# Rotational Equilibrium of a Rotating Titled Rod about a Vertical Axis

Consider:

Here a rod AB is rotating about a vertical axis with a constant angular velocity and is supposed to not fall down but maintain its angle with the vertical axis. Here the point of suspension is frictionless.

Here I have two arguments:

1. The rod has an unbalanced torque about the axis of rotation due to gravity. The Normal reaction passes through axis of rotation and hence has no torque about the fixed axis. So the rod must move closer to the axis with time.

2. It may happen that the horizontal component of Normal provide for centripetal acceleration and vertical component of Normal balances gravity. Hence the centre of mass may undergo circular motion in a fixed plane.

Cleary 1 and 2 contradict each other. But I also have a feeling that 1 and 2 should be consistent and I am missing something big here.

What am I missing here? How can the rod stay in equilibrium ?

## Edit:

I have developed a way to explain equilibrium in the rod by taking elementary masses and treating each part as a conical pendulum.

However, when I take the rod as whole:

Here the angular momentum of the rod is constant about Y axis even though a net torque is acting.

What am I missing here?

• Please don't post photos of text. It is always better to quote your question rather than posting a photo. – lee Jan 14 at 5:24
• Well that assumption that you took in 1st point was irrelevant to it. Look at it this way the normal force is definitely along the axis of motion but when you say that you don't mean to say it for the rotation that would be caused by the torque due to gravity. Try resolving the normal force parallel and perpendicular to the rod, you'll see it, and one more thing to note that you don't need to have the forces individually balancing the torque you're mention. – buddy001 Jan 16 at 6:04
• The torque vector is perpendicular to the angular momentum vector in your last query why would you expect it to change the angular momentum? – JustJohan Jan 17 at 8:48

to reach this steady state configuration , you can obtain from the equation of motion at the center of mass, that the magnitude of the angular velocity $$~\omega~$$ must be:

$$\omega^2=2\,{\frac {mgL}{\cos \left( \theta \right) \left( {L}^{2}m+4\,{\it I_x }-4\,{\it I_y} \right) }}$$

$$~\theta~$$ is the constant angle between the rod and the vertical axis $$~L~$$ is the rod length and $$~I_x~,I_y~$$ are the rod inertias about the local x and y axis . with $$~I_x=I_y$$

$$\Rightarrow$$

$$\omega^2=\frac{2\,g}{L\cos(\theta)}$$

The equations

Translation \begin{align*} &\text{Position vector to the CM }\\ &\vec{R}=S_y(\omega\,\tau)\,\frac{L}{2}\left[ \begin {array}{c} \sin \left( \theta \right) \\ \cos \left( \theta \right) \\ 0\end {array} \right]\\ &\text{with}\\ &S_y= \left[ \begin {array}{ccc} \cos \left( \omega\,\tau \right) &0&\sin \left( \omega\,\tau \right) \\ 0&1&0 \\ -\sin \left( \omega\,\tau \right) &0&\cos \left( \omega\,\tau \right) \end {array} \right]\\ &\Rightarrow\\ &\text{kinetic energy}\\ &T_T=\frac{m}{2}\vec{\dot{R}}^T\,\vec{\dot{R}}\\ &\text{potential energy}\\ &U=-m\,g\,\vec{R}_y \end{align*} rotation \begin{align*} &\text{transformation matrix}\\ &S=S_y(\omega\,\tau)\,S_z(\theta)\\ &S_z=\left[ \begin {array}{ccc} \cos \left( \theta \right) &-\sin \left( \theta \right) &0\\ \sin \left( \theta \right) &\cos \left( \theta \right) &0\\ 0&0&1\end {array} \right]\\ &\Rightarrow\\ &\vec{\Omega}= \left[ \begin {array}{c} \sin \left( \omega\,\tau \right) \dot\theta \\ \omega\\ \cos \left( \omega\, \tau \right) \dot\theta \end {array} \right]\\ &\text{kinetic energy}\\ &T_R=\frac{1}{2}\vec{\Omega}^T\,S^T\,\Theta\,S\,\vec{\Omega}\\\\ &\text{where ~\Theta~ the rod inertia tensor }\\ &\Theta=\left[ \begin {array}{ccc} J_{{x}}&0&0\\ 0&J_{{y}}&0 \\ 0&0&J_{{z}}\end {array} \right]\\ &\text{the lagrangian }\\ &\mathcal{L}=\frac 18\,{\dot\theta }^{2} \left( {L}^{2}m+4\,J_{{z}} \right) +\frac 12\,mgL\cos \left( \theta \right)\\& -\frac 18\,{\omega}^{2} \left( -{L}^{2}m+{L}^{2}m \left( \cos \left( \theta \right) \right) ^{2}-4\,J_{{x}}+4\,J_{{x}} \left( \cos \left( \theta \right) \right) ^{2}-4\, \left( \cos \left( \theta \right) \right) ^{2}J_{{y}} \right) \end{align*} from here you can obtain the equation of motion \begin{align*} & \left( \frac 14\,{L}^{2}m+J_{{z}} \right) \ddot\theta \\&+ \left( -\frac 14\,m{ L}^{2}\cos \left( \theta \right) -J_{{x}}\cos \left( \theta \right) + \cos \left( \theta \right) J_{{y}} \right) \sin \left( \theta \right) {\omega}^{2}\\&+\frac 12\,mgL\sin \left( \theta \right)=0\tag 1 \end{align*} steady state means that $$~\ddot{\theta}=0~$$ solve Eq. (1) for $$~\omega^2$$ and you obtain the above solution

