Assume a rigid, cylindrical bar which rotates around $\Omega$ (axis of a engine).

The bar is immersed in a fluid of kinematic viscosity $\nu$ and has a plane motion.

What is the torque applied to the axis $\Omega$ needed to rotate the bar with an angular velocity $\omega$?

Attempt: For a viscous fluid the force $F$ applied by the fluid on the bar of length $d$ is $F=-k v$ where $v=d\omega$ and $k$ is a constant (characteristic of ?).

So the torque would be $$ \tau = F \cdot d = -k v^2. $$

EDIT: More precisely, I am at low Reynold's number because $\text{Re}=d^2\omega/\eta\ll 1$

About the constant $k$: it is equal to $\ell*\nu$ where $\ell$ is a coefficient taking into account the geometry of the object (here we can take $\ell=d$ the length of the bar?) and $\nu$ is the dynamic viscosity of the fluid $\nu=\rho \eta$ where $\rho$ is the volumetric mass.

Finally I get that the expression of the torque is $$ \tau = -\nu \omega \ell^3 $$


  • $\begingroup$ The formula you have mentioned is valid only for low Reynolds number flow. Since you mentioned engine, the flow is likely characterised by high Reynolds number, and is also turbulent. For a rough estimate you may assume that drag force is proportional to velocity-squared, but the required constants of proportionality must be determined from experiments. $\endgroup$
    – Deep
    May 31, 2018 at 10:09
  • $\begingroup$ I am indeed at low Reynolds number because my fluid is very viscous (kinematic viscosity = $\eta$=10000 mm^2/s), altought my engine has very low angular velocity. So I can use this method ? About the constant k : it is equal to l*$\nu$ where l is a coefficient taking into account the geometry of the object (here we can take l=d the length of the barre ?) and $\nu$ is the dynamic viscosity of the fluid $\nu=\rho \eta$ ? $\endgroup$
    – Smilia
    May 31, 2018 at 10:44
  • $\begingroup$ I am at low Reynolds number because $Re=d^2\omega/\eta \ll 1$. $\endgroup$
    – Smilia
    May 31, 2018 at 10:58

1 Answer 1


I don't know if this question still matters, but I will answer it. Basically, the idea is very general and one can derive all the equations of motion from first principles.

