I have some questions about this exercise:

In an horizontal plane, a $OA$ bar with mass $m$ and length $a$ moves, with another bar $AB$ (same mass, double length) attached in the point A. In the point B, there is a force $F=K\frac {K} {r^2} \frac {B-O} {r}$ Find the equations of motion.

To get the equations, I used the balance of angular momentum. But my problems started when I tried to get the inertia tensor (I know that just with the inertia momentum it would be enough, but I want to find the tensor in order to understand more the physics behind all of this).

The inertia tensor of the first bar $OA$ is easy, just use the Steiner's Theorem to move every inertia momentum of a bar in his center off mass, to a momentum in one of its extremes.

But When I try to get every inertia momentum of the second bar $AB$, some difficulties appear:

I use the formula $I_a=I_b + M(R\times n)^2$ With $I_b$ as the inertia momentum of the bar in its center of mass, $M$ the mass of the body, $R$ the distance between the origin and the center of mass, and $n$ as a normal vector parallel with the principal axis I want to calculate.

But when I use this formula, I get results of the momentum which depends on the module of $R$, which depends on time. ¿Is this ok?

I had the idea of make a "double Steiner", translating the inertia momentum's axis to the extreme of the bar, and then use it again with an axis which crosses the origin. That way, it wouldn't depend on time. ¿Is this possible?

Which of the both ways is the correct one?

  • $\begingroup$ You have not specified the orientations of the bars. $\endgroup$ Commented Feb 3, 2014 at 16:41


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.