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 $m$ placed 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 positive $x$-axis with a constant acceleration $a$. The distance of new equilibrium position of the bead from $y$-axis, where the bead can stay at rest with respect to the wire will be:

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

Answer 1

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

Since wire is accelerated along +ve X-axis with an acceleration $a$. Hence to solve from the frame of reference of wire (i.e wire will remain stationary with respect to this frame), we have to apply a pseudo-force $F_{pseudo}=ma$ in the -ve X-direction on the bead.

FBD of bead is shown in figure below:

Applying the condition for bead, to remain stationary. at rest with respect to wire.

$\Rightarrow$ Equilibrium equation for bead in X and Y directions respectively given as:

$N\sin \theta =ma$........$i$
$N\cos \theta =mg$.........$ii$

Dividing these two equations, we get:

$\tan \theta =\frac{a}{g}..........iii$​

Also, $\tan \theta =\text{slope}=\frac{dy}{dx}......iv$​

where $y=k{x}^2$, differentiating w.r.t $x$ and putting in Eq. $iv$ then equating with Eq. $iii$:

$\Rightarrow \tan \theta =2kx$

Or, $\frac{a}{g}=2kx$​

$\therefore x=\frac{a}{2gk}$​