• I am new to this. So is moment of inertia a vector here? – Tony Stark Jan 17 at 10:57
• I will write you the equations – Eli Jan 17 at 11:30

Let there be a ball of mass 'M' tethered to a massless string you will see we run into the same problems encountered with the rod but this is more simplified.

The angular momentum about the topmost point will not be along the axis of rotation but rather like this.(For an instant)

Here the ball moves with velocity into the plane and the torque due to gravity will also be into the plane. You would expect that this torque would cause the ball to move closer to the axis but this will not happen. Yes quite counter-intuitive.As there is angular momentum the angular momentum vector will "follow" the torque due to gravity vector.

This video by Eugene demonstrates how the angular momentum follows the torque vector and it is quite similar to this scenario as you would expect the gyroscope to topple over but it does not due to the reasons discussed above

"Here the angular momentum of the rod is constant about Y axis even though a net torque is acting." This is not correct.

Note that the angular momentum is not a constant. It depends on the origin. Taking the origin at the supporting point, on the top of the rod. The resultant angular momentum is perpendicular to the rod, as shown in Fig. $$\vec{L} = \int \vec{r} \times \vec{v} \rho dr,$$ Figure-1.

where $$\rho$$ is the linear density of the rod.

Therefore, the angular momentum is cycling around on the top of a cone, fig-2. The torque on the center of mass $$\vec{\tau} = \vec{r} \times \vec{f}_g$$ becomes a vector in the horizontal plane. As the angular momentum precesses around the cone, the torque is the driving force to make $$\vec{L}$$ turning. $$\frac{\Delta \vec{L}}{\Delta t} = \vec{\tau}.$$

Thinking in the equation of $$\vec{L} = I \vec{\omega}$$, be reminded that the inertial moment $$I$$ is a symmetric matrix of $$3\times 3$$, and in this case, the matrix elements are functions of time. This time-dependent $$I(t)$$ results in a prcessing angular momentum, $$\vec{L}(t)$$, even though the $$\vec{\omega}$$ is a constan vector.

$$\vec{L} (t)= \begin{bmatrix} I_{xx}(t) & I_{xy}(t) & I_{xz}(t) \\ I_{xy}(t) & I_{yy}(t) & I_{yz}(t) \\ I_{xz}(t) & I_{yz}(t) & I_{zz}(t) \end{bmatrix} \times \begin{bmatrix} 0\\ 0 \\ \omega \end{bmatrix}$$ The elements of inertial moment matrix being functions of time is clear as we set up a fixed coordinate frame and calculate the matrix elements, e.g. $$I_{xy} = -\int x y \rho d l$$. It is a general situation when rotating not around the center of gravity and not on a principle axis. The precession of angular momentum.