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 stays at rest with respect to the wire, from the $y-$axis is

• A. $\frac{a}{gk}$
• B. $\frac{a}{2gk}$
• C. $\frac{2a}{gk}$
• D. $\frac{a}{4gk}$

Answer 1

The correct option is B $\frac{a}{2gk}$

$\text{Step 1: Choosing the Frame Of Reference and Drawing the FBD}$

Solving problem from the frame of wire which is accelerating towards the right.

So, Pseudo force on mass $m$ will act on left direction.

$\text{Pseudo force}=\text{Mass of body}\times \text{Acceleration of Observer}$

$=ma$

$\text{Step 2: Equilibrium conditon}$

At equilibrium position, the acceleration of the block will be zero in the chosen frame.

Therefore, Balancing the forces in x and y direction, we get:

$N\sin \theta =ma$ $....\left(1\right)$

$N\cos \theta =mg$ $....\left(2\right)$

Dividing these two equations, we get

$\tan \theta =\frac{a}{g}$ $....\left(3\right)$

$\text{Step 3: Slope of curve}$

Equation of curve, $y=k{x}^2$

From figure, we see that tangent to the curve at equilibrium point also makes the angle $\theta$ with the x axis.

Therefore, Slope of curve $=\tan \theta$

$\Rightarrow \frac{dy}{dx}=\tan \theta$

$\Rightarrow 2kx=\tan \theta$

$\Rightarrow 2kx=\frac{a}{g}$ (Using Equation $3$)

$\therefore x=\frac{a}{2gk}$

Hence, Option B is correct.