A small body is kept at a distance of 7 cm from the centre of a gramophone disc. The disc starts rotating with gradually increasing speed and the body is just on the verge of being thrown off when the disc rotates at 60 rpm. What will be the rate of rotation of the disc when the body, kept at a distance of 12 cm from the centre, is just being thrown off?

Explanation

The magnitude of the limiting frictional force between the coin and the table is the maximum value of centripetal force. So, if the coin of mass m is rotating along the turntable in a circular path of maximum radius r with angular velocity `omega`, limiting frictional force, f = m`omega^(2)`r.
In the given problem,
f = m`omega_(1)^(2) r_(1),` in the first case
and f = m`omega_(2)^(2) r_(2)`, in the second case
`therefore " "m omega_(1)^(2) r_(1) = m omega_(2)^(2) r_(2) " " or, omega_(2)^(2) = omega_(1)^(2) cdot (r_(1))/(r_(2))`
or, `" " omega_(2) = omega_(1) sqrt((r_(1))/(r_(2))) = 60 xx sqrt((7)/(12)) = 45.8` rpm.