Question

A piece of wire is bent in the shape of a parabola $y=k{x}^2$ (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 the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

• A. $\frac{a^2}{2gk}$
• B. $\frac{a}{2gk}$
• C. $\frac{a^3}{2gk}$
• D. none of these

Answer 1

Answer: (B)

$\frac{a}{2gk}$

When the wire is accelerated to the right, the bead experiences a pseudo force ma to the left. The tangent makes an angle $\theta$ with a (pseudo).

The force $mgcos({90}^0-\theta )$ i.e. $mgsin\theta$ is acting down the tangent, opposing $macos\theta$

$\therefore macos\theta =mgsin\theta$ or $tan\theta =a/g$

For a function, $y=k{x}^2$ (the parabola-given)

$tan\theta =\frac{dy}{dx}=2kx=\frac{a}{g}\Rightarrow x=\frac{a}{2gk}.$

The distance of the new equilibrium position is $\frac{a}{2gk}$.