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.

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

  2. 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

  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/216871/2451 $\endgroup$
    – Qmechanic
    Nov 7, 2015 at 18:54
  • $\begingroup$ Note that for a moving body the instant center of rotation is always uniquely defined. See physics.stackexchange.com/a/215165/392 $\endgroup$ Nov 7, 2015 at 23:00
  • $\begingroup$ @ja72 I have to ask this. I take it the definition of the instant center of rotation you mention is like here: 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. $\endgroup$
    – udrv
    Nov 8, 2015 at 8:37
  • $\begingroup$ Yes, as the author of the section on the relative centers (en.wikipedia.org/wiki/…) 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. $\endgroup$ Nov 8, 2015 at 16:25
  • $\begingroup$ 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. $\endgroup$
    – udrv
    Nov 8, 2015 at 20:08

3 Answers 3


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


  1. Find the moment arm $c$ of the force through A. $$ c = r \cos \theta$$
  2. Find the radius of gyration about the center of mass C $$\rho = \sqrt{ \frac{I_C}{m} }$$
  3. 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.


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.

  • $\begingroup$ 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? $\endgroup$ Nov 8, 2015 at 17:25
  • $\begingroup$ 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? $\endgroup$ Nov 8, 2015 at 17:48
  • $\begingroup$ $c$ 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 $\vec{r}_{CA} \times \vec{F}$ $\endgroup$ Nov 8, 2015 at 21:54
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2015 at 21:58
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2015 at 22:01

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.

  • $\begingroup$ 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. $\endgroup$ Nov 7, 2015 at 23:05
  • $\begingroup$ @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. $\endgroup$
    – udrv
    Nov 8, 2015 at 8:23
  • $\begingroup$ 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 $$c = \frac{ \| \vec{F} \times \vec{M} \| }{ \| \vec{F} \|^2} $$ $\endgroup$ Nov 8, 2015 at 16:21

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.


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