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 ware 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. $a/g$ $k$
• B. $a/2$ $gk$
• C. $2a /gk$
• D. $a/4$ $gk$

Answer 1

Answer: (B)

$ N \sin \theta =m g \Rightarrow N \cos \theta=m a $
$ \tan \theta =\frac{g}{a} \Rightarrow \cot \theta=\frac{a}{g}=\tan \left(90^{\circ}-\theta\right) $
$=\frac{d y}{d x}=2 k x \Rightarrow \therefore x=\frac{a}{2 k g} $