A bead, under the influence of gravity, slides along a frictionless wire whose height is given by the function y(x). Assume that at position (x,y) = (0,0), the wire is horizontal and the bead passes this point with a given speed v0 to the right. What should the shape of the wire be (that is, what is y as a function of x) so that the horizontal speed remains v0 at all times? One solution is simply y = 0. Find the other.

Question

anonymous · Answer

I started out this problem by considering the initial and final energies of the bead, Ei=12mv20 and Ef=mgy+12mv2. From there, I can find the velocity of the bead at a point on the wire to be v=√(v(o)^2−2gy). This is the part where I start to get tripped up. I know that the horizontal component of the velocity (v0) would be equal to vcosθ. I already have an expression for v, but I'm not quite sure how to represent the cosine in terms of variables I already have. Once I have that, I'm pretty sure I'd be in the clear for solving this one. Any help would be appreciated! Thanks.

anonymous · Answer

anyone??

perl · Answer

zuccini

anonymous · Answer

hurray! got it took a while but yeah. Answer if anyone is interested is y(x)= (v0 -1)/2g