A body is kept at rest at a distance of 10 cm from the centre of a gramophone disc. If the coefficient of friction between the body and the disc is 0.3, then for what maximum rps of the disc, will the body not be thrown off the disc?

Explanation

The magnitude of the limiting frictional force between the body and the gramophone disc is the maximum value of the centripetal force. If the mass of the body is m and it rotates with the disc along a circular path of radius r with angular velocity `omega` such that the body will not be thrown off the disc, then
limiting frictional force, f = m`omega^(2)`r
or, `mu mg = m omega^(2) r `[ `omega = 2pi `n, n = number of revolutions of the disc per second (rps), `mu` = coefficient of friction between the body and the disc ]
or, `" " mu g = (2pi n )^(2) r = 4 pi^(2) n^(2) r `,
or,`" " n^(2) = (mu g)/(4 pi^(2) r)`
or, `" " n = sqrt((mu g )/(4 pi^(2) r )) = sqrt((0.3 xx 980)/(4 xx pi^(2)xx 10 )`
= 0.863 rps