Question

A disc rotates at 30 rev/min around a vertical axis. A body lies on the disc at the distance of 20 cm from the axis of rotation. What should be the minimum value of the coefficient of friction between the body and the dise, so that the body will not slide off the disc ? what is the minimum distance from centre without rotates at 60 rev/min​

Answer 1

$\Large\bf {\orange {\underline { \underline {Solutíon}} : - }}$

As the disc rotates, the body will tend to slip away from axis.

Due to this tendency to slip, force of static Friction arises towards the centre.

The centripetal force required for the circular motion is provided by this friction force.

At the point of slipping, Friction = $\sf μ_{s}N$

$\sf μ_{s}N = m {ω}^{2} r \: \: \: \: [where \: N = mg]$

$\: \sf μ_{s} = {ω}^{2} r/g$

where ω = 2π(30/60) = π rad/s and r = 0.2m.

$\Longrightarrow \sf u_{s} = 0.2$

$\sf f_{max} \: will \: remain \: same$

$\Longrightarrow \sf { ω_{1}}^{2} r_{1} = { ω_{2} }^{2} r_{2}$

$\Longrightarrow \sf {30}^{2} \times 20 = {60}^{2} r_{2}$

$\Longrightarrow \sf r_{2} = \frac{ {30}^{2} \times 20}{ {60}^{2} }$

$\Longrightarrow \sf r_{2} = 5cm$

Thus,the minimum distance from centre without rotates at 60 rev/min is 5cm.