Question

a wire shaped in a parabola has a bead of mass m at the bottom most point.The bead is Initially at rest.Now the wire frame is acclerated along the wire with an acceleration of a.Locate the new equilibrium position of the bead.

Equation:y=kx²

refer the figure..​

Answer 1

$ma \cos( \alpha ) = mg \sin( \alpha ) \\ \tan( \alpha ) = \frac{a}{g} \\ is \: the \: equation \: of \: this \: trajectory \\ \\ now \: as \: we \: know \: that \: we \: calculate \: the \: angle \: on \: the \\ x - axis \: anticlockwise \: \\ the \: equation \: of \: trajectory \: will \: be \\ tan(\pi - \alpha ) = \frac{a}{g} \\ \\ hope \: that \: helps \: mate \:$

Answer 2

Explanation:

macos( θ)=mgsin( θ)

tan(θ) = a/g

y = kx^2

dy/dx = 2kx

tanθ= a/g = 2kx

So, x = a/2gk