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TopNewestNeat. I like this. I have to go now, but will come back and check this.

The way to approach this is to think equilibrium. If the thin rod is not moving, then it is in equilibrium. All torques (taken about any convenient point must sum to zero. And all forces must sum to zero.

Let mu the coefficient of friction be u... So imagine that u is large and the rod is horizontal. The CCW rotating axle gives the hoop a leftward pointing force. Take torques about the center of the hoop. That force is mu* m * g and it makes a torque out of the page equal to

r * u * m * g The bob makes a torque = m * g * (l+r) sine (theta) into the page and that is equal to the first torque.

So sine (theta) = r * u * m * g / ( m * g * ( l+r) ) = r * u / ( l+r)

Thus theta = arcsine ( r * u / ( l+r ) ) This is true for the hoop touching the axle at the top of the hoop.

it will touch at an angle though, and that will make the mg break into components, in the r * u * m * g part. So that is messier. I will come back to this. – Sandy Roman · 5 days, 16 hours ago

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Torque applies here?? – Elethelectric Penguin · 1 month ago

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Tan(Equlibruim angle)= (ur)/(l+r) Is that right ?? – Ayush Sharma · 2 months ago

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Any ideas guyzzz???? – A E · 2 months, 2 weeks ago

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Has anybody solved this yet? – A E · 2 months, 2 weeks ago

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Thanks for the q.let me try. – Spandan Senapati · 2 months, 3 weeks ago

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@Harsh Shrivastava – A E · 2 months, 3 weeks ago

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@Spandan Senapati – A E · 2 months, 3 weeks ago

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