trebuchet optimizing problem
|Aug28-06, 11:43 AM||#1|
trebuchet optimizing problem
hello everyone, i'm new here and i have a few questions about the parameters of a trebuchet. sorry for such a long post.
with a simple see saw trebuchet i'm given the projectile weight and counterweight, the length and mass density of the beam, and the height of the support. i am supposed to find the optimal distance of the counterweight to the pivot and the optimal starting angle that gives the longest range when the projectile is released. first i'm supposed to find the angle made when the projectile separates from the beam (when normal force with the beam equals zero) and with that i'm supposed to find the angular velocity at the time of separation. all energy is conserved and i can ignore friction, damping, and wind.
i've calculated the inertia for the whole system along with the initial torque and i've attempted to come up with an equation which equates the final kinetic energy with the change in potential energies.
here are my questions:
a) is there any way to solve for this problem without involving time?
b) for the total final kinetic energy, would this be Iw^2 + mp*a*(distance to pivot)? with w being the final velocity taken around the pivot, mp being the mass of the projectile, and a being its acceleration? i understand that the whole system has translational motion, not just the projectile, but i don't know how to structure this.
c) in order to find the angle where the projectile is launched (normal force with beam = 0), am i to assume that at this instance there is no angular acceleration? if there is no angular acceleration at this point, the angular acceleration would have been decreasing, how do this factor in?
d) when summing moments around the pivot, do i take into account the normal forces made by the counterweight and projectile? currently i've only included the two weights and the weight of the beam.
the most important question is solving for the separation angle, i'm very confused by this. once i have an equation for that, i think everything else will fall into place.
thank you for your time, i know this post is long and this looks bad as a first post but i'm in desperate need of help. any and all help is appreciated. thanks in advance.
|Aug29-06, 12:46 PM||#2|
I think it will break away when static friction is overcome. It is needed in order to give the projectile the necessary centripetal acceleration while it is accelerated tangentially by the beam. One component of the weight assists also to keep it on the beam. Even if the beam is not accelerating the system will rotate at a constant speed if friction at the pivot is ignored. So there is no reason for the projectile to leave the system unless it falls off under the influence of gravity due to the beam rotating too slow. Is there an outside shoulder for the projectile to keep it on the beam?
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