Recent content by Buck268

Less mass means faster acceleration. Less rotational mass means faster roational acceleration. And since F=ma, the force required accelerate the pulley to the same speed would be less for a less massive pulley. The difference of a few ounces on your crank puley isn't going to do anything at...

If you are referencing your x-axis perpendicular to g, then no there is no force in the x direction. Personally, I would set the axis such that x is along the average of the function of the path of decent, y being perpendicular, g will be a resultant of two vectors, Gx and Gy. I'm not really...

Run "skinnies" up front (single disc CD wheels) and "slicks" out back (multi disc wheels) :D

Excellent, thank you for the assistance :biggrin: ...now on to providing a general proof that for a given speed there are two angles between 0 and 90 degree which have the same range... Yay...

Ahhh well I've came up with \gamma = \frac{1}{2}ArcCot(Tan \theta)

Ahhh I'm retared... I think that everything is supposed to be a constant except gamma... Maybe that would make it easier

the equation for the range (r(t) or s if you prefere) seems to work out to: r(t) = \frac{2Vo^2 * Sin\gamma * Cos\gamma}{g*Cos\theta} - \frac{2Vo^2 * Sin^2 \gamma}{gCos\theta}*Tan\theta hmmmm...

OK, well I've never seen that form, but I'm using R(t) = do + Vox*t - .5Axt^2. Looks to me like s = r(t), u = Vox, and of course Do = 0... Anyways, The problem I'm running into is simplifiing to a deferentiable form, since I'm trying to find the maximum gamma (probably should solve for gamma...

Well, right now I'm working on one helluva a problem... Basically, a projectile is given a velocity of V sub "o" (Vo). The launch angle is gamma degrees above an surfaced which is inclined theta degrees above the horizontal. I'm tasked with finding its range along the inclined surface as well...