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aerowenn
aerowenn is offline
#6
Jan21-13, 12:43 AM
P: 19
Quote Quote by haruspex View Post
That's relatively easy, then.
Consider an element ds of an arm. At time t it will be (in polar relative to hub centre) at some position (r=r(t), θ=θ(t)). The tangential acceleration is ##r\ddot{\theta}+2\dot{r}\dot{\theta}##. See http://en.wikipedia.org/wiki/Polar_c...ector_calculus
If the element has mass dm (=ρds if uniform density) then the reactive torque is ##-(r\ddot{\theta}+2\dot{r}\dot{\theta})r.dm##

I see, this makes sense. Will this account for the flexibility of the arm as well? I suppose I would specify that motion somehow. Also, this gives torque from the arm, so I can translate this directly to where the arm is mounted on the central hub correct?

The hub will have a moment of inertia as well, and the end result I'm trying to reach is its angular motion around the axis "O" (in the center). In the end I would have a torque from the arm on one side of the equation and the torque applied to the hub on the other.

Solving for the motion on the hub would it come out to angular acceleration?

I really appreciate the help, this seems to be a problem that's not easily found by searching or flipping through text books.