Question

A penny is placed at the outer edge of a disk (radius = 0.179 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.60 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk

Answer 1

Answer:

$\mu = 0.28$

Explanation:

Time period of the revolution of the disc is given as

$T = 1.60 s$

now the angular speed of the disc is given as

$\omega = \frac{2\pi}{T}$

so we will have

$\omega = \frac{2\pi}{1.60}$

$\omega = 3.92 rad/s$

now at the rim position of the disc the net force on the penny is due to friction force

So we will have

$\mu mg = m\omega^2 r$

here we will have

$\mu g = \omega^2 r$

$\mu = \frac{\omega^2 r}{g}$

$\mu = \frac{3.92^2 (0.179)}{9.81}$

$\mu = 0.28$