If the distribution of mass in a shaft is such that it is balanced around its center, then the shaft may still generate a couple force if the weight is not balanced around its axis of rotation. For example:

enter image description here

In this diagram the weight, assuming $x=y$ is balanced around the center of mass, produces no centrifugal force, but does produce a couple around the center of mass, illustrated by the arrow C. Note Cd = Fl.

Let's imagine that we remove weight Y. Now, there will be a centrifugal force G. Does a couple still exist? How is it computed?

  • $\begingroup$ If you remove weight Y, you will have different forces at each end of the rod, not the equal and opposite forces F. Calculating the forces is exactly the same as any other mechanics problem - for example, take moments about one end of the rod to find the reaction force at the other end. $\endgroup$
    – alephzero
    Aug 20, 2016 at 13:58

2 Answers 2


Yes, a single rotating mass will produce a couple reaction. Why? (read to the bottom)

General Case

A rotating shaft with various eccentric masses attached to it has the following combined inertial properties about the center of the shaft.

  • Total mass (scalar) $$ m = \sum_{i=1}^n m_i $$
  • Center of mass (3×1 vector) $$ \vec{r}_C = \frac{1}{m} \sum_{i=1}^n \vec{r}_i m_i $$
  • Mass moment of inertia tensor about the center of mass (3×3 matrix) $$ \mathrm{I}_C = \sum_{i=1}^n m_i \left| \matrix{ y_i^2 + z_i^2 & -x_i y_i & -x_i z_i \\ -x_i y_i & x_i^2+z_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2} \right| = \left[ \matrix{ I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} } \right] $$ where $$ \pmatrix{x_i \\ y_i \\ z_i} = \vec{r}_i - \vec{r}_C $$


Suppose this axis is rotating with constant speed about the fixed x-axis $\vec{\omega} = \pmatrix{\omega & 0 & 0}$ through the origin. The total momentum of the shaft is

  • Linear Momentum $$ \vec{p} = m \vec{v}_C = m (\vec{\omega} \times \vec{r}_C) $$
  • Angular Momentum $$ \vec{L}_C = \mathrm{I}_C \vec{\omega} $$

Support Forces and Moments

  • The shaft is supported on two points $\vec{r}_A = \pmatrix{ -\frac{\ell}{2} & 0 & 0}$ and $\vec{r}_B = \pmatrix{ +\frac{\ell}{2} & 0 & 0}$.
  • The two reaction forces are $\vec{F}_A = \pmatrix{0 & Fy_A & Fz_A}$ and $\vec{F}_B=\pmatrix{0 & Fy_B & Fz_B}$
  • The two reaction moments are $\vec{M}_A = \pmatrix{0 & My_A & Mz_A}$ and $\vec{M}_B=\pmatrix{0 & My_B & Mz_B}$

Equations of Motion

The time derivative of momentum equals the net force, or net torque for the angular case.

$$ \begin{aligned} \vec{F}_A + \vec{F}_B & = \frac{{\rm d}}{{\rm d}t} \vec{p} = \vec{\omega} \times m \vec{v}_C = m \vec{\omega} \times ( \vec{\omega} \times \vec{r}_C ) \\ \vec{M}_A + \vec{r}_A \times \vec{F}_A + \vec{M}_B + \vec{r}_B \times \vec{F}_B & = \frac{{\rm d}}{{\rm d}t} \vec{L}_C = \vec{\omega} \times \mathrm{I}_C \vec{\omega} \end{aligned}$$


After doing all the math, and assuming the lever rule on what proportion of the reaction goes to which support according to the location of the center of mass $x_C$

  • Support Forces $$ \begin{aligned} Fy_A & = \left( \frac{1}{2} - \frac{x_C}{\ell} \right) (-m \omega^2 y_C) \\ Fy_B & = \left( \frac{1}{2} + \frac{x_C}{\ell} \right) (-m \omega^2 y_C) \\ Fz_A & = \left( \frac{1}{2} - \frac{x_C}{\ell} \right) (-m \omega^2 z_C) \\ Fz_B & = \left( \frac{1}{2} + \frac{x_C}{\ell} \right) (-m \omega^2 z_C) \end{aligned} $$

  • Support Moments $$ \begin{aligned} My_A & = \left( \frac{1}{2} - \frac{x_C}{\ell} \right) (-\omega^2 ( I_{xz} + m x_C z_C)) \\ My_B & = \left( \frac{1}{2} + \frac{x_C}{\ell} \right) (-\omega^2 ( I_{xz} + m x_C z_C)) \\ Mz_A & = \left( \frac{1}{2} - \frac{x_C}{\ell} \right) (+\omega^2 ( I_{xy} + m x_C y_C)) \\ Mz_B & = \left( \frac{1}{2} + \frac{x_C}{\ell} \right) (+\omega^2 ( I_{xy} + m x_C y_C)) \\ \end{aligned} $$

In this case a single rotating mass does not have any mass moment of inertia components $I_{xy}$ and $I_{xz}$ but if the center of mass is not at the center of the shaft axially $x_C \neq 0$, then a reaction couple proportional to $m \omega^2 x_C y_C$ exists. This is essentially the torque required to balance out the centrifugal force $m \omega^2 y_C$ at the supports.

PS. I hope you didn't expect 3D dynamics to be easy. Overall the above is straightforward but there are a lot of details to account for in order to get it right.


The way the diagram is drawn only makes sense if you treat the system in the rotating coordinate system, but I'll discuss it in both inertial and non-intertial frames.

Inertial frame

By assumption the ends of the rod are held fixed in space and the rod turns at steady angular frequency $\omega$. We use the frame of a nearby observer at rest with respect to the rod.

The two masses X and Y move in circles. That means they accelerate which means they are subject to a non-zero net force. This (centripetal) force is supplied by a tension where they are connected to the rod which means that the masses pull back on the rod, and because of their positions this amounts to a couple of size $d m_x r_x \omega^2$ that spins as the positions of the masses rotate. The bearing must supply a couple to compensate.

In this view the picture doesn't make much sense because the forces acting on the masses X and Y are inward.

Rotating frame

We use a frame of reference fixed in the body of the rod. This is a non-inertial frame, so their can be inertial-pseudoforces to deal with.

In particular the masses X and Y are subject to centrifugal pseudoforce pointing outward, and they are connected to rod, so the couple develops directly from the pseudoforces, and again the bearings must resist this couple.

The drawing makes sense in this context, but you have to understand that the couple is in a constant direction in the rotating frame, which means the direction of the couple turns at angular velocity $omega$.


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