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Zolol [24]
3 years ago
10

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:
Roman55 [17]3 years ago
8 0

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}})

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