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This question already has an answer here:

For example : A rod is held stationary in vertical position on a smooth horizontal ground and then released. Now the center of mass has velocity and acceleration and every point of the rod has some acceleration and velocity which are different from center of mass'. How can one prove that the rod is rotatating purely about centre of mass?

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marked as duplicate by ja72, ACuriousMind, Kyle Kanos, John Rennie newtonian-mechanics Aug 19 '15 at 5:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ From the problem statement you have given it seems like the rod is just going to be balanced when you release it and undergo no change. $\endgroup$ – user263399 Aug 18 '15 at 14:15
  • $\begingroup$ Related: physics.stackexchange.com/a/199709/392 $\endgroup$ – ja72 Aug 18 '15 at 14:16
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    $\begingroup$ possible duplicate of Do objects rotate around the torque vector or its center? $\endgroup$ – ja72 Aug 18 '15 at 14:17
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    $\begingroup$ The answer is not. If net forces are zero then the only motion possible is for the center of mass to move on straight line. Is the bottom of the rod in contact still, or is it in free fall? $\endgroup$ – ja72 Aug 18 '15 at 14:26
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    $\begingroup$ This question is poorly defined. What does "smooth" mean here? Frictionless or not? A body does not always rotate about it's centre of mass. Imagine a uniform horizontal bar attached firmly at a third of its length to a vertical bar, which is made to rotate on its vertical centre line. The horizontal bar will obviously not rotate on its centre of gravity. $\endgroup$ – Gert Aug 18 '15 at 15:25
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Does a body always rotate purely about its center of mass?

Well, that depends. The first assumption you need is that the body is rigid. Violate this assumption and all bets are off the table because you can't even necessarily classify all motions as "rotations": for example if a long thin board starts twisting sinusoidally into/out-of a helix shape, back and forth, is that vibration mode a "rotation" about some "axis?" Positive, or negative? And what axis?

The formal way to make an object rigid is to fix all of the distances which means any motion of the object is, technically speaking, an isometry. Isometries come in three flavors: an isometry is a composition of reflections, translations, and rotations. Reflections are not valid for motion because they are not "continuous" isometries. Every other motion can indeed be viewed as a translation of a point and a rotation about that point, for any point on (or off!) the object. So, choose the point to be the center of mass: then any motion must be a translation of the center of mass plus a rotation about that center of mass.

Notice that there's a lingering nasty problem: Suppose we have a ball on a tether, which we swing around our head. Assume that the ball is not spherically symmetric: then yeah, we can describe the ball's motion between two points as a translation followed by a rotation about the center of mass: however, this is not what we mean when we talk about what the ball is "rotating around", which is us, holding the tether.

For that we have to turn not just to the abstract mathematics of the groups $E^+(3)$ and $O(3)$ and $SO(3),$ but also the dynamics of the system.

In these cases we can say that if all forces are central (the system can be modeled as a bunch of point masses which have forces directed at each other, satisfying Newton's third law), then as long as there are no external forces, the dynamics decouples into a continuous translation of the center of mass in a straight line (the center of mass does not accelerate) and a continuous rotation about some axis $\vec \omega$.

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A body in free motion does not necessarily rotate about the center of mass. The center of mass might have straight linear motion in addition any rotation. The general motion is a screw motion with a rotation about some instantaneous axis and parallel translation at the same time.

Consider an arbitrary body rotating by $\vec{\omega}$ and at some instant the center of mass (point C) has linear velocity $\vec{v}_C$.

I can prove that there is always a point A where the linear velocity of the extended rigid body is only parallel to the rotation axis defined by $\vec{\omega}$. The combined motion is a rotation about A with a parallel translation of $\vec{v}_A = h \vec{\omega}$. The scalar $h$ is called pitch. If the body is purely rotating without translation then $h=0$ and if the body is purely translates then $h=\infty$ and $\|\vec{\omega}\|=0$.

The motion of an arbitrary rigid motion is decomposed as such:

  • Speed of rotation $$\omega = \| \vec{\omega} \|$$
  • Direction of rotation $$\hat{e} = \frac{\vec{\omega}}{\omega}$$
  • Location of rotation axis $$\vec{r}_A = \vec{r}_C + \frac{\vec{\omega}\times \vec{v}_C}{\omega^2}$$
  • Screw pitch $$h = \frac{\vec{\omega}\cdot\vec{v}_C}{\omega^2}$$

NOTES: $\cdot$ is the vector inner product, and $\times$ is the vector cross product.

Proof

The linear velocity at A is found by the frame transformation law $$\vec{v}_A = \vec{v}_C + \vec{\omega} \times (\vec{r}_A-\vec{r}_C)$$ Using the location expression from above is

$$ \vec{v}_A = \vec{v}_C + \frac{\vec{\omega} \times(\vec{\omega}\times \vec{v}_C)}{\omega^2}$$

Using the Vector Triple Product

$$ \vec{v}_A = \vec{v}_C + \frac{\vec{\omega}(\vec{\omega}\cdot\vec{v}_C)-\vec{v}_C (\vec{\omega}\cdot\vec{\omega})}{\omega^2}$$

With the simplification that $(\vec{\omega}\cdot\vec{\omega}) = \omega^2$ and the definition for screw pitch $\vec{\omega}\cdot\vec{v}_C = h \omega^2$ the above is

$$ \vec{v}_A = \vec{v}_C + \frac{\vec{\omega}(h \omega^2)-\vec{v}_C (\omega^2)}{\omega^2} = h \vec{\omega}$$

So the velocity at A is parallel to the rotation $\vec{\omega}$

Reverse Proof

You can start from a general screw motion at a known point A, with direction $\hat{e}$, speed $\omega$ and pitch $h$ (two parameters for the point as the location along the line does not count, two parameters for the direction as the magnitude does not mater, one for the speed and one for the pitch equals six independent parameters that describe the motion. These will be transformed to the more familiar six motion parameters $\vec{\omega}$ and $\vec{v}_C$ below:

  • Rotational Vector $$\vec{\omega} = \omega \hat{e}$$
  • Linear Vector $$\begin{align} \vec{v}_C &= \vec{v}_A - \vec{\omega} \times (\vec{r}_A-\vec{r}_C) \\ & = h \vec{\omega} + (\vec{r}_A-\vec{r}_C) \times \vec{\omega} \end{align}$$

All six motion parameters are defined now at the center of mass. Sometimes the above is combined into one expression

$$ \begin{bmatrix} \vec{v}_C \\ \vec{\omega} \end{bmatrix} = \omega \begin{bmatrix} h \hat{e} + \vec{r}\times \hat{e} \\ \hat{e} \end{bmatrix} $$ which clearly decomposes the motion into the magnitude (speed) in the first part, and the screw axis (geometry) in the second part. The two vectors defining the screw axis are called plücker line coordinates.

Example

A cylindrical body is rotating about its axis, and translating perpendicular to the axis (red vectors). This motion is described by a pure rotation about the screw axis (purple vectors).

pic

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