Given a system of mass points with mass $m_j \, :\, j =1 ... n$ and position vectors $\vec{r}_j \, :\, j =1 ... n$ in an inertial coordinate system they satisfy Newton's equations of mottion $$m_j \, \frac{d^2\vec{r}_j}{dt^2} = \vec{f}_j\Big(\, \vec{r}_1, ..., \vec{r}_n, \frac{d\vec{r}_1}{dt}, ..., \frac{d\vec{r}_n}{dt}, t\, \Big) \text{ for } j = 1...n$$ where $$\vec{f}_j = \vec{f}_j\Big(\, \vec{r}_1, ..., \vec{r}_n, \frac{d\vec{r}_1}{dt}, ..., \frac{d\vec{r}_n}{dt}, t\, \Big)$$ are the forces acting on each particle. We can cross-product multiply both sides of each equation as $$m_j\left( \vec{r}_j \times \frac{d^2\vec{r}_j}{dt^2} \right) = \vec{r}_j \times \vec{f}_j$$ Due to the properties of the cross-product $$\frac{d}{dt}\left( \vec{r}_j \times \frac{d\vec{r}_j}{dt} \right) \, =\, \frac{d\vec{r}_j}{dt} \times \frac{d\vec{r}_j}{dt} \, + \, \vec{r}_j \times \frac{d^2\vec{r}_j}{dt^2}\, = \, \vec{r}_j \times \frac{d^2\vec{r}_j}{dt^2} $$ Thus, we can rewrite the equations above as $$ \frac{d}{dt}\, \left(\, m_j \Big( \vec{r}_j \times \frac{d\vec{r}_j}{dt} \Big)\, \right) \, =\, \vec{r}_j \times \vec{f}_j $$ Finally, we can sum them together to obtain $$ \sum_{j=1}^{n} \, \frac{d}{dt}\, \left(\, m_j \Big( \vec{r}_j \times \frac{d\vec{r}_j}{dt} \Big)\, \right) \, =\, \sum_{j=1}^{n} \, \vec{r}_j \times \vec{f}_j $$ Now, let us focus on the rotating bar. Fix an inertial coordinate system $O\,\vec{e}_x \,\vec{e}_y\,\vec{e}_z $ with origin $O$ coinciding with the point of rotation of the bar. We represent the rotating rod as a system of continuum many particles, each of which is represented by the position vector $\vec{r} = \vec{r}(t) = x(t)\, \vec{e}_x + y(t)\, \vec{e}_y + z(t)\, \vec{e}_z $ pointing from $O$ to the point on the bar which represents the particles at time $t$. By $\mu(\vec{r})$ we represent the mass distribution (mass-density) of the bar. The important part is that the bar is a rigid body, so we can consider a coordinate system $O\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$ attached firmly to the bar, which means it rotates together with the bar and the bar is at rest with respect to $O\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$. The position of a point on the bar can be expressed as $\vec{R} = X\, \vec{E}_X + Y\, \vec{E}_Y + Z\, \vec{E}_Z $ wher $\vec{R}$ doesn't change with time with respect to $O\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$. Then, there is a time dependent rotation matrix $U = U(t)\, \in \, \text{SO}(3)$ such that $$\vec{r}(t) = U(t)\,\vec{R}$$ where $$\vec{r}(t) = \begin{bmatrix} x(t)\\y(t)\\z(t)\end{bmatrix} \, \text{ and } \, \vec{R} = \begin{bmatrix} X\\Y\\Z\end{bmatrix}$$ Thus, as already explained before, for each point from the rigid bar in the inertial coordinate system $O\,\vec{e}_x \,\vec{e}_y\,\vec{e}_z$ we have $$\frac{d}{dt}\, \left(\, \mu(\vec{r}) \Big( \vec{r} \times \frac{d\vec{r}}{dt} \Big)\, \right) \, =\, \vec{r} \times \vec{f}\Big(\vec{r}, \frac{d\vec{r}}{dt}, t\Big) $$ where $\vec{f} = \vec{f}\Big(\vec{r}, \frac{d\vec{r}}{dt}, t\Big)$ is the force acting on the bar at the point $\vec{r}$. The idea is to sum all the points on