How to find the center of rotation? (2D)

Question

First off, I'm assuming that a free floating polygon doesn't always rotate around its center of mass unless the net force is zero (based on the points below). If this isn't correct please tell me.

A pure force through the center of gravity (with no net torque) will purely translate a rigid body (any point on the body).

A pure torque any point on the body (with no net force) will purely rotate a rigid body about its center of gravity

With that out of the way, how do I find the center of rotation for a polygon with center of mass $C$, force $F$, and the point where the force is applied $A$? Also, how much of the force translates to translational force and how much of it translates to rotational force? How do the forces add up when there's more than one?

Force on a polygon

Related question by OP: http://physics.stackexchange.com/q/216871/2451 — Qmechanic, Nov 07 '15 at 18:54
Note that for a moving body the instant center of rotation is always uniquely defined. See http://physics.stackexchange.com/a/215165/392 — John Alexiou, Nov 07 '15 at 23:00
@ja72 I have to ask this. I take it the definition of the instant center of rotation you mention is like here: https://en.wikipedia.org/wiki/Instant_centre_of_rotation. Is this something that is used in engineering? Personally I haven't seen it used in physics. The standard procedure I am familiar with uses the COM, for the reasons I outlined in my answer. So if the OP considers this problem in a physics context I would guess the COM approach would be preferred and expected. But if it is related to an engineering context, then expectations may differ. — udrv, Nov 08 '15 at 08:37
Yes, as the author of the section on the relative centers (https://en.wikipedia.org/wiki/Instant_centre_of_rotation#Relative_centre_of_rotation_for_two_contacting_planar_bodies) the obvious engineering use is to find the gear ratio of two contacting bodies. But, in physics the instant center of rotation must be important too as it reveals the geometry behind mechanics. — John Alexiou, Nov 08 '15 at 16:25
The geometry is indeed intuitive once you get a hang of it. Still can't recall seeing it applied in physics, although this may well be a limitation of my experience/interests. But then there's always something new to learn. Thanks for explaining it. — udrv, Nov 08 '15 at 20:08
I recently learned that this is rather more subtle than I had previously appreciated, see my answer below. — Kyle Oman, Apr 21 '16 at 07:34
Should the points mentioned by OP be centre of mass ? Because COM and Cg are not always the same — Habibullah Khan, Aug 10 '22 at 01:57

score 2 · Answer 1 · edited Apr 13 '17 at 12:39

2

There is a easy semi-geometrical way of finding the center of rotation due to a force.

Find the moment arm $c$ of the force through A. $$ c = r \cos \theta$$
Find the radius of gyration about the center of mass C $$\rho = \sqrt{ \frac{I_C}{m} }$$
Measure the distance $\ell$ away from the center of mass and mark point R $$\ell = \frac{\rho^2}{c}$$

Point R is the instance center of rotation.

See also https://physics.stackexchange.com/a/212939/392 for more details.

Example

A plank of length $a$ and and width $b$ is struck on one end with a force $F$. Place a coordinate system at the center of mass and measure the force moment arm as $c=\frac{a}{2}$. The mass moment of inertia of a plank is $$I_C = \frac{m}{12} \left( a^2+b^2\right)$$ and hence the radius of gyration about the center is $$\rho = \sqrt{ \frac{a^2+b^2}{12} }$$

With the method above the center of rotation is at a distance $$\boxed{\ell = \frac{\rho^2}{c} = \frac{ \frac{a^2+b^2}{12} } {\frac{a}{2} } = \frac{a^2+b^2}{6 a} }$$

Lets find the same answer using the equations of motion.

The sum of the forces of the body are $$\sum \boldsymbol{F} = (0,F,0)$$
The sum of moment about the center are $$\sum \boldsymbol{M} = (0,0,\frac{a}{2} F)$$
The linear acceleration of the center is $$\boldsymbol{a} = \frac{\sum \boldsymbol{F}}{m} = (0,\frac{F}{m},0)$$
The angular acceleration of the body is $$\boldsymbol{\alpha} = \frac{\sum \boldsymbol{M}}{I_C} = (0,0,\frac{6 F a}{m (a^2+b^2)})$$
Find the center of rotation R such that $\boldsymbol{a}_y=\ell \boldsymbol{\alpha}_z$ $$ \boxed{\ell = \frac{ \frac{F}{m} }{ \frac{6 F a}{m (a^2+b^2)} } = \frac{a^2+b^2}{6 a}}$$

Same answer!! No need to do equations of motion to get the center of rotation. The point is purely a result of the inertial properties and geometry.

