Question

a piece of wire is bent in the form of a parabola y=kx^2(y-axis vertical) with a bead of mass m on it.The bead can slide on the wire without friction.It stays at the lowest point of the parabola when the wire is at rest.The wire is now accelerated parallel to x-axis witn a constant acceleration a.The distance of the new equlibrium position of the bead,where the bead can stay at rest w.r.t the wire,from y-axis

Answer 1

dy/dx=2kx
angle b/w a and g is tan inverse g/a
g/a =2kx solve