An equilibrium problem -- Spinning a hinged rod and a ball

[Brilli]

Homework Statement

This is a practice olympiad problem
A light rod with length l is
hinged in such a way that the hinge folds
in one plane only. The hinge is spun with
angular speed ω around a vertical axis. A
small ball is fixed to the other end of the
rod. (a) Find the angular speeds for which
the vertical orientation is stable. (b) The
ball is now attached to another hinge and,
in turn, to another identical rod; the upper
hinge is spun in the same way. What is
now the condition of stability for the ver-
tical orientation?

Homework Equations

The Attempt at a Solution

balancing torques of lower ball, got
w=g (sina)/(sina+sinb) a and b are deviations from verical of upper and lower ball respectively. Then for the upper ball equating torques of thread tension of lower thread and mass and centridugal force. But i am stuck here. The solution says to derive expression for potential energy and get for what w is it minimum because forequilibrium the potential energy isminimum. How to derive this?

 

[haruspex]

I think I understood the arrangement until part b. It says the ball is "attached to another hinge". I thought the ball was attached to the rod, and the rod attached to a hinge.

If it means the ball is attached to another rod and in turn to another hinge that makes a bit more sense, but now I am unclear on the relationship between the two rods and hinges. Are the rods the same length? Are they both above the ball? Do the hinges flex in the same plane?
If the answer is yes to all those then I do not see how it matters whether there is one rod or two.

Is there a diagram?

 

[Dr.D]

Just to muddy the water a bit, is this really an equilibrium problem? Sounds like parts of the system are in accelerated motion, and by definition, that would seem to exclude equilibrium. Would it be better to use the term "Steady State" as opposed to "Equilibrium"? I'm curious as to what others think about this question.

 

[Brilli]

Yes

I think I understood the arrangement until part b. It says the ball is "attached to another hinge". I thought the ball was attached to the rod, and the rod attached to a hinge.

If it means the ball is attached to another rod and in turn to another hinge that makes a bit more sense, but now I am unclear on the relationship between the two rods and hinges. Are the rods the same length? Are they both above the ball? Do the hinges flex in the same plane?
If the answer is yes to all those then I do not see how it matters whether there is one rod or two.

Is there a diagram?

Yes.

Thank you for responding
The mass m1 and m2 are same. And l1 and l2 are also same.

 

[Brilli]

Just to muddy the water a bit, is this really an equilibrium problem? Sounds like parts of the system are in accelerated motion, and by definition, that would seem to exclude equilibrium. Would it be better to use the term "Steady State" as opposed to "Equilibrium"? I'm curious as to what others think about this question.

It was from a mechanics booklet. I think it as an equilibrium problem. If you have got another approach you are welcome to guide me through it.

 

[haruspex]

Yes

Yes.View attachment 221797
Thank you for responding
The mass m1 and m2 are same. And l1 and l2 are also same.

Ok, that's clear. The text did not mention the second ball.
You seem to be asking about part b. Did you solve part a?
I think I have solved both parts, but not using potential energy.

 

[Brilli]

Ok, that's clear. The text did not mention the second ball.
You seem to be asking about part b. Did you solve part a?
I think I have solved both parts, but not using potential energy.

Yes i have solved part a. I am stuck at part b.

 

[haruspex]

Yes i have solved part a. I am stuck at part b.

I think the potential energy hint would be using the rotating frame of reference, giving you a centripetal force. I used that frame, but just looked at forces.
Please post your equations as far as you get.

 

[Brilli]

I think the potential energy hint would be using the rotating frame of reference, giving you a centripetal force. I used that frame, but just looked at forces.
Please post your equations as far as you get.

For the lower ball balancing torque about point of attachment to upper ball
mw^2rcos (a)<=mgsin (a)
Thus as r=lsin (a)+lsin (b)
w^2 <=gsina/(lsina+lsinb)

For the upper ball balancing torque about attachment point with roof
Tension due to lower ball=mg (1+sin^2a)
I wanted to balance torque for this but it does not get to any end. How can we do it properly?

 

[haruspex]

For the lower ball balancing torque about point of attachment to upper ball
mw^2rcos (a)<=mgsin (a)

I assume your a and b are the θ₁ and θ₂ of the diagram.
How far is the lower ball from the axis of rotation?

 

[Brilli]

I assume your a and b are the θ₁ and θ₂ of the diagram.
How far is the lower ball from the axis of rotation?

l sinθ1+lsinθ2

 

[haruspex]

l sinθ1+lsinθ2

Right, so what is the centrifugal force?
By the way, we are only concerned with infinitesimal perturbations, so you can use x instead of sin(x) and 1 instead of cos(x) nearly everywhere. There can be exceptions, but I'll point it out if you fall into one.

 

[Brilli]

Right, so what is the centrifugal force?
By the way, we are only concerned with infinitesimal perturbations, so you can use x instead of sin(x) and 1 instead of cos(x) nearly everywhere. There can be exceptions, but I'll point it out if you fall into one.

Centrifugal force=mω^2l (sina + sinb)

 

[haruspex]

Centrifugal force=mω^2l (sina + sinb)

I'm sorry, I did not read your post #9 correctly.. you already have this part.
Next, what is the tension in the lower rod?

 

[Brilli]

Tension in lower rod=mg(sin^2a)+mgcosa≈mg (sin^2a)+mg=mg (sin^2+1)

Please send the solution as i am not able to access pf everytime because of tecnical issues

 

[Brilli]

Also torque balance for upper ball
mω²lsina+tsin (b-a)=mgsina

I am stuck here

 

[haruspex]

w^2 <=gsina/(lsina+lsinb)

Tension in lower rod=mg(sin^2a)+mgcosa≈mg (sin^2a)+mg=mg (sin^2+1)

Also torque balance for upper ball
mω²lsina+tsin (b-a)=mgsina

All your equations are good, but as I posted you can make small angle approximations.
The above reduce to:
$\frac {l\omega^2}g=\frac a{a+b}$
$T=mg$
$m\omega^2la+T(b-a)=mga$
You can solve for ω.

Edit: missed a factor a on the left in the third equation. Corrected.

 

[Brilli]

All your equations are good, but as I posted you can make small angle approximations.
The above reduce to:
$\frac {l\omega^2}g=\frac a{a+b}$
$T=mg$
$m\omega^2l+T(b-a)=mga$
You can solve for ω.

But the answer is not matching. Fiven answer is 2-√2

 

[haruspex]

But the answer is not matching. Fiven answer is 2-√2

Typo: I missed a factor a on the left in the third equation. Does that help?

 

[Brilli]

Typo: I missed a factor a on the left in the third equation. Does that help?

No sir. This does not yield the answer. I did it this way too but did not get it. The final given answer is (2-√2) (g/l)>ω^2

 

[haruspex]

No sir. This does not yield the answer. I did it this way too but did not get it. The final given answer is (2-√2) (g/l)>ω^2

Ok, found it. The right hand side here is using the wrong angle:

mw^2rcos (a)<=mgsin (a)

 

[Brilli]

Yes. I get it. Thank you