Question

A small particle is placed at the edge of a phonograph turntable of diameter 30 cm. If the turntable rotates at 45 revolutions per minute, what must be the coefficient of friction so the particle will bot slide out?

Answer 1

$45\ rev/min=4.71\ rad/s$

$F_f=m\frac{v^2}{r}\to \mu mg=m\frac{v^2}{r}\to\mu g=\frac{v^2}{r}\to\mu=\frac{v^2}{gr}$

$\mu=\frac{\omega^2r^2}{gr}=\frac{\omega^2r}{g}=\frac{4.71^2\cdot0.15}{9.81}=0.339$ . Answer