Homework Help: Olympiad Mechanics problem (torque)

1. May 16, 2015

Ceva

Hello, there is a not-hard-at-all problem i am struggling with and i am must be missing something.

1. The problem statement, all variables and given/known data

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 coef- ficient of friction between the shaft and the hoop is µ. Find the equilibrium angle between the rod and the vertical.

2. Relevant equations

3. The attempt at a solution
I am going to work with the torque of forces. I am taking the torques about O, where the shaft and the hoop are in contact, so that the torque of friction is zero. Also, the torque of the tension of wire S is zero as it passes through O.
The condition where slipping begins on an inclined plane is tanθ = μ. (where θ here is i guess the angle between the horizontal plane and Friction).

My diagram:

If i am to work only with the torque of mg, how can i reach a solution for μ? The torque equation just gives me:
μmg(l+2r)cosθ=0.

Last edited: May 16, 2015
2. May 16, 2015

Alettix

Hi!
This looks like a nice problem! :)

I haven't tried to solved it myself, but I wonder: Is it really right to use tanθ = μ in this case?
In this kind of problem you might sometimes be supposed to have μ as a part of your answer.

I would start with setting Fr = μ * FN , and setting the point O at the point where the rod and the circle intersect.

3. May 16, 2015

TSny

Yes, the torques must sum to zero about O. The only torque about O is due to mg. But in your drawing mg gives a nonzero torque about O.

See if you can re-draw the figure with the wire positioned so that mg does not produce any torque about O.

Also think carefully about the direction of the friction force. I can't quite make out the direction of the angular velocity of the shaft. Is it counterclockwise?

Last edited: May 16, 2015
4. May 16, 2015

Ceva

Yes mg is the only force to produce a nonzero torque about O.The equation i wrote is for the equilibrium state, where the torques should result to zero. But the other forces give zero torque. How can equilorium happen if there is only one torque about O?

The torque of mg is zero only if the angle θ is 0 ie the rod is vertical.

Yes it's rotating counterclockwise. Isn't the friction correctly drawn? If the rod is to slip, it's slipping cockwise.

5. May 16, 2015

TSny

Not necessarily. If mg is to have zero torque about O, what can you say about the line of action of mg?

The shaft is rotating counterclockwise and slipping against the wire. Try to picture the surface of the shaft rubbing against the wire.

For example, if you slide your hand to the right across a table, what is the direction of the friction force from your hand on the table?

6. May 16, 2015

Ceva

Ah yes, i think i can see it. It's like the top of the shaft (point of contact) "pushes" the hoop as it rotates.

Anyway, what other torque do we have except of mg's when taking moments abouy O? Any idea?

7. May 16, 2015

TSny

There are no other torques about O. So, at the equilibrium angle, the torque about O due to mg must be zero. You will need to think about how to draw the figure at the equilibrium angle so that O is located on the shaft such that mg does not produce any torque about O.

8. May 16, 2015

Ceva

How is that possible for positions of M different than the two on the vertical axis (lowest and highest point)?

9. May 16, 2015

TSny

I don't want to give the answer away too easily. But note that in order for mg to produce zero torque about O, the line of action of mg must pass through O. Try to position O on the line of action of mg when m is at a small angle to the vertical.

10. May 16, 2015

Ceva

Sure, it's better to guide first!
Maybe if i treat it like it doesn't actually pass through O but for a very small angle, you can assume it does?

11. May 16, 2015

TSny

Consider the wire as shown below. The rotating shaft is not shown. Draw the line of action of mg and see if that helps determine the location of O. Then you can add the rotating shaft to the drawing.

Attached Files:

• wire.png
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12. May 16, 2015

Ceva

Directly above m, inside the hoop, but can i do that? Will the rotating shaft and hoop-wire come in that position? I thought that they would remain in the same positions relative to each other.
And anyway, now we have mg = Friction, as expected and for the torques i cannot produce any equation as we've proved that all of them are zero.

13. May 16, 2015

TSny

The system will adjust itself until it finds the equilibrium position. Note that your first drawing shows a shift in relative positions from when the mass was hanging vertically.

I don't see how you conclude that mg = Friction.

You won't need a torque equation since each of the individual torques is zero about O. That is, the torque equation Στ = 0 would just give 0 = 0.

14. May 17, 2015

Ceva

Equilibrium for forces gives that. mg downward and the friction directly upward? Isn't it so?

15. May 17, 2015

haruspex

You're forgetting the normal force.

16. May 18, 2015

Ceva

And also a tension from the wire itself?
Friction, weight and the normal force give a resultant to the right (normal) so i need another one to the left. Right?

17. May 18, 2015

haruspex

If you have drawn the diagram correctly (now), the frictional force and the normal force should balance the weight of the mass. Do you have the mass vertically below the point of contact?

18. May 19, 2015

Ceva

Here it is. Now i find mg = N(μ2 + 1)1/2 .

19. May 19, 2015

TSny

OK. Try to find θ by working with the geometry of triangle OCB shown below. Point C is at the center of the circular portion of the wire.

Attached Files:

• ring2.png
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20. May 19, 2015

haruspex

That's much closer, but you seem to have drawn the rod from the mass meeting the ring at its lowest point. It needs to point at the centre of the ring. Compare with TSny's diagram.

21. May 17, 2017

fedecolo

Sorry, but can anyone explain me how to find the final result?

22. May 17, 2017

TSny

Hi fedecolo. Welcome to PF.

Please show the work that you've done so far on the problem and state specifically where you are getting stuck.

23. May 17, 2017

fedecolo

Ok TSny. I came to the expression $mg= N \sqrt { \mu^2 + 1 }$ but i don't know how to relate this expression to an angle

24. May 17, 2017

TSny

What principle did you use to obtain this expression? Can you apply a very similar principle to find another relation?

In addition, post #19 might be of help.

25. May 17, 2017

fedecolo

Thank you! By the law of the sines I have that $$sin \theta = \frac {sin \phi R}{l +R}$$

Then $\mu N = mg sin \phi = N \sqrt {\mu ^2 + 1} sin \phi$

And by substitution $sin \theta = \frac {R \mu}{(l +R) \sqrt {\mu ^2 +1}}$ that si the correct result!