Solve Suspended Compass Problem from Physics Olympiad

[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

 

[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? ​

 

[benf.stokes]

Hi

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

 

[tiny-tim]

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

 

[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

 

[tiny-tim]

I can't find an expression for the height of the pivot

Start with something easier …

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

 

[benf.stokes]

I draw the triangle but I see no usable angle

 

[tiny-tim]

Well, it's very simple …

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

 

[benf.stokes]

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

 

[tiny-tim]

(have a theta: θ and a degree: º )

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

Where did 45º come from? And how did 2l get into the answer?

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

 

[benf.stokes]

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

 

[tiny-tim]

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

What do you mean by the centre of the triangle?

 

[benf.stokes]

The point where all the triangle bisectors join; the circumcenter

 

[tiny-tim]

The point where all the triangle bisectors join; the circumcenter

Yes, but where exactly is that?

How far along what line?

 

[benf.stokes]

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

 

[tiny-tim]

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

(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?

 

[benf.stokes]

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

 

[tiny-tim]

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

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

we're only here to help. ​

 

[benf.stokes]

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

 

[tiny-tim]

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

 

[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

 

[tiny-tim]

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

"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

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.

 

[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?

 

[tiny-tim]

Thanks but just one more thing: why must PB be horizontal for the maximum angle?

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 also found out that by ussing the law of sines it get's even easier. How was your solution?

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.

 

[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

 

[tiny-tim]

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

hmm … no need to apologise … I'm not the one who's inconvenienced …

you'll never learn this stuff if you get methods from someone else that you don't understand …

and one of the best ways of using this forum is by admitting in the first place that you can't do it, and asking for hints or explanations.

ok, let's start again: we'll leave the circles method till later (because it's not the standard way of doing it, and it probably won't work with most problems … I'm quite impressed that your friend managed to get it! ).

Forget the circles and the point X …

just call the angle APB 2θ (so APC is θ), and remember that you know that LC = CM, and so PC is perpendicular to LCM.

Calculate either PC or LC, and then apply the sine and/or cosine rule to triangle APC or ALC to find angle PAC (which you will then have to maximise).