A small coin is placed at a distance r from the centre of a gramophone record. The rotational speed of the record is gradually increased. If the coefficient of friction between the coin and the record is μ , the minimum angular frequency of the record, for which the coin will fly off, is given by

Question

A small coin is placed at a distance r from the centre of a gramophone record. The rotational speed of the record is gradually increased. If the coefficient of friction between the coin and the record is μ , the minimum angular frequency of the record, for which the coin will fly off, is given by (A) √(2 μ g/r) (B) √(μ g/2 r) (C) √(μ g/r) (D) 2√(μ g/r)

Tardigrade Admin · Accepted Answer

The coin performs circular motion in a circle of radius

r

. It will fly off when centrifugal force is greater than the centripetal force provided by friction
Hence, at the moment when the coin is about to fly off,

μ N = m ω^{2} r

or

μ m g = m ω^{2} r

ω = \frac{μg}{r}

$\frac{2 μg}{r}$

$\frac{μg}{2 r}$

$\frac{μg}{r}$

$2 \frac{μg}{r}$

The coin performs circular motion in a circle of radius $r$ . It will fly off when centrifugal force is greater than the centripetal force provided by friction
Hence, at the moment when the coin is about to fly off,
$μ N = m ω^{2} r$
or $μ m g = m ω^{2} r$
$ω = \frac{μg}{r}$

r2μg​​

2rμg​​

rμg​​