Question

A coin kept on a horizontal rotating disc has its center at a distance of 0.1 m from the axis of the rotating disc. If the coefficient of friction between the coin and the disc is 0.25; find the angular speed of disc at which the coin would be about the slip off (Given: g = 9.8 m/s)

Answer 1

$N=mg$

$fs=mgr$

$=\frac{mg}{4}$

$f_s=mr{w}^2$

$\frac{mg}{4}=\left(m\right)\left(0.1\right)\left(w^2\right)$

$w^2=\frac{g}{0.4}$

$w=\frac{100}{4}$

$w^2=5rads/sec$