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Neat. 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.

I'm facing a problem regarding which books to refer, A Das Gupta or Cengage (As subjective might make me slow and objective might not ensure conceptual clarity). Could you kindly comment?

i actually use cengage for both physics and maths. Cengage has quite a good number of subjective questions in each chapter,so you should not worry about conceptual clarity.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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Thanks!!!

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@Spandan Senapati

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@Harsh Shrivastava

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Thanks for the q.let me try.

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Has anybody solved this yet?

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Any ideas guyzzz????

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Tan(Equlibruim angle)= (ur)/(l+r) Is that right ??

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Torque applies here??

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Neat. 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.

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@A E Hi, I wanted an advice from your side.

I'm facing a problem regarding which books to refer, A Das Gupta or Cengage (As subjective might make me slow and objective might not ensure conceptual clarity). Could you kindly comment?

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i actually use cengage for both physics and maths. Cengage has quite a good number of subjective questions in each chapter,so you should not worry about conceptual clarity.

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Thanks!

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