# Forces that exert torque on a rigid body in rotation when angular momentum is not parallel to angular velocity

I'm confused about the rotation of a rigid body, when the angular momentum $\vec{L}$ is not parallel to the angular velocity $\vec{\omega}$. Consider a barbell with two equal masses that rotates around a vertical axis $z$ not passing through its center with angular velocity $\vec{\omega}$.

Taking a generic point $P$ on the $z$ axis as pivot point to calculate momenta, the total angular momentum $\vec{L}=\vec{L_1}+\vec{L_2}$ is not parallel to the rotation axis $z$, thus $\vec{L}$ follows a precession motion and, from the theorem of angular momentum, there must be a torque $\vec{\tau}$ on the system, exterted by external forces: $\vec{\tau}=\frac{d \vec{L}}{dt}\neq 0$.

What are the forces exerting this torque?

Weight has non zero torque $\vec{P}$ and it is an external force, but there is also the reaction of the support that must exert an opposite torque $\vec{R}$, since the barbell stays in this position during the rotation. $\vec{P}$ and $\vec{R}$ are opposite but not equal, in particular

$$\vec{P}+\vec{R}=\frac{d\vec{L}}{dt}\neq 0$$

Is this correct?

• The weight does not apply a moment on the center of mass, but it affects the support which does apply a moment on the center of mass. Thus a top will not precess at microgravity. – ja72 Apr 18 '16 at 17:57

## 2 Answers

You have to approach problems systematically, and not intuitively. Like I stated in a previous (accepted) answer, resolve everything on the center of mass, and only in the end transfer the quantities to a different point (like P) to get the results you want.

I start with the kinematics. Use $\ell_1$ and $\ell_2$ for the horizontal distances and $h$ for the vertical height above point P.

\begin{aligned} \vec{r}_{1C} & = \begin{pmatrix}-\ell_1 & h & 0 \end{pmatrix} & \vec{r}_{2C} &= \begin{pmatrix} \ell_2 & h & 0 \end{pmatrix}\\ \vec{\omega} & = \begin{pmatrix} 0&0& \Omega \end{pmatrix} & \vec{\alpha} & = \begin{pmatrix} 0&0& 0 \end{pmatrix}\\ \vec{v}_{1C} & = \begin{pmatrix} 0&0&\Omega \ell_1 \end{pmatrix} & \vec{v}_{2C} & = \begin{pmatrix} 0&0&-\Omega \ell_2 \end{pmatrix} \\ \vec{a}_{1C} & = \begin{pmatrix}\Omega^2 \ell_1&0&0 \end{pmatrix} & \vec{a}_{2C} & = \begin{pmatrix} -\Omega^2 \ell_2&0&0 \end{pmatrix} \end{aligned}

Now find the momentum at the center(s) of mass

\begin{aligned} \vec{p}_{1C} & = \begin{pmatrix} 0 & 0 & m \ell_1 \Omega \end{pmatrix} & \vec{p}_{2C} &= \begin{pmatrix} 0 & 0 & -m \ell_2 \Omega \end{pmatrix} \\ \vec{L}_{1C} & = \vec{0} & \vec{L}_{2C} & = \vec{0} \end{aligned}

Note point masses do not have angular momentum.

The support for each mass consists of two forces and one torque. These are defined at the support(s) and need to be transferred to the center(s) of mass

\begin{aligned} \vec{F}_1 & = \begin{pmatrix} R_{1x} & R_{1y} & 0 \end{pmatrix} & \vec{F}_2 & = \begin{pmatrix} -R_{2x} & R_{2y} & 0 \end{pmatrix} \\ \vec{M}_1 & = \begin{pmatrix} 0 & 0 & -\tau_1 \end{pmatrix} & \vec{M}_2 & = \begin{pmatrix} 0&0& \tau_2 \end{pmatrix} \\ \vec{M}_{1C} & = \begin{pmatrix} 0& 0 & \ell_1\,R_{1y}-\tau_1 \end{pmatrix} & \vec{M}_{2C} & = \begin{pmatrix} 0 & 0 & \tau_2 - \ell_2\,R_{2 y} \end{pmatrix} \end{aligned}

This is the moment that rotates the momentum vectors.

