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If a rod is turned around it's own axis, force/torque is needed.

Let's assume the rod it self has a diameter $d$, a length $l$, the mass $m$ and should be turned by a motor on one side. The other side is laying on a support, that prevents it from moving sideways which is not frictionless $\mu_{rod-supp}$.

Sketch

How strong must this motor be?!

So far I tried around a bit with Moment of Inertia. But I don't know how to go from there.

It is a real life problem I need to solve. The rod needs to turn once in 3-8 minutes and I have to find a fitting stepper motor for that, that's why I need the torque.

Wasabi
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Auge
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  • Can you add a diagram? – Jonathan R Swift Aug 17 '18 at 09:59
  • Welcome to Engineering! This looks like a " [homework question](http://meta.engineering.stackexchange.com/q/121/1832)" (even if it isn't homework). In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please [edit] your question to include this information. – Wasabi Aug 17 '18 at 11:26
  • I have added a small sketch and a few infos on what I did so far. – Auge Aug 18 '18 at 07:51
  • Possible duplicate of [Understanding required torque for a motor lifting a weight](https://engineering.stackexchange.com/questions/2826/understanding-required-torque-for-a-motor-lifting-a-weight) – hazzey Aug 22 '18 at 14:34

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If you assume that 100% of the weight is carried by the support, then the normal force is just $mg$, so the friction force is $mg\mu$, and the torque required to overcome friction is $mg\mu d/2$. If you want to be able to acceleration the rod quickly, then the polar moment of inertia will come into play. However, since you indicated a rate of 1 rev every few minutes, the accelerations involved here will be trivial. The friction force will dominate. Take the calculation for friction and add a little margin and call it a day.

Daniel K
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