Question

An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attached to the other end of the rod. The pendulum thus formed is hung by the hoop onto a revolving shaft. The coefficient of friction between the shaft and the hoop is µ. Find the equilibrium angle between the rod and the vertical.

Answer 1

Answer:

$arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

Explanation:

By the Law of Sines,

$sin \theta = \frac{sin \phi R}{ l + R}$

From Newton's Law,

$mg = N\sqrt{\mu^2+1}$

And the last equation again from Newton's Law,

$\mu N = mgsin\phi$

Then if we collect all equations together,

$\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi$

$sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}$

Thus,

$\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$