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I have a motor and gearbox that is rated for a certain amount of torque in newton meters. Its primary load $M_1$ is connected by a rigid rod (assumed massless) length $L_1$. So I reckon the (max) torque required is $L_1 {\times} M_1 {\times} g$, achieved at the rotation angle that's directly perpendicular / furthest out from the motor+gearbox.

However that's a bit outside of my motor+gearbox's specs. So my plan was to put a "counterweight" directly opposed on the other side to oppose this torque, and therefore get my requirements within range. The counterweight is just a mass $M_2$ attached a distance $L_2$ in the same axis of the first load, but going on the other side of the motor+gearbox

Check the image below:
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Questions:

  1. Does this scheme physically work, as in lower my (max) torque required on the motor+gearbox to $M_1 {\times} L_1 {\times} g - M_2 {\times} L_2 {\times} g$?

  2. If so this seems pretty sweet, but what other physics quantities might I be sacrificing in this setup? Speed? Forces on the shaft? Any other practical considerations I should think about?

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    $\begingroup$ With an asymmetric load you will induce extra stresses on the shaft of the motor and hence the motor itself which will cause the motor to vibrate. You really want the centre of mass of your rod assembly to be at the shaft to reduce/eliminate the vibration. Adding a counterweight will move the centre of mass of the rod assembly towards the shaft. $\endgroup$ – Farcher May 23 '18 at 7:10
  • $\begingroup$ So are you saying that my counterweight setup is advantageous in this sense? $\endgroup$ – JDS May 24 '18 at 8:36
  • $\begingroup$ A counterweight will help “balance” the system. $\endgroup$ – Farcher May 24 '18 at 9:19
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Torque $\tau$ is related to angular acceleration $\alpha$ by the equation:

$\tau = I\alpha$

where $I$ is the moment of inertia of the rotating body. The moment of inertia in the case of a point mass $m$ is:

$I = mr^2$

The moment of inertia opposing acceleration by the motor's torque will therefore increase due to the additional mass $M_2$ by the amount: $I_{incr} = M_2r_2^2$

Practically, this means that while you will now be able to get the motor to lift $M_1$ because it is counterbalanced by $M_2$, applying torque to lower $M_1$ from any position from which it can be lowered, and for an equivalent torque, will now result in less angular acceleration. Your system will be more 'sluggish', i.e. it will be less responsive.

Yes, your torque formulation is correct.

Further about 'sluggish':

In the figure below we analyse motion of $M_1$, in the right hemisphere. The analysis applies for the mirror image case in the left hemisphere. The torque required to lift $M_1$ is proportional to $cos\theta$, where $\theta$ is the angle between the vertical (mass weight) and the direction of action of the torque. Without $M_2$, the motor would not be able to lift $M_1$ between some points $A$ and $C$, since $cos\theta$ would be high, close to $1$ near $\theta = 0^o$ (horizontal shaft position). The addition of $M_2$ will facilitate upward motion of $M_1$ between $A$ and $C$, a positive thing.

The negative effect of adding $M_2$ is that all acceleration to lower $M_1$ (red arrows in figure) will necessary require longer acceleration time since $\alpha = \tau / I$ and $I$ is now larger. Changes in rotational direction will also take longer because of the higher angular momentum to be overcome. There will be a larger time lag between changing torque direction in the motor (by electrical control), and the masses changing direction. This delayed response is called sluggishness.

enter image description here

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  • $\begingroup$ Thanks for your answer. I didn't look specifically at moment of inertias, but you would say my formulation of the torque is correct? And can you follow up on what you mean by more "sluggish" / less responsive? $\endgroup$ – JDS May 23 '18 at 6:17
  • $\begingroup$ @JDS Yes, your torque formulation is correct. I have added to the question to affirm that and on sluggishness. $\endgroup$ – Dlamini May 23 '18 at 8:28
  • $\begingroup$ Hey fantastic update on your answer, thanks so much! $\endgroup$ – JDS May 24 '18 at 1:36

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