Olympiad Mechanics problem (torque)

In summary, the student is trying to find a solution for the equation where the torque due to friction is zero. The student has found that the equilibrium angle is 0 when the rod is vertical, and that the torque due to mg must be zero in order for mg to have zero torque about O.
  • #36
Dr.D said:
He did not discuss it in detail, but only offered it as a homework problem in such a way that the person working the problem would develop a rather comprehensive discussion of the several possible cases.
So what are the several possible cases?
 
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  • #37
Several different friction models.
 
  • #38
It does not make sense to consider friction models whenever non slipping condition has already been imposed.

The phase portraits of described above system essentially differ from each other for the following four cases ##\epsilon r/g>1,\quad \epsilon r/g=1,\quad 0<\epsilon r/g<1,\quad \epsilon=0##

The center of the ring can rotate about the shaft; it can oscillate near its stable equilibrium; it can even perform asymptotic motions
 
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  • #39
I'll leave you to discuss it with Den Hartog.
 
  • #40
I understand the fact that on an incline the slipping begins at a angle ## \alpha ## s.t tan ## \alpha =
\mu ## , but how can you assocciate that fact to this problem?
 
  • #41
RoloJosh16 said:
I understand the fact that on an incline the slipping begins at a angle ## \alpha ## s.t tan ## \alpha =
\mu ## , but how can you assocciate that fact to this problem?
Are you referring to the original problem in post #1 or to the very different problem zwierz introduced (in violation of forum guidelines!) in post #29?
 
  • #42
haruspex said:
Are you referring to the original problem in post #1 or to the very different problem zwierz introduced (in violation of forum guidelines!) in post #29?

To the original problem.
 
  • #43
haruspex said:
Are you referring to the original problem in post #1 or to the very different problem zwierz introduced (in violation of forum guidelines!) in post #29?

Do you understand what I am asking?
 
  • #44
RoloJosh16 said:
To the original problem.
In the original problem the shaft is turning at constant speed and the system is in steady state, so we are only concerned with kinetic friction.

Edit: but, interestingly, if we define α as tan(α)=μk then the answer can be written as (R+L)sin(θ)=Rsin(α ).
 
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  • #45
RoloJosh16 said:
I understand the fact that on an incline the slipping begins at a angle ## \alpha ## s.t tan ## \alpha =
\mu ## , but how can you assocciate that fact to this problem?
The original problem is similar to having a conveyer belt inclined at angle ##\alpha## with the belt moving upward at a constant speed. Suppose a box is placed on the moving belt and the box happens to stay at rest relative to the earth. Then you can ask for the relation between ##\alpha## and the coefficient of kinetic friction ##\mu_k##.
 
  • #46
TSny said:
The original problem is similar to having a conveyer belt inclined at angle ##\alpha## with the belt moving upward at a constant speed. Suppose a box is placed on the moving belt and the box happens to stay at rest relative to the earth. Then you can ask for the relation between ##\alpha## and the coefficient of kinetic friction ##\mu_k##.

Ahhh, thank you. I was thinking about that relation but didn' t think about a moving surface. The box in that case would be the tiny part of the hoop in contact with the shaft?
 
  • #47
RoloJosh16 said:
Ahhh, thank you. I was thinking about that relation but didn' t think about a moving surface. The box in that case would be the tiny part of the hoop in contact with the shaft?
Yes, that's right.
 

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