Question

A particle is projected along a horizontal field whose coefficient of friction varies as $\mu = {\rm A}{\text{ }}{r^{ - 2}}$ , where $r$ is the distance from the origin in metres and $A$ is a positive constant. The initial distance of the particle is $1m$ from the origin and its velocity is radially outwards. The minimum initial velocity at this point so the particle never stops is:
A. $\infty $
B. $2\sqrt {gA} $
C. $\sqrt {2gA} $
D. $4\sqrt {gA} $

Answer 1

Hint: To solve this question, one must know how friction acts and we will use the basics of Newton’s law of motion to find the acceleration and then we will integrate the obtained equation to get the minimum initial velocity at this point so the particle never stops.

Formula used:
$F = ma$
Where, $F$ is the force,
$m$ is the mass and
$a$ is the acceleration.

Complete answer:
As we know that at any distance $r$ the value of friction coefficient is $\mu = {\rm A}{\text{ }}{r^{ - 2}}$ .
According to the question,
$u = 1m$
As we know that from Newton’s law of motion,
$
  F = ma \\
   \Rightarrow a = \dfrac{F}{m} \\
 $
Here, we know that friction force is working so we can say that,
$
  F = ma \\
   \Rightarrow \mu mg = ma \\
   \Rightarrow a = \dfrac{{\mu mg}}{m} \\
   \Rightarrow a = \mu g \\
 $
$ \Rightarrow a = - \mu g$ (Here, we take the negative sign because frictions act in the opposite direction.)
Now, putting the value of $\mu $ in the above equation,
$
  a = - \mu g \\
   \Rightarrow a = \dfrac{{ - Ag}}{{{r^2}}} \\
 $
And we know that we can write acceleration as $\dfrac{{dv}}{{dt}}$ .
$\dfrac{{dv}}{{dt}} = \dfrac{{ - Ag}}{{{r^2}}}$ ,
And we can write, $\dfrac{{dv}}{{dt}} = \dfrac{{dv}}{{dr}} \times \dfrac{{dr}}{{dt}}$
And $\dfrac{{dr}}{{dt}}$ is the velocity of a particle.
$
   \Rightarrow \dfrac{{vdv}}{{dr}} = \dfrac{{ - Ag}}{{{r^2}}} \\
   \Rightarrow vdv = - Ag\dfrac{{dr}}{{{r^2}}} \\
 $
Now, integrating the above equation to get the minimum initial velocity,
$
  \int\limits_{{v_0}}^v {vdv = - \int\limits_1^\infty {Ag\dfrac{{dr}}{{{r^2}}}} } \\
  \dfrac{{{v^2} - v_0^2}}{2} = + Ag\left[ {\dfrac{1}{r}} \right]_1^\infty \\
  {v^2} = 2Ag \\
  v = \sqrt {2Ag} \\
 $
So, the minimum initial velocity at this point so the particle never stops is $\sqrt {2Ag} $ .
Hence, the correct option is C.

Note:
In this type of question, one must note that the value of acceleration that is obtained is negative because the force of friction that works is in the opposite direction and the force of friction depends on two factors: the coefficient of friction and the normal force. For any two surfaces that are in contact with one another, the coefficient of friction is a constant that depends on the nature of the surfaces.