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I am trying to solve a problem involving a motor attached to an object. The motor is below the centre of gravity. When the motor is running it will apply a torque off axis to the object.

What will the effect of this be on the object?

Does the distance between the motor and the C.O.G have any effect on the magnitude of the response?

Object-Motor arrangement


marked as duplicate by Aaron Stevens, ja72, Jon Custer, John Rennie newtonian-mechanics Nov 5 '18 at 16:04

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  • 1
    $\begingroup$ FYI - A torque does not have a point of application, so it is always applied to the center of mass. When an object is forced to rotate about an point other than the center of mass, a reaction force must be present which accelerates the center of mass. $\endgroup$ – ja72 Nov 5 '18 at 1:06

The answer is that the torque will not have an effect on the response, but the geometry of the object will.

The equations of motion state that if a net zero force is acting on the body then the center of mass will move with constant velocity (or not move at all). In your case, if the body is free floating without any external forces, but a pure torque applied then the body will rotate about the center of mass.

But the amount of angular acceleration does depend on the mass moment of inertia (MMOI) of the object about the center of mass. It looks like from your diagram than the further away the motor is from the object, the higher the MMOI is, and hence the slower the angular acceleration.

Looking at your diagram you can see how increasing $\ell$ also changes where the center of mass id $c$ relative to the object geometric center. If you want the response of the object, then consider the velocity vector of the center of mass $\boldsymbol{v}_C$ and its derivative the acceleration $\boldsymbol{a}_C$ in the equations of motion. Additionally the rotational velocity vector $\boldsymbol{\omega}$ and its derivative $\boldsymbol{\alpha}$:

$$ \require{cancel} \begin{aligned} \boldsymbol{F}_B & = m \boldsymbol{a}_C \\ \boldsymbol{\tau}_B + (\boldsymbol{r}_B - \boldsymbol{r}_C) \times \boldsymbol{F}_B &= \mathrm{I}_C \boldsymbol{\alpha} + \boldsymbol{\omega} \times \mathrm{I}_C \boldsymbol{\omega} \end{aligned} $$

Where $\boldsymbol{F}_B$ is any force acting on the motor center B, along with the corresponding torque $\boldsymbol{\tau}_C$. The location of the motor is $\boldsymbol{r}_B$ and the location of the center of mass is $\boldsymbol{r}_C$, The value $m$ is the total mass, and $\mathrm{I}_C$ is the 3×3 mass moment of inertia matrix at the center of mass.

Here for a planar problem $\boldsymbol{\omega}\times \mathrm{I}_C \boldsymbol{\omega}=0$, and all the rotational vectors are "out-of-plane".


PS. In real life, it is very hard to apply a pure torque, without any reaction forces. If the motor in your case is entirely internal, then nothing is going to happen. But if the motor is rotating against any air resistance, then it is the reaction to the air resistance that will rotate the object.

Please look at this post (and links therein) for more details on this situation.


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