the bar up, but it is difficult because they change position with time. Therefore, switching to the system $O\,\vec{E}_X \,\vec{E}_Y\,\vec{E}_Z$ the position vectors $\vec{R}$ do not change with time. Thus, we calculate $$\frac{d\vec{r}}{dt} = \frac{d U}{dt}\, \vec{R} = U \big(\vec{\Omega} \times \vec{R}\big)$$ because for any time-dependent orthogonal matrix $U = U(t)$, there exists a time-dependent vector $\vec{\Omega} = \vec{\Omega}(t)$, called angular velocity, such that $$U^{-1}\frac{d U}{dt} \vec{R} = U^T\frac{d U}{dt} \vec{R} = \vec{\Omega} \times \vec{R}$$ Consequently, \begin{align}\frac{d}{dt}\, \left(\, \mu(\vec{r}) \Big( \vec{r} \times \frac{d\vec{r}}{dt} \Big)\, \right) \, =& \, \frac{d}{dt}\, \left(\, \mu(\vec{R}) \Big( U\,\vec{R} \times U\, \big(\vec{\Omega} \times \vec{R}\big)\,\Big)\, \right) = \mu(\vec{R})\, \frac{d}{dt}\, \left(\, U \,\Big(\vec{R} \times \big(\vec{\Omega} \times \vec{R}\big)\,\Big)\, \right) \end{align} To simplify the notation, let us set the following linear transformation acting linearly on $\vec{\Omega}$ and changing quadratically with respect to $\vec{R}$ $$A\big(\vec{R}\big)\, \vec{\Omega} = \vec{R} \times \big(\vec{\Omega} \times \vec{R}\big) = |\vec{R}|^2\, \vec{\Omega} - \big(\vec{R} \circ \vec{\Omega}\big)\, \vec{R}$$ However, recall that $\vec{R}$ does not change with time $t$. Thus \begin{align} \frac{d}{dt}\, \left(\, \mu(\vec{r}) \Big( \vec{r} \times \frac{d\vec{r}}{dt} \Big)\, \right) \, =& \, \mu(\vec{R})\, \frac{d}{dt}\, \left(\, U \,\Big(\vec{R} \times \big(\vec{\Omega} \times \vec{R}\big)\,\Big)\, \right) \, = \, \mu(\vec{R})\, \frac{d}{dt}\, \left(\, U \, A\big(\vec{R}\big)\, \vec{\Omega}\,\right)\\ =& \,\mu(\vec{R})\, \left(\, \frac{dU}{dt} \, A\big(\vec{R}\big)\, \vec{\Omega} \, + \, U \, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt} \, \right)\\ =& \,\mu(\vec{R})\, \left(\, U \,\Big[ \vec{\Omega} \times \Big(\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big)\,\Big]\, + \, U \, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt} \, \right)\\ =& \, U\, \left(\, \Big[ \vec{\Omega} \times \Big(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big)\,\Big]\, + \, \mu(\vec{R})\, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt} \, \right)\\ =&\, U\, \left(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt}\, + \,\Big[ \vec{\Omega} \times \Big(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big)\,\Big] \,\right) \end{align} On the other hand, the torque is \begin{align}\vec{r} \times \vec{f}\Big(\vec{r}, \frac{d\vec{r}}{dt}, t \Big) \, =& \, \Big[U\,\vec{R}\Big] \times \Big[ \, U\, U^{-1}\,\vec{f}\Big(\, U\vec{R}, \, U\, \big(\vec{\Omega}\times \vec{R}\big), \, t \,\Big)\, \Big] \\ =& U \Big[\, \vec{R} \times U^{-1}\,\vec{f}\Big(\, U\vec{R}, \, U\, \big(\vec{\Omega}\times \vec{R}\big), \, t \,\Big) \, \Big]\\ =& \, U \Big[\, \vec{R} \times U^T\vec{f}\Big(\, U\vec{R}, \, U\, \big(\vec{\Omega}\times \vec{R}\big), \, t \,\Big) \, \Big]\\ =& \, U \Big[\, \vec{R} \times \vec{F}\Big(\, \vec{R},\, U,\, \, \vec{\Omega}, \, t \,\Big) \, \Big] \end{align} Where $ \vec{F}\Big(\, \vec{R},\, U,\, \, \vec{\Omega}, \, t \,\Big) = U^T\vec{f}\Big(\, U\vec{R}, \, U\, \big(\vec{\Omega}\times \vec{R}\big), \, t \,\Big)$. Consequently, the equations of motion $$\frac{d}{dt}\, \left(\, \mu(\vec{r}) \Big( \vec{r} \times \frac{d\vec{r}}{dt} \Big)\, \right) \, = \, \vec{r} \times \vec{f}\Big(\vec{r}, \frac{d\vec{r}}{dt}, t \Big)$$ can be written as $$ U\, \left(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt}\, + \,\Big[ \vec{\Omega} \times \Big(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big)\,\Big] \,\right) \, = \, U \Big[\, \vec{R} \times \vec{F}\Big(\, \vec{R},\, U,\, \, \vec{\Omega}, \, t \,\Big) \, \Big]$$ When we multiply both sides of the equations with the inverse orthogonal matrix $U^{-1} = U^T$ we obtain the equations written in the rotating frame $O\,\vec{E}_X \vec{E}_Y \vec{E}_Z$, firmly attached to the rigid bar $$ \mu(\vec{R})\, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt}\, + \,\Big[ \vec{\Omega} \times \Big(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big)\,\Big] \, = \, \vec{R} \times \vec{F}\Big(\, \vec{R},\, U,\, \, \vec{\Omega}, \, t \,\Big) $$ However, these are the equations of motion for only one point $\vec{R}$ from the bar. Summing them up means to integrate them, obtaining $$ \int_{B}\, \left(\, \mu(\vec{R})\, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt}\, + \,\Big[ \vec{\Omega} \times \Big(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big)\,\Big] \, \right) \, dR \, = \, \int_{B}\, \left(\, \vec{R} \times \vec{F}\Big(\, \vec{R},\, U,\, \, \vec{\Omega}, \, t \,\Big)\, \right) \, dR $$ where $dR$ is the Lebesgue measure on the rigid body we are dealing with. The linearity of the integral yields $$ \int_{B}\, \left(\, \mu(\vec{R})\, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt}\, \right) \, dR \, + \, \vec{\Omega} \times \int_{B}\, \left(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big) \, \right) \, dR \, = \, \int_{B}\, \left(\, \vec{R} \times \vec{F}\Big(\, \vec{R},\, U,\, \, \vec{\Omega}, \, t \,\Big)\, \right) \, dR $$ Then we can define the linear transformation (linear matrix) $J$ as follows $$J\, \vec{\Omega} = \int_{B}\, \left(\,\mu(\vec{R})\, A\big(\vec{R}\big)\, \,\vec{\Omega}\,\Big) \, \right) \, dR$$ $$J\,\frac{d\vec{\Omega}}{dt} = \int_{B}\, \left(\, \mu(\vec{R})\, A\big(\vec{R}\big)\, \frac{d\vec{\Omega}}{dt}\, \right) \, dR $$ and the expression for the total torque is $$\vec{T}\big(U, \vec{\Omega}, t\big) \, = \, \int_{B}\, \left(\, \vec{R} \times \vec{F}\Big(\, \vec{R},\, U,\, \, \vec{\Omega}, \, t \,\Big)\, \right) \, dR$$ Thus, we have arrived at the standard equations of motion for the rigid body, which are \begin{align} J \, &\frac{d\vec{\Omega}}{dt} + \vec{\Omega} \times J\, \vec{\Omega} \, = \,\vec{T}\big(U, \vec{\Omega}, t\big)\\ & \frac{dU}{dt} = U \, (\vec{\Omega} \times \cdot) \end{align} where by $(\vec{\Omega} \times \cdot)$ I have denoted the skew-symmetric matrix that acts on any vector $\vec{R}$ as $$(\vec{\Omega} \times \cdot) \,\vec{R} = \vec{\Omega} \times \vec{R}$$ The matrix $J$ is what is called the inertia tensor. These equations are very general equations of motion for any rigid body rotating around a fixed point $O$.

