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I'm trying to determine a suitable motor in order to reciprocate a small wooden table that is hanging from the ceiling for an art installation. I've made a basic diagram that illustrates how I think I might be able to model the motion.

I'm assuming I can model the small table as a hanging (from length, $L$, from a ceiling) pendulum mass ($m= 7kg$) that requires a force, $F_p$, in order to start the motion. Regarding the motor setup: it's basically an offset cam attached to an upright motor that will apply a force $F_p$ to push the table in a horizontal circular motion at an angle, $\theta$, resulting in a displacement of $\Delta$x. I'm assuming $\theta$ is small with respect to $L$ so that $\sin(\theta) \approx tan(\theta) $. With this approximation, I'm deriving the following expression for the maximum required pushing force $F_p = mgsin(\theta) = mg \frac{\Delta x}{L}$.

Let's say I want to merely move the hanging table ($\Delta x$) 2 cm and the length of the hanging string is, $L= 0.5 m$. I'm getting $F_p = 7 kg *9.41 m/s^2*(0.02/0.5)$ = 2.74 N-m. Which means I would need a motor capable of providing at least 2.74 N-m holding torque. I should maybe add that I'm wanting to reciprocate the table at a speed between 0.5 - 2 Hz. NB: I'm neglecting the need to accelerate the mass within a certain time frame and also the mass of the string. I'm also neglecting the other supporting strings that may be holding up the table from the ceiling.

Is my thinking on this correct? Intuitively I feel like that's a lot of torque for such a small displacement so I just wanted to check.

Many thanks in advance for any help you may be able to provide!

enter image description here

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  • $\begingroup$ Just for clarification: is the motion planar or does it rotate around a vertical axis? I understood it's the former $\endgroup$
    – NMech
    Nov 17 '21 at 11:45
  • $\begingroup$ @NMech yeah the motor is fixed to a structure on the ground (it doesn't rotate itself at all) and just rotates around a vertical axis. $\endgroup$
    – melonhead
    Nov 17 '21 at 11:52
  • $\begingroup$ @NMech sorry perhaps that wasn't clear: so the motor itself is fixed into place on the ground but there's an offset cam attached to the motor shaft which is inserted into a notch on the bottom side of the table. The table traces our a circle of radius = $\Delta$x on the horizontal plane $\endgroup$
    – melonhead
    Nov 17 '21 at 11:58
  • $\begingroup$ I think I have a clearer picture now, however I still think that the table will wobble/rotate around the axis of the motor. (of course it depends on the coupling of the motor pin and the slot on the table. $\endgroup$
    – NMech
    Nov 17 '21 at 12:07
  • $\begingroup$ @NMech yeah I was planning on creating enough clearance for the vertical and horizontal tilt from the small displacement. $\endgroup$
    – melonhead
    Nov 17 '21 at 12:32
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Hopefully I will find some time later to put up some drawings, that will make this easier to understand.


IMHO, you are correct 2.74 N-m seems too high a torque for this, (if I have the correct idea). As I mentioned in the comments the table will rotate around the axis motor. Therefore the force that you are calculating ($F_p$) will always be in the radial direction (i.e. pushing the pin toward the direction of motor axis).

enter image description here

Figure: top view of the motor/table

That force is not indirectly related to the resistance experienced by the motor. I.e. $F_p$ will create friction and the friction will be opposing the movement of the element.

So what I'd do, is I'd use a rolling element there (like a self adjusting rolling bearing). That will allow rolling of the element inside the slot and reduce greatly the load.

If you use the rolling bearing then there is also friction (albeit much smaller), however in that case the calculations can become a bit more complicated, because then the inertial forces start to become important. You need to change the direction of a 7 kg mass which can be somewhat demanding.

In this case, the natural vibration of the structure will also play in. The good thing is that with the choice of length -- 0.5 m -- the natural frequency of the pendulum is about 0.7Hz (although this is a very rough approximation for this structure) which means that the natural frequency and the excitation frequency are pretty close, so it might be possible to adjust the length and end up with a resonant structure that would not need much force to move.


Finally, another thing you need to worry about is the fact that if you fix the motor, then all the load will be carried by the motor bearings. That can be devastating for the motor, so you might want to reconsider the fixing of the motor.

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  • $\begingroup$ Hey, thanks so much for your response! Yeah, good idea on using a self-adjusting bearing - also, I didn't even think about the resonant frequency of the pendulum mass! When you say consider the fixing of the motor, do you mean how it's attached to the ground? $\endgroup$
    – melonhead
    Nov 20 '21 at 15:18
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I think you have made a mistake in the calculation.

$F_p = 7 kg *9.81 m/s^2*\frac{0.02}{0.5}$ = 2.75 N (not Nm!) is the amount of force required to hold the table. $\frac{0.02}{0.5}$ is the approximate angle, using the approximation that $\tan\theta\approx\theta$.

You specified gravity as $9.41 m/s^2$ but 9.81 is typical and your result agrees with 9.81 so I assume that was just a typo.

This force is again multiplied by the length of the lever arm to get torque. $2.75N*0.02m = 0.055Nm$. This is your torque.

I think you mixed up the formula because the number 0.02 appears twice but you only used it once.

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  • $\begingroup$ ah yes you’re right, typo in g and I forgot about factoring in the lever arm length. So 2.75 N would be force required if lever arm was 1 m. Thanks for your response! $\endgroup$
    – melonhead
    Nov 18 '21 at 12:18

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