The equations of motion are

\begin{aligned} \vec{F}_{1C} - m g \hat{j} & = m \vec{a}_{1C} & \vec{F}_{2C} - m g \hat{j} & = m \vec{a}_{2C} \\ \vec{M}_{1C} &= I_{1C} \vec{\alpha} + \vec{\omega} \times I_{1C} \vec{\omega} = \vec{0} & \vec{M}_{2C} &= I_{2C} \vec{\alpha} + \vec{\omega} \times I_{2C} \vec{\omega} = \vec{0} \end{aligned}

Each mass moment of inertia about the center of mass is zero for a point mass. Since $\vec{\alpha}=\vec{0}$ the right hand side of the torque equation is zero. The result is the support forces as

\begin{aligned} R_{1x} &= m \ell_1 \Omega^2 & R_{1y} & = m \ell_2 \Omega^2 \\ R_{1y} & = m g & R_{2y} &= m g \\ \tau_1 &= m g \ell_1 & \tau_2 &= m g \ell_2 \end{aligned}

As you can see these forces are non-zero. The horizontal forces keep the masses going in circles, the vertical forces react to the weight, and the torques support the weight also. From all of this you see that the location of P does not matter. The value of $h$ does not appear in any result.

• Thanks for the answer! If I may ask, why did you impose $M_{1C}$ and $M_{2C}$ equal to zero? The change in angular momentum is not zero – Sørën Apr 18 '16 at 22:13
• It came from $I \vec{\alpha} + \vec{\omega} \times I \vec{\omega}$ which is zero for a constant rotation with a point mass. At each center of mass, the mass moment of inertia is zero. Even if you give the masses a radius $r$, due to symmetry this would be zero. – ja72 Apr 19 '16 at 1:18
• Also, I fixed a few typos in the equations above. – ja72 Apr 19 '16 at 1:23

In the example cited in the question, the center of mass of the barbell rotates about the pivot axis. Some external force needs to be applied to the barbell to make that happen. Assuming a constant angular velocity, this force is directed radially inward from the center of mass toward the axis. From the perspective of a point $P$ on the pivot axis but some distance $d$ from the barbell axis, this force also results in a torque.

So where does this force come from? Obviously, it comes from the axle that keeps the barbell rotating off-center.

This is why you need to ensure that the tires on your car are balanced.

• Thanks for the answer! There also the torque by weight, isn't it? The only explanation I came up with is that the torque by weight $\vec{P}$ and the one by the support $\vec{R}$ are opposite and their sum gives the change in angular momentum (following precession motion), $\vec{P}+\vec{R}=\frac{d\vec{L}}{dt}\neq 0$. Is this correct? – Sørën Apr 18 '16 at 13:41
• You can do the same without gravity :) then you have a force $F_0$. You can take a not-rotating barbell with gravity, and get some force $F_1$. If you rotate it in gravity, the force is $F_0+F_1$ – Ilja Apr 18 '16 at 13:46
• Thanks for the reply! But would both the torques due to the two forces add up to give the change in angular momentum? – Sørën Apr 18 '16 at 14:14
• yes, of course. You have part of the torque from the pivot that compensates the torque from gravity and you have the rest of the torque from the pivot that does the necessary change in the angular momentum. – Ilja Apr 22 '16 at 11:07
• It's like the centripetal force in normal circular motion. This is a good analogy, I find. You need some radial force to rotate momentum (which has always to be tangential) in a circular motion; and here you need some tangential torque to make the angular momentum (which will always follow the radial direction, if you ignore the constant component along the axis of rotation) circulate. – Ilja Apr 22 '16 at 11:09