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 attac

Question

2 months ago

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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 attac

hed 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.

Physics

1 answer:

ValentinkaMS [3.4K] · Answer 1 · 2026-04-07T13:27:02+03:00

ValentinkaMS [3.4K]2 months ago

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Answer:

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

Explanation:

By applying the Law of Sines,

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

Based on Newton's Law,

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

And the final equation also derived from Newton's Law,

$\mu N = mgsin\phi$

Then by consolidating all the 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,

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