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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 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.

Respuesta :

Answer:

[tex]arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})[/tex]

Explanation:

By the Law of Sines,

[tex]sin \theta = \frac{sin \phi R}{ l + R}[/tex]

From Newton's Law,

[tex]mg = N\sqrt{\mu^2+1}[/tex]

And the last equation again from Newton's Law,

[tex]\mu N = mgsin\phi[/tex]

Then if we collect all equations together,

[tex]\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\[/tex]

[tex]sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}[/tex]

Thus,

[tex]\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})[/tex]