Bead on wire with constant velocity

[treynolds147]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I started out this problem by considering the initial and final energies of the bead, E_{i} = \frac{1}{2}mv^{2}_{0} and E_{f} = mgy + \frac{1}{2}mv^{2}. From there, I can find the velocity of the bead at a point on the wire to be v = \sqrt{v^{2}_{0} - 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 v\cos{\theta}. 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.

 

[PeterO]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I started out this problem by considering the initial and final energies of the bead, E_{i} = \frac{1}{2}mv^{2}_{0} and E_{f} = mgy + \frac{1}{2}mv^{2}. From there, I can find the velocity of the bead at a point on the wire to be v = \sqrt{v^{2}_{0} - 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 v\cos{\theta}. 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.

If the wire ended at (0,0), the bead would become a projectile, and the horizontal component of the velocity would remain the same, so perhaps if the wire just followed the appropriate parabolic path down that the projectile would have followed?.