Question

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Answer 1

Given Question

A piece of wire is bent in the shape of a parabola y = kx^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 –

$\large\underline{\sf{Solution-}}$

Let assume that, the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is 'x'.

Now, Given equation of Parabola is

$\rm :\longmapsto\:y = {kx}^{2}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} {kx}^{2}$

$\bf\implies \:\dfrac{dy}{dx} = 2kx$

As we know, if a line makes an angle p with the positive direction of x - axis, then,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{dy}{dx} = tanp \: \: }}$

So,

$\bf\implies \:2kx = tanp - - - - (1)$

Now,

It is given that, The wire is now accelerated parallel to the x- axis with a constant acceleration a.

So, the two forces acted on the parabola.

1. Normal force, due to acceleration

2. Gravitational force.

Now,

Force F, can be resolved in to two components, F sinp and F cosp

So,

$\boxed{ \tt{ \: Fsinp = ma}}$

and

$\boxed{ \tt{ \: Fcosp = mg}}$

So, on dividing these two above equations, we get

$\rm :\longmapsto\:\dfrac{Fsinp}{Fcosp} = \dfrac{ma}{mg}$

$\bf\implies \:tanp = \dfrac{a}{g} - - - - (2)$

So, from equation (1) and (2), we get

$\rm :\longmapsto\:2kx = \dfrac{a}{g}$

$\bf\implies \:\boxed{ \tt{ \: \:x = \dfrac{a}{2kg} \: \: }}$

Answer 2

Answer:

Given Question

A piece of wire is bent in the shape of a parabola y = kx^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 –

$\large\underline{\sf{Solution-}}$

Let assume that, the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is 'x'.

Now, Given equation of Parabola is

$\rm :\longmapsto\:y = {kx}^{2}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx} {kx}^{2}$

$\bf\implies \:\dfrac{dy}{dx} = 2kx$

As we know, if a line makes an angle p with the positive direction of x - axis, then,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{dy}{dx} = tanp \: \: }}$

So,

$\bf\implies \:2kx = tanp - - - - (1)$

Now,

It is given that, The wire is now accelerated parallel to the x- axis with a constant acceleration a.

So, the two forces acted on the parabola.

1. Normal force, due to acceleration

2. Gravitational force.

Now,

Force F, can be resolved in to two components, F sinp and F cosp

So,

$\boxed{ \tt{ \: Fsinp = ma}}$

and

$\boxed{ \tt{ \: Fcosp = mg}}$

So, on dividing these two above equations, we get

$\rm :\longmapsto\:\dfrac{Fsinp}{Fcosp} = \dfrac{ma}{mg}$

$\bf\implies \:tanp = \dfrac{a}{g} - - - - (2)$

So, from equation (1) and (2), we get

$\rm :\longmapsto\:2kx = \dfrac{a}{g}$

$\bf\implies \:\boxed{ \tt{ \: \:x = \dfrac{a}{2kg} \: \: }}$