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Suppose I'm sitting on a sort of office chair without back and I'm rotating a bicycle wheel connected to a stick. The moments of inertia of these two object are known. Also let's assume my body is absolutely symmetric so the chair doesn't get inclined because of my non-uniform mass distribution. According to the law of conservation of angular momentum if the whole system is at rest and then I start to rotate the wheel, the chair will have to rotate too in order to keep constant the value of angular momentum. Unless there's friction between parts of the chair. I wonder how can I find angular momentum of the wheel that will be enough to cause a tiny motion of the chair. Even if I knew the force of friction chair has to overcome I don't know how it is related to angular momentum that the wheel is trying to cause. Hope my question is clear.

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The friction is related to a force by something like $\textbf{F} = \mu m g$, where $\mu$ is the adhesion coeff. What you are interested is a torque $$\frac{d}{dt}\textbf{L} = \textbf{M} = \textbf{r} \times \textbf{F} = d_\perp \cdot F_\perp$$ Therefore, you will have to find out the distance $d_\perp$ of the force with respect to the center of the rotation -- here $d_\perp$ is orthogonal to the direction of the force.

Let's assume that the force acts in z-direction, $\textbf{F} = F_\perp \textbf{e}_z= \mu m g \;\textbf{e}_z$, and that we know $d_\perp = \textbf{r} \cdot \textbf{e}_\rho$. Then the chair starts moving, if $\frac{\Delta L}{\Delta t} \ge \mu m g \, d_\perp$.

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  • $\begingroup$ if we know force then finding torque is a trivial thing, but even given its value how do we get the value of wheel's angular momentum that will force the chair to rotate? $\endgroup$ – Leo Lucini Sep 4 '17 at 16:27
  • $\begingroup$ I edited my answer. It's not the angular momentum, but the change in angular momentum which has to overcome the minimal torque given by $d_\perp \cdot F_\perp$. $\endgroup$ – Semoi Sep 4 '17 at 17:27

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