# Suspended compass

1. Dec 30, 2009

### benf.stokes

Hi

A friend of mine showed this problem he translated from a physics olympiad but I can't solve it:
"Suspended compass
Consider a perfectly symmetrical compass. It is constituted by two rigorously equal connecting rods, that join in a vertex the one that we call “pivot”. The opening angle is regulable. Imagine that we suspend the compass for the tip of one of the connecting rods, attaching it to a wire whose other end is attached at the ceiling.
a) Sketch the position of the bar at various angles of opening.
b) What is the opening angle of the compass, so that the pivot is the highest possible?
Hint: The center of mass of the bar, assuming that the rods have densities
uniform is the point where the bisector of the angle of the bar crosses the line
passing through the midpoint of both rods. The mass center has to be, for any angle of opening, in the vertical line of the point of suspension of the wire."

Sorry for the english in the translation. My question is how do you solve this using only knowledge available to a 11th grader??
Thanks

2. Dec 30, 2009

### tiny-tim

Welcome to PF!

Hi benf.stokes! Welcome to PF!

You can do this with just elementary trigonometry and Pythagoras' theorem.

Hint: draw the compass as a triangle ABC …

where is the centre of mass?

3. Dec 30, 2009

### benf.stokes

Hi

Thanks for the reply, but how would that work out exactly? I couldn't do it.

4. Dec 30, 2009

### tiny-tim

Show us how far you get, and where you're stuck, and then we'll know how to help!

5. Dec 30, 2009

### benf.stokes

I draw the triangle and everything but I just can't get to a expression that relates the height of the pivot to the opening angle.
Thanks

6. Dec 30, 2009

### tiny-tim

how far is the centre of mass from the top (for an angle 2θ at the pivot)?

7. Dec 30, 2009

### benf.stokes

I draw the triangle but I see no usable angle

8. Dec 30, 2009

### tiny-tim

Well, it's very simple …

go to sleep :zzz:, and try again tomorrow.

9. Dec 30, 2009

### benf.stokes

Is it (2lcos(theta))/(cos(45+theta))?? Being 2l the total length of the rod

10. Dec 30, 2009

### tiny-tim

(have a theta: θ and a degree: º )
Where did 45º come from? And how did 2l get into the answer?

Tell us, in words, where the centre of mass is.​

11. Dec 30, 2009

### benf.stokes

The centre of mass is in the centre of the triangle. Right?

12. Dec 30, 2009

### tiny-tim

What do you mean by the centre of the triangle?

13. Dec 30, 2009

### benf.stokes

The point where all the triangle bisectors join; the circumcenter

14. Dec 30, 2009

### tiny-tim

Yes, but where exactly is that?

How far along what line?

15. Dec 30, 2009

### benf.stokes

It is l*cos(theta) along the bisector of the opening angle

16. Dec 30, 2009

### tiny-tim

(what happened to that θ i gave you? )

No, that's the end of the bisector (the centre of AC).

Call the centre of AC "D" … where is the centre of mass along BD?

17. Dec 31, 2009

### benf.stokes

Please tell me how to get there. I never was much of a geometer

18. Dec 31, 2009

### tiny-tim

Sorry, on this forum you have to do the work yourself …

we're only here to help.

19. Jan 2, 2010

### benf.stokes

Hi, I think I got it. Can I use this sketch to get to the answer? Is it arccos(1/3)?

20. Jan 2, 2010

### tiny-tim

Sorry, I've no idea what that sketch is supposed to do.

21. Jan 3, 2010

### benf.stokes

"In the picture, the arms of the compasses, suspended from the point A, are AP and PB, with the pivot at P. The midpoints of AP and BP are L and M, and the centre of mass is at C, half way between L and M.

If the length of AP is a, then PL = PM = a/2, so L and M both lie on a circle of radius a/2 centred at P. The point C is half way from L to M, so lies on circle of radius a/4 centred at the point X half way between P and L.

The pivot point P will be highest when the angle between AP and the vertical line AC is greatest. That will happen when AC is tangent to the circle through P, C and L. Then ACX is a right angle, and the angle AXC is \cos^{-1}1/3 (because XC = a/4 and XA = 3a/4). But the angles AXC and APB are equal (because they are both equal to twice the angle LPC), so the answer to the problem is that the pivot is highest when the opening angle of the compasses is arccos1/3." This isn't my answer but does it work?
Thanks

22. Jan 4, 2010

### tiny-tim

Hi benf.stokes!

Yes, that's very neat!

(But since your examiner/teacher won't be expecting it, I suggest you expand slightly on why XC is parallel to PB, and on why AC must be a tangent.)

You've used ordinary geometry, and completely avoided having to differentiate (or to complete the square) … which is what I would have done.

23. Jan 5, 2010

### benf.stokes

Thanks but just one more thing: why must PB be horizontal for the maximum angle? I also found out that by ussing the law of sines it get's even easier. How was your solution?

24. Jan 5, 2010

### tiny-tim

uhh? Because it's parallel to XC, and that's perpendicular to the tangent.

You do know why AC has to be the tangent, don't you (or did you just guess, or get it from someone else)?
I considered triangle PCA (without using any circle), and I think (I did it yesterday, so I'm not sure) I used both the sine and cosine rules, and then differentiated, and got cos2APC = 2/3, which is the same.

25. Jan 5, 2010

### benf.stokes

I got the solution from somebody else, sorry!! And i don't know why AC is tangent to PB in the highest angle