edited Apr 13 '17 at 12:39

Community

1

answered Nov 07 '15 at 22:30

John Alexiou

38,341

I want to make sure I understand everything; $c$ is $CB$, $B$ is collinear with the net $F$'s vector, and $CB$ and $BA$ are perpendicular? I'm building a physics engine, so I need all the little details. When there are multiple forces where would $A$ be? – TakingItCasual Nov 08 '15 at 17:25
Also, when the center of rotation isn't the center of mass, is the linear acceleration applied to the vertexes with respect to the center of rotation or the center of mass? – TakingItCasual Nov 08 '15 at 17:48
$c$ is CB. $B$ is the closest point to the center of mass on the action line of $F$. $CB$ and $BA$ are perpendicular. Each applied force will have their own location, so each will have it's own $A$. To find the torque of a force about the center of mass do $\vec{r}_{CA} \times \vec{F}$ – John Alexiou Nov 08 '15 at 21:54
If you know the linear velocity and linear acceleration of the center of mass $\vec{v}C$ and $\vec{a}_C$ then these quantities at each node are $$\begin{align} \vec{v}_i &= \vec{v}_C - \vec{r}_i \times \vec{\omega} \ \vec{a}_i &= \vec{a}_C - \vec{r}_i \times \vec{\alpha} + \vec{\omega}\times\vec{\omega} \times \vec{r}_i \end{align} $$ where $\vec{r}_i$ is the position of node _i relative to the center of mass, $\vec{\omega}$ and $\vec{\alpha}$ the rotational velocity and acceleration. – John Alexiou Nov 08 '15 at 21:58
For 2D problems it is easier to start from the 3D equations, and project to the plane by placing zeros for the z axis for linear quantities and zeros in the x and y axes for rotational quantities. – John Alexiou Nov 08 '15 at 22:01
See http://physics.stackexchange.com/a/80449/392 for the derivation of the equations of motion for rigid body dynamics. – John Alexiou Nov 08 '15 at 22:02
So I need to calculate the center of rotation for each individual force? How would I add them together then, since they have different distances and angles from the center of mass? – TakingItCasual Nov 09 '15 at 03:11
No you don't. You add up all the forces and moments about the center of mass and use the equations of motion to find the motion of the center. Instant center is not necessary to calculate in order to simulate dynamics. It is more of a post-process nice to have. – John Alexiou Nov 09 '15 at 06:09
You said “Measure the distance $\ell$ away from the center of mass and mark point R” in step 3. How does one know which point on the circle of radius $\ell$ about point R is the center of rotation? Is it the point that would maximize the net torque that is rotating the object? – gen-ℤ ready to perish Jun 17 '17 at 03:06
Point R lies on line BC which is perpendicular to the line of action of the force, and through the center of mass. Point R is always "on the other side" of the center of mass from the force. – John Alexiou Jun 22 '17 at 00:42
@JohnAlexiou So R is the instantaneous center of rotation due to force F? If the force is removed wouldn't the center of rotation be the COM? How does one describe the transition from the instantaneous center of rotation to rotation about the COM? – Bob D Mar 27 '23 at 18:34
@BobD - A force would cause the COM to translate, and if the force is offset also rotate about the COM. So combine the translation and rotation to find the point about it purely rotates with no translation. That is R. – John Alexiou Mar 28 '23 at 12:40
@JohnAlexiou Thanks for getting back. I'm trying to process what you're saying, so please bear with me. First, are you saying that the plank in your example will rotate under the influence of F about the COM or about R? – Bob D Mar 28 '23 at 15:00
About R. Chasle's theorem says that the general motion can be decomposed as a translation of an arbitrary point plus a rotation about that point. If you choose R as the arbitrary point to describe the motion, then there will be no translational component, but just a pure rotation. If you choose the COM there will be both a translation component (from the force) and a rotation (from the torque). – John Alexiou Mar 28 '23 at 17:17
@JohnAlexiou I find two things troubling me. (1) Since the moment of inertia is a measure of resistance to change in rotation and the moment of inertia is greater at R than the COM per the parallel axis theorem, I would think rotation about the COM where I is a minimum would be favored. (2) How can R purely rotate without translation when, for a rigid body, all points must undergo the same translation? – Bob D Mar 28 '23 at 21:00
@BobD - Newton's law states that if a non-zero force is applied then the COM accelerates. Just this fact means the body will not rotate about the COM unless the net force is zero. This is evident by my equation for $\boldsymbol{a}$, the linear acceleration. In addition there is $\boldsymbol{\alpha}$ the rotational acceleration since there is non-zero net torque about the COM. – John Alexiou Mar 28 '23 at 21:04
@JohnAlexiou "Newton's law states that if a non-zero force is applied then the COM accelerates". Yes, but doesn't that apply whether or not the line of action of the net force is through the COM or offset from the COM? "Just this fact means the body will not rotate about the COM unless the net force is zero" Huh? If the net force is offset from the COM will there not be a torque about the COM? – Bob D Mar 28 '23 at 21:13
Yes a net torque about COM means rotation in addition to any motion of the COM. I don't see where you disagree with me. https://physics.stackexchange.com/a/750758/392 – John Alexiou Mar 28 '23 at 23:00
@JohnAlexiou I'm not disagreeing with you, just trying to understand you. Do you agree with urdv answer, specifically the statement "Take the center of rotation as (C), regardless of where the force is applied."? – Bob D Apr 02 '23 at 17:35
