laterally away from the impact. While the rod is turning and laterally moving it also retains some of the energy by vibrating like a snake.

If we simplify and ignore impact effects that are of smaller orders, the energy of impact to the rod is divided between its angular momentum and kinetic energy of linear movement, proportional to the inverse of those quantities.

Angular momentum: $$I. \alpha$$ and $$I = 1/12 ml^2 $$ and its Kinetic energy $$ E = 1/2 mv^2$$ For example, a long rod will absorb most of the energy accelerating laterally with smaller angular acceleration.


When you say that the rod rotates about its centre of mass I believe you mean that the centre of rotation of the rod is at its centre of mass. This is not necessarily true. In fact the point on the rod which temporarily has a zero velocity depend on the point of hit as well as the magnitude of the hit.

Please refer to this solution that I had provided in Quora: https://www.quora.com/In-space-a-uniform-rod-of-length-l-is-initially-at-rest-It-is-tapped-at-one-end-perpendicular-to-its-length-How-far-does-its-CM-translate-while-it-rotates-once-completely-Is-this-problem-valid-considering-that-there-are-no-given-parameters/answer/Rajan-Pal-23

The solution here can be easily manipulated to find the point where the velocity is zero. Hope this helps.


A rod hit offset from the center of mass will not only rotate about the center of mass but also translate at the same time.

There is a simple geometrical relationship between the impulse offset $a$ from the center of mass, and the location of the center of rotation $b$ away from the center of mass. The impulse happens in a direction $\hat{j}$ that is perpendicular to the offset $a$.


The rotation center is always on the other side of the center of mass, at a distance $$b = \frac{I_{\rm CM}}{m\, a} $$ as shown above.

The larger $a$ is, the smaller $b$ is and vise versa. At the two limits when $a=0$ and the impulse is through the center of mass, the rotation center is at infinite (designating a pure translation). When $b=0$ and the body is rotating about the center of mass then $a$ is at infinity designating an impulsive couple (short-lived torque).

All of the above is a consequence of the equations of motion of a rigid body. Actually, it is a consequence of the definition of momentum.


Rigid body momentum (translational and rotational) is defined as follows

$$\begin{aligned} \vec{p} & = m\, \vec{v}_{\rm CM} \\ \vec{L}_{\rm CM} & = \mathrm{I}_{\rm CM}\, \vec{\omega} \\ \end{aligned} $$

Now if the perpendicular line from the rotation center to the impulse has direction $\hat{i}$ and goes through the center of mass. The kinematics of this situation are such that

$$ \vec{v}_{\rm CM} = \vec{\omega} \times (b\, \hat{i}) $$

and if the impulse has magnitude $J$, the impulse vector is decomposed into magnitude and direction as $$\vec{J} = J\,\hat{j}$$

Finally, the momentum of the object is solely due to the impulse and thus

$$ \begin{aligned} \vec{p} & = J\, \hat{j} \\ \vec{L}_{\rm CM} & = (a\,\hat{i}) \times (J\, \hat{j}) \end{aligned} $$

Together these are combined into

$$\begin{aligned} J\,\hat{j} & = m\, \vec{\omega} \times (b \hat{i}) \\ (a\,\hat{i}) \times (J\, \hat{j}) & = \mathrm{I}_{\rm CM}\, \vec{\omega} \\ \end{aligned} $$

and use the first into the second, as well as define the direction $\hat{k} = \hat{i} \times \hat{j}$ as where rotation happens with $$ \vec{\omega} = \omega \hat{k}$$

$$ (a\,\hat{i}) \times \left( m\, (\omega \hat{k}) \times (b \hat{i}) \right) = \mathrm{I}_{\rm CM}\, (\omega \hat{k}) $$

or in scalar form

$$ \left( a \, m (\omega b) \right) \hat{k} = ( I_{\rm CM} \omega ) \hat{k} $$ recognizing that $\mathrm{I}_{\rm CM}$ acts along $\hat{k}$ and that $\hat{i} \times ( \hat{k} \times \hat{i}) = \hat{i} \times \hat{j} = \hat{k}$.

The above is simplified by eliminating $\omega$ into

$$ I_{\rm CM} = m a b $$ or $$ b = \frac{I_{\rm CM}}{m a} $$


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