In your case, the body $B$ is a bar, so one dimensional segment of length $l$. We can assume that the coordinate system $O\,\vec{E}_X \vec{E}_Y \vec{E}_Z$ is attached to the bar so that the bar is positioned along the $X$ axis only, so any vector $\vec{R} = X \, \vec{E}_X$. Furthermore, the rotation happens so that the $z-$axis stays fixed which means that $\vec{e}_z = \vec{E}_Z$ is fixed. The rotations that keep the $z-$axis fixed have the form $$U = U(\theta) = \begin{bmatrix} \cos(\theta) & - \, \sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ where $\theta = \theta(t)$ determines the matrix's change with respect to time, and consequently the angular velocity is $$\vec{\Omega}\times \vec{R} = \Big(U(\theta)^T \frac{d}{dt}U(\theta) \Big)\, \vec{R} = \frac{d\theta}{dt}\, \vec{E}_Z \times \vec{R}$$ for any vector $\vec{R}$, i.e. $\vec{\Omega} = \frac{d\theta}{dt}\, \vec{E}_Z = \omega\, \vec{E}_Z $ where $\omega = \frac{d\theta}{dt}$.

First, let us calculate the torques. If you have to account for gravity, say pointing down the $y-$axis of the inertial coordinate system $O\, \vec{e}_x\vec{e}_y\vec{e}_z$, then the gravitational force is $$\vec{f}_{gr} = -\, mg\,\vec{e}_y$$ and transforms to the rotating system $O\, \vec{E}_X\vec{E}_Y\vec{E}_Z$ as $$\vec{F}_{gr} = U^{-1}\, \vec{f}_{gr} = U^T\, \vec{f}_{gr} = -\, mg\,U^T\,\vec{e}_y = -\, mg\, \big(\sin(\theta)\, \vec{E}_X + \cos(\theta)\, \vec{E}_Y\big)$$ because $$\vec{e}_y =\begin{bmatrix} 0\\ 1 \\0 \end{bmatrix} \text{ and } \, U^T\,\vec{e}_y = \begin{bmatrix} \cos(\theta) & - \, \sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}^T \begin{bmatrix} 0\\ 1 \\0 \end{bmatrix} = \begin{bmatrix} \cos(\theta) & \sin(\theta) & 0 \\ -\, \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0\\ 1 \\0 \end{bmatrix} = \begin{bmatrix} \sin(\theta)\\ \cos(\theta) \\0 \end{bmatrix}$$ Thus $$\vec{R} \times \vec{F}_{gr} = - mg\, X \, \vec{E}_X \times \big(\sin(\theta)\, \vec{E}_X + \cos(\theta)\, \vec{E}_Y\big) = - \, mg\, X \, \cos(\theta) \, \vec{E}_X \times \vec{E}_Y = - \, mg\, X \, \cos(\theta) \, \vec{E}_Z$$ Thus, the total gravitational torque is $$\vec{T}_{gr} = - \, mg\, \int_{0}^{l} \, \Big(\, X \,\cos(\theta) \, \vec{E}_Z \,\Big)dX = - \, mg\, \Big( \int_{0}^{l} \, X \, dX \Big)\, \cos(\theta) \, \vec{E}_Z = -\, \frac{mgl^2}{2}\, \cos(\theta) \, \vec{E}_Z$$ Assume, the fluid resistance force is modeled in the inertial coordinate system $O\, \vec{e}_x\vec{e}_y\vec{e}_z$ as $$\vec{f} = - k \, \left|\frac{d\vec{r}}{dt}\right|^{\beta}\,\frac{d\vec{r}}{dt} $$ Having in mind that $$\frac{d\vec{r}}{dt} = U \big(\vec{\Omega} \times \vec{R}\big) = U \Big(\big[\omega\, \vec{E}_Z\big] \times \big[X\,\vec{E}_X\big]\Big) = \omega\,X \, U \big(\vec{E}_Z \times \vec{E}_X\big) = \omega\,X \, U \,\vec{E}_Y $$ $$\vec{f} = - k \, \left|\frac{d\vec{r}}{dt}\right|^{\beta}\,\frac{d\vec{r}}{dt} = - k \, \left|\omega\,X \, U \,\vec{E}_Y\right|^{\beta}\,\omega\,X \, U \,\vec{E}_Y = - k \,U \left( \left|\omega\,X \,\vec{E}_Y\right|^{\beta}\,\omega\,X \,\vec{E}_Y\right) $$ so $$\vec{F} = U^{-1}\vec{f} = - k \, \left|\omega\,X \,\vec{E}_Y\right|^{\beta}\,\omega\,X \,\vec{E}_Y = -\,k\, \omega^{\beta+1}\, X^{\beta+1} \big|\vec{E}_Y\big|^{\beta}\,\vec{E}_Y = -\,k\, \omega^{\beta+1}\, X^{\beta+1}\,\vec{E}_Y$$ Thus $$\vec{R} \times \vec{F} = \big[ X \, \vec{E}_X\big] \times \big[ -\,k\, \omega^{\beta+1}\, X^{\beta+1}\,\vec{E}_Y \big] = -\,k\, \omega^{\beta+1} \, X^{\beta+2} \, \big(\vec{E}_X \times \vec{E}_Y\big) = -\,k\, \omega^{\beta+1} \, X^{\beta+2} \, \vec{E}_Z$$ Finally the total resistance torque is $$\vec{T} = -\,k\, \int_{0}^{l}\Big(\, \omega^{\beta+1} \, X^{\beta+2} \, \vec{E}_Z \,\Big)dX = -\,k\, \Big(\int_{0}^{l} \, X^{\beta+2}dX \Big)\, \omega^{\beta+1}\, \vec{E}_Z = -\,\frac{k\,l^{\beta + 3}}{\beta+3}\, \omega^{\beta+1}\, \vec{E}_Z $$ To calculate the inertia tensor, one goes back to the expression \begin{align} A\big(\vec{R}\big)\,\vec{\Omega} =& |\vec{R}|^2\, \vec{\Omega} - \big(\vec{R} \circ \vec{\Omega}\big)\, \vec{R} = |X\, \vec{E}_X|^2\, \omega\, \vec{E}_Z - \big(X\,\vec{E}_X \circ \omega\,\vec{E}_Z\big)\, X\,\vec{E}_X\\ =& X^2\,\omega\,|\vec{E}_X|^2\, \vec{E}_Z - X\,\omega\,\big(\vec{E}_X \circ \vec{E}_Z\big)\, X\,\vec{E}_X\\ =& X^2\,\omega\, \vec{E}_Z \end{align} because the $X-$ and $Z-$axes are orthogonal and therefore $\big(\vec{E}_X \circ \vec{E}_Z\big) = 0$ as well as $|\vec{E}_X|^2 = 1$. Analogously $$A\big(\vec{R}\big)\,\frac{\vec{\Omega}}{dt} = X^2\,\frac{d\omega}{dt}\, \vec{E}_Z$$ Moreover, if we assume that the bar has length $l$ and mass $m$ which is distributed homogeneously along the bar, $\mu(\vec{R}) = \frac{m}{l}$. Thus $$ J\, \vec{\Omega} = \int_{B}\, \mu(\vec{R})\, A(\vec{R})\,\vec{\Omega}\, dR = \int_{0}^{l}\, \Big(\frac{m}{l}\, X^2\, dX \Big)\, {\omega}\, \vec{E}_Z = \frac{m}{l}\, \frac{l^3}{3}\, {\omega}\, \vec{E}_Z = \frac{m\,l^2}{3} \, \omega\, \vec{E}_Z$$ $$ J\, \frac{d\vec{\Omega}}{dt} = \int_{B}\, \mu(\vec{R})\, A(\vec{R})\,\frac{d\vec{\Omega}}{dt}\, dR = \int_{0}^{l}\, \Big(\frac{m}{l}\, X^2\, dX \Big)\, \frac{d\omega}{dt}\, \vec{E}_Z = \frac{m}{l}\, \frac{l^3}{3}\, \frac{d\omega}{dt}\, \vec{E}_Z = \frac{m\,l^2}{3} \, \frac{d\omega}{dt}\, \vec{E}_Z$$ Since in this case $$\vec{\Omega} \times J\,\vec{\Omega} = \omega\, \vec{E}_Z \times \Big(\frac{m\,l^2}{3} \, {\omega}\, \vec{E}_Z \Big) = \Big(\frac{m\,l^2}{3}\, \omega^2\Big)\, \vec{E}_Z \times \vec{E}_Z =\vec{0}$$ the general equations of motion of the system \begin{align} J \, &\frac{d\vec{\Omega}}{dt} + \vec{\Omega} \times J\, \vec{\Omega} \, = \,\vec{T}\big(U, \vec{\Omega}, t\big)\\ & \frac{dU}{dt} = U \, (\vec{\Omega} \times \cdot) \end{align} where by $(\vec{\Omega} \times \cdot)$ reduce to \begin{align} \frac{m\,l^2}{3} \, &\frac{d\omega}{dt}\, \vec{E}_Z \, = \, - \,\left(\frac{k\,l^{\beta+3}}{\beta+3}\right)\, \omega^{\beta+1}\, \vec{E}_Z \,-\,\frac{m\, g\, l^2}{2}\,\cos(\theta)\, \vec{E}_Z \\ & \frac{d\theta}{dt} = \omega \end{align} which, after equating the coefficients in front of the vector $\vec{E}_Z$, yield \begin{align} \left(\frac{m\,l^2}{3}\right) \, &\frac{d\omega}{dt}\, = \, - \,\left(\frac{k\,l^{\beta+3}}{\beta+3}\right)\, \omega^{\beta+1}\,-\,\left(\frac{m\, g\, l^2}{2}\right)\,\cos(\theta) \\ & \frac{d\theta}{dt} = \omega \end{align} or as one equation \begin{align} &\left(\frac{m\,l^2}{3}\right) \, \frac{d^2\theta}{dt^2}\, = \, - \,\left(\frac{k\,l^{\beta+3}}{\beta+3}\right)\, \left(\frac{d\theta}{dt}\right)^{\beta+1}\,-\,\left(\frac{m\, g\, l^2}{2}\right)\,\cos(\theta)\end{align} Now, if the bar is horizontal and the gravity doesn't apply, then the equations are \begin{align} \left(\frac{m\,l^2}{3}\right) \, &\frac{d\omega}{dt}\, = \, - \,\left(\frac{k\,l^{\beta+3}}{\beta+3}\right)\, \omega^{\beta+1}\\ & \frac{d\theta}{dt} = \omega \end{align} and if you want the bar to rotate at (or very near to) a fixed angular velocity $\omega_0 \, \vec{E}_Z$, you can choose the controlling torque $\vec{T}_{c}$ that cancels the fluid resistance to be say $$\vec{T}_{c} = \left(\,- \, K \, \big(\omega - \omega_0\big) \, + \, \left( \frac{k\,l^{\beta+3}}{\beta+3}\right)\, \omega^{\beta+1}\, \right) \, \vec{E}_Z$$ where $K>0$. This torque also adds Lyapunov asymptotic stability around $\omega_0$

  • $\begingroup$ Thanks, I have to take time to understand your answer, I will validate it later. $\endgroup$
    – Smilia
    Jan 14, 2019 at 8:59
  • $\begingroup$ where does the expression for the fluid resistance force come from: $\vec{f}=-k\left|\frac{d \vec{r}}{d t}\right|^\beta \frac{d \vec{r}}{d t}$? It introduces the parameters $k$ and $\beta$. To they have names? $\endgroup$ May 8, 2023 at 21:54
  • $\begingroup$ @DouglasJamesBock $\vec{f}$ is drag force. The parameter $\beta$ is traditionally equal to 1, and so the force $\vec{f}$ is proportional to the usual drag force that depends on the magnitude of the velocity squared and opposes the direction of motion. I just wrote $\beta$ to generalize it to different models of drag. However, for practical purposes just set $\beta = 1$. The constant coefficient $k$ is expected to be related to the density of the fluid, drag coefficient, some cross-sectional area... Maybe an empirically defined constant. $\endgroup$ May 8, 2023 at 23:12

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