No. The center of mass is the center of rotation only when there is zero net force applied. As soon as there is a net force and the COM moves you combine the translation and rotation to find a new center of rotation that is displaced $\rho^2/c$ from the COM on the other side of the force. The diagram in my answer has this relationship, and it is actually a geometric relationship since the quantities involved ($\rho$ the radius of gyration and $c$ the offset from the COM) are both length quantities. The center of rotation by definition cannot translate and hence point C is not it. – John Alexiou Apr 02 '23 at 19:16
@JohnAlexiou Thanks for the clarification. The reason I asked is you commented on the answer but you were silent on the statement I quoted. So I guess, in this case, silence did not mean consent. Regarding your plank example, you described the force as being "struck" on the plank. I assume by that you meant an impulse of force. – Bob D Apr 03 '23 at 14:02
So would I be correct to say the ICR at R only applies during the impulse and that afterwards rotation will be around the COM, since there are then no external forces? If so, then what is the nature of the transition from the ICR about R to rotation about the COM? Does it occur instantaneously? – Bob D Apr 03 '23 at 14:02
Once the motion is set due to an impulse it will continue on its path with constant translational and rotation speed. Hence the ICR will always be at a distance $\rho^2/c$ from the COM. This is due to Newton's 1st law. For example, a helicopter flying at a constant speed $v$ and the rotor rotating with speed $\omega$. At any instant, the ICR of the rotor is going to be at an offset off $v/\omega$ perpendicular to the line of motion when viewed from above. The situation above is similar with $$ \frac{v}{\omega} = \frac{\rho^2}{c} $$ – John Alexiou Apr 03 '23 at 14:17
@JohnAlexiou So, just to be clear, you are saying that a rigid body that is translating and rotating without any external force will not be rotating about its COM? – Bob D Apr 03 '23 at 14:30
Yes, exactly that. The key is that it is translating. By definition, the ICR is the point with no translation. Similar to a wheel rolling. The center of the wheel is translating and the body is rotating. The ICR of the wheel is on the contact point if there is no slipping, or anywhere along the vertical line connecting the contact to the center when slipping is present. The no-slip condition means the contact point is not translating which is also the definition of the ICR. – John Alexiou Apr 03 '23 at 15:02
@JohnAlexiou But John, every point on that wheel is rotating about the COM while the COM is translating, including the points on the wheel that make contact with the surface. The point of contact on the surface is constantly changing. Let's put the wheel in outer space in contact with no external forces or contact with any surface. Where is the ICR now? And what is its relevance? – Bob D Apr 03 '23 at 15:57
The ICR is the center of a rotational field which describes the velocity of every point around it. For a wheel this point is at the contact point. The wheel is not rotating about the COM. The wheel is rotating about the ICR. – John Alexiou Apr 03 '23 at 16:39
Also, note that the ICR is not a material point but a geometric point in space. It does not follow any particle attached to a body and can move about in space smoothly or abruptly. – John Alexiou Apr 03 '23 at 16:40
@JohnAlexiou OK, so the ICR is not a material point. But every material point of the wheel is rotating about the COM, correct? And if the ICR is not a material point, then what is the physical relevance of the ICR? – Bob D Apr 03 '23 at 17:25
"But every material point of the wheel is rotating about the COM". No. In general, every material point has some instantaneous velocity and the whole body is rotating. This situation is decomposed as a pure rotation about a point in space (the ICR). There is nothing special about the COM when it comes to the kinematics (the motion), and the body does not in general rotate about the COM. Only under the influence of a pure torque (zero net force) would the ICR coincide with the COM. The answer by @udrv is incorrect BTW. – John Alexiou Apr 03 '23 at 21:26
@JohnAlexiou John, you’ve got me going around in circles (pun intended). Consider a rigid body consisting of two spherical masses one at each end of a rod of negligible (compared to the two spheres) mass with no externally applied force. The rod is length $d$ between the centers of the two spherical masses. The center of mass is $d/2$. If the center of rotation is, say, a distance of $d/4$ from one of the masses instead of $d/2$, there will be unequal centripetal forces acting on each mass along the bar and passing thru the COM. – Bob D Apr 03 '23 at 22:23
That would mean a net force acting along the rod and thru the COM, which should be impossible since there is no net external force acting in the direction of the bar. It would be like two planets of equal mass orbiting one another (with no other gravitating bodies influencing them) with a barycenter different than half the distance between the orbiting bodies. – Bob D Apr 03 '23 at 22:23
@BobD - If the rod has no mass there is an equal and opposite force that can act on the two masses by definition. The magnitude of these forces are equal resulting in zero net force and the center of mass moving with steady velocity as the masses rotate. But you are confused because you are imagining a system pivoted about a fixed point (the ICR) which would require a pin reaction force to act on the rod since the COM is accelerating (as it is moving in an arc around the pivot). Although the two scenarios at some instant might have the same velocity field they are not the same over time. – John Alexiou Apr 04 '23 at 00:27
@BobD I found this simulation of a force applied on a flexible rod and you can observe the resulting motion. You see the COM displace, and the ICR not displace. – John Alexiou Apr 07 '23 at 12:10
@JohnAlexiou Very interesting and thanks. While the stick as a whole rotates about the fixed ICR, it looks to me like all the particles of the stick rotate about the translationally moving COM. So don't we have two centers of rotation, one about which the body as a whole rotates (ICR) and another about which the individual particles rotate (COM)? – Bob D Apr 07 '23 at 12:43

