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Question: A gramophone record is revolving with angular velocity . A coin is placed at a distance r from the c...

A gramophone record is revolving with angular velocity . A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ\mu . The coin will revolve with the record if
A. r=μgω2r = \mu g{\omega ^2}
B. r<ω2μgr < \dfrac{{{\omega ^2}}}{{\mu g}} .
C. rμgω2r \leqslant \dfrac{{\mu g}}{{{\omega ^2}}}
D. rμgω2r \geqslant \dfrac{{\mu g}}{{{\omega ^2}}}.

Explanation

Solution

In this question we will use the concept of static friction. While solving this problem, we have to remember the concept of static friction. The force of friction which comes into play between two bodies before one body actually starts moving over the other is called static friction (fS)({f_{\text{S}}}), fS{f_S} is defined as, fS(static friction)μSmg{f_{\text{S}}}{\text{(static friction)}} \leqslant {\mu _S}mg.

Formula used: F = mrω2mr{\omega ^2}, Force of friction = μmg\mu mg, fS(static friction)μSmg{f_{\text{S}}}{\text{(static friction)}} \leqslant {\mu _S}mg.

Complete step-by-step answer:
Given that, the gramophone record is revolving with angular velocity = ω\omega and the static coefficient of friction =μ\mu .
We know that a force required to make a body move along a circular path with uniform speed is called centripetal force.
Centripetal force = mass ×\times centripetal acceleration.
F = mrω2mr{\omega ^2} …………………………(i)
And, whenever a body moves or tends to move over the surface of another body, a force comes into play which acts parallel to the surface of contact and opposes the relative motion. This opposing force is called the force of friction.
Force of friction = μmg\mu mg……….(ii)
Here, the centripetal force provided by the revolving record is the static friction.
So, the coin will revolve with the record of centripetal force \leqslant force of friction.
Therefore,
\Rightarrow centripetal force \leqslant force of friction.
\Rightarrow mrω2μmgmr{\omega ^2} \leqslant \mu mg
\Rightarrow rμgω2.r \leqslant \dfrac{{\mu g}}{{{\omega ^2}}}.
Hence, we can see that the coin will revolve with the record if rμgω2.r \leqslant \dfrac{{\mu g}}{{{\omega ^2}}}.
Therefore, the correct answer is option (C).

Note: Whenever we ask such types of questions, we have to remember some basic points of static friction. First we have to find out the given details in the question. Then we will use the relationship of static friction with the coefficient of static friction and normal reaction(mg), i.e. fSμSmg{f_{\text{S}}} \leqslant {\mu _S}mg. then by using this relation, we can easily find out the required condition, that has been asked in the question.