Question

A Coin kept at a distance of 5cm from the center of a turntable of radius 1.5 m just begins to slip when the turntable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable. $[g=9.8m/s^2]$

Answer 1

Given that, $r=5cm=0.05m$,

$n=90r.p.m.=1.5Hz$

Limiting force of static frictions =Centrifugal force

${\mu }_{s}mg=mr{\omega }^2$

$\therefore {\mu }_s=\frac{r{\omega }^2}{g}=\frac{r(2\pi n{)}^2}{g}$

$\therefore {\mu }_s=\frac{4{\pi }^2{n}^{2}r}{g}$

$=\frac{4\times 0.05\times (3.14{)}^2\times (1.5{)}^2}{9.8}$

$\therefore {\mu }_s=0.4527$