You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Friction and Vertical Circular Motion Problems

A piece of wi...

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

\frac{a}{g k}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{a}{2 g k}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2 a}{g k}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{a}{4 g k}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{a}{2 g k}$
Let us consider the observer in the accelerated wire frame.
Let the coordinates of the bead be (x, y).
Slope of the tangent $= \frac{d y}{d x} = 2 k x$
Slope of normal $= - \frac{1}{2 k x} = t a n θ$
$N c o s θ = m a$
$N s i n θ = m g$

$t a n θ = \frac{g}{a} = - \frac{1}{2 k x}$
$\Rightarrow x = - \frac{a}{2 k g}$
x is in the second quadrant and hence is negative.
Hence, (B).

Suggest Corrections

Similar questions

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:

Q. 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

Q. 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 bad can slide on the wire without friction. It stays at the lowest point of the parabola when the wire at the 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 bead of mass ‘m’ is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is $x^{2}$ =4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point $y = 4 a$ on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by $y = a$ is (Acceleration due to gravity is g)

Join BYJU'S Learning Program