score 1 · Accepted Answer · answered Nov 07 '15 at 18:39

I think what you are asking about is answered by the fundamental theorem in the mechanics of rigid bodies, which states that the motion of any rigid body can be decomposed into the motion of its center of mass (not necessarily rectilinear) and a rotation about its center of mass (COM). The two statements you emphasize are direct corollaries. Please see details in the suggested link or any text on the subject.

However, the idea is not that there is some intrinsic center of rotation that sometimes happens to coincide with the center of mass, or that we necessarily decompose the forces acting on the body into some component translating the COM and another generating the torque about the COM. It is just that the equations of motion acquire the simplest form when expressed this way.

So, to answer your questions directly:

How do I find the center of rotation for a polygon when the center of mass is (C), the force is (F), and the point where the force is applied is (A)?

Take the center of rotation as (C), regardless of where the force is applied.

Also, how much of the force translates to translational force and how much of it translates to rotational force?

All of it, in both cases. The equations of motion for the COM of your planar polygon are just $$ M \vec{a} = \vec{F}\\ I\alpha \vec{k} = \vec{T} $$ where $M$ is the total polygon mass, $\vec{a}$ is the linear acceleration of the COM (and of the rigid body), I is the moment of inertia relative to the COM, $\alpha$ is the angular acceleration about the COM, $\vec{k}$ is the unit vector normal to the polygon plane, and $\vec{T}$ is the torque of $\vec{F}$ about the COM. In 3D the torque equation would contain an additional term on the left hand side.

How do the forces add up when there's more than one?

Vector-wise as always. The total force is the vector sum of individual forces, and for 2D the total torque is the sum of individual torques about the COM or the torque of the total force about the COM, whichever you prefer.

Actually forces add up using equipollent force/moment pairs. The sum of forces is indeed $\sum \boldsymbol{F} = F_1+F_2+\ldots$ but the sum of moments is $\sum \boldsymbol{M} = c_1 F_1 + c_2 F_2 + \ldots$ where $c_i$ are the moment arms for each force. — John Alexiou, Nov 07 '15 at 23:05
@ja72 My statement "the total torque is the sum of individual torques about the COM" coincides with yours on $\sum{\vec{M}}$, assuming the arms $c_i$ you mention are wrt to the COM. I guess you object to my statement about the "torque of the total force about the COM". It involves defining the equivalent arm of the total force, which may be cumbersome, but doable. — udrv, Nov 08 '15 at 08:23
I agree. BTW If you have a force/moment pair $\vec{F}$ and $\vec{M}$ you can find the moment arm it corresponds to by $$c = \frac{ | \vec{F} \times \vec{M} | }{ | \vec{F} |^2} $$ — John Alexiou, Nov 08 '15 at 16:21

score 1 · Answer 3 · answered Apr 21 '16 at 07:34

The centre of rotation of a rigid body is actually rather poorly defined. One sensible definition (indeed, probably the one you want) is to pick the point that has zero velocity once you subtract the mean motion of the object. However, this is only a unique point once you pick a frame of reference (e.g., with respect to the "immobile" ground). If you pick a new (inertial) reference frame, you'll find a new centre of rotation that's just as well defined as the first one!

This is easy to show, consider the vector field of a solid body rotator (in cylindrical coordinates): $$\vec{v} = 0\,\hat{r}+r\,\hat{\phi}+0\,\hat{z}$$ Now pick a new reference frame moving with respect to the old one, you can do this by adding a constant-velocity vector field to the one above (this one in cartesian coordinates): $$\vec{v} = 1\,\hat{x}+0\,\hat{y}+0\,\hat{z}$$ If you add those two fields (I leave all the coordinate transformations to you), you'll find a new centre for your solid body rotator. You can apply a simple translation and get the same solid body rotator vector field back.

How to find the center of rotation? (2D)

3 Answers3

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