1
$\begingroup$

I have a pretty basic question but I would like to be sure of what I think.

I have a setup like the one in the figure. I have two motors which can generate two forces $F_1$ and $F_2$ perpendicular to the plane of the drawing (coming out of the plane of the drawing). The black box is a rigid body of mass $m$ which has the Center of Gravity (CoG) in the middle (the white circle). We can think about two forces coming from two propellers as on a quadrotor UAV or a helicopter.

enter image description here

My question is: regarding $F_1$, since the corresponding arm is aligned with one of the axes of symmetry of the body (specifically y), if it will be activated it will contribute in torque terms only to a rotation $\phi$ with respect to the x-axis and therefore the acceleration around the x-axis will have this shape:

$I_x\ddot{\phi}=...+L \cdot F_1$

where $I_x$ is the moment of inertia with respect to the $x$ axis.

Instead, if I activate the motor 2, corresponding to the force $F_2$, since it is displaced from the axes of symmetry, but still parallel to one of them, I should have:

$I_x\ddot{\phi}=...+L \cdot F_2 - \underbrace{f(I_y,d_2,a,mg)}_{M_d}$

where $I_y$ is the moment of inertia with respect to the $y$ axis, $m$ the total mass of the rigid body and $g$ the acceleration due to gravity. The term $M_d$ should take into account the torque coming from the motor and the counter-torque coming from the fact that the point of attachment is not aligned with the $y$ axis. I would like to know how to write the expression of $f(\cdot)$ which should be quite easy to do.

Thanks for the help!

$\endgroup$
1
$\begingroup$

You would need to calculate $ \ I_{xy} \ for \ the \ angle\ \alpha $

Then you can apply the torque $F2*(d2^2+L^2)^{1/2}$, which is F2 working around the new axis for $I_{xy}$

One easy way of calculating $I_{xy} \ $is by constructing the Mohr circle.

On the x axis you draw the circle at the centre of:

$ Ix - (Ix - Iy)/2 \ and\ radius\ of: \ (Ix- Iy)/2 $

Then you draw an angle equal to $ 2*\alpha \ $ counter clockwise from Iy on x axis of the Mohr circle and draw the diameter by continuing this angle to intersect the other side of the circle. The projections of these two points on y and x axis are your new $I_{xy}$.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for the answer! I think you have got a typo (andradiuso) in the last formula. Could you check? $\endgroup$ – desmond13 Dec 17 '18 at 23:31
  • $\begingroup$ @kalmanIsAGameChanger, yes actually there is a typo but not where you think. basically in your case Ix is> Iy and the second part is correct, but the first part should be corrected by changing the + sign to minus. Basically Mohr circle is a circle with the diameter of max I - min I, and any other angle I can be determined by constructing the central angle of 2*alpha. $\endgroup$ – kamran Dec 18 '18 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.