Homework Help: Rotational Equilibrium and Dynamics

1. Mar 14, 2006

Jasonn

I am currently studying Rotational Equilibrium and Dynamics in Physics. Our class was given a competitive lab project for extra credit, where the group with the lowest margin of error would win. It is set up as following:

You will be given a meter stick and one weight (approximate value of .1kg, but this will be measured for absolute accuracy). You are to calculate the weight of the meter stick. (Correct weight - Your weight)/Correct weight will be used to determine margin of error.

I thought about this pretty much all day with my partner and decided to come here for help. To determine the weight of the meterstick, we will need to suspend the weight from a fixed point (axis), but wouldn't we need a support force acting in the opposite direction?

Any starting points on how to solve this problem would be helpful.

2. Mar 14, 2006

nrqed

Hint: you could put the meter stick on your finger, off centered and then .... (I am assuming that the meter stick has uniform mass distribution, in which case there is an obvious way to get its mass)

Pat

3. Mar 14, 2006

Jasonn

I was thinking sort of the same thing...maybe hanging it off a table at a fixed point. And yes, the meter stick is uniform.

4. Mar 14, 2006

nrqed

Don't hang it from a table (there will be anormal force acting on it and that will complicate things). It has to be at equilibrium on a sharp object (let's say the tip of a finger). Then what?

5. Mar 14, 2006

Jasonn

Should the axis be placed at the very end of the stick?

Fw1 = ? at .5m
Fw2 = ~.981N at .25m (example)
L = 1m

Sum of y: -Fw1 - Fw2 = 0
-Fw1 = Fw2
...that can't be right.

Shouldn't I be solving for tourqe though?

Maybe this is incredibly too easy and I'm not on the right page. :tongue:

6. Mar 14, 2006

nrqed

What are you Fw?? You mean forces? No! You must work with torque!! And no, the axis (your fingertip) can't be at the very end!!! There is no way to have equilibrium then!! For the trick to work, the axis can't be at one very end. There is another point where it can't be....

Do you see the trick? (I won't tell you but hopefully these hints should get you there)

7. Mar 14, 2006

Jasonn

K, so the axis will have to be placed above the mass...

Fw1 is the force of weight of the meter stick
Fw2 is the force of weight of the .1kg mass (.981N)
L = .25m (placement of mass)

Torque = (-Fw1)*.5m - .981N*.25 = 0
= -Fw1*.5 - .736 = 0
= -Fw1*.5 = .736
Fw = 1.4715N
m = .15kg

This look right?

8. Mar 14, 2006

nrqed

well, with your equations, the mass will come out negative!

I can't answer without knowing exactly what is the physical situation you are thinking about. Before plugging any number in, tell me what steps you would take (what do you place where, what do you then do, what do you measure). Then we can make up some numbers and see what mass would come up. But right now, I have no idea what you have in mind...

9. Mar 14, 2006

Jasonn

[T = F*distance (perpendicular)]
[T = (negative force of weight * distance) - (negative force of weight * distance) = 0]
Torque = (-Fw1)*.5m - .981N*.25 = 0
[Solve.]
= -Fw1*.5 + .736 = 0
= -Fw1*.5 = - .736
Fw = 1.4715N
m = .15kg

...should have been what I posted.

10. Mar 14, 2006

nrqed

Ok, but I still don't know what the physical situation is...Just from the number you are using, I can tell you that there is something wrong. Tell me exactly what you have in mind (what do you with th eruler, what do you with the mass, the distances are measured from where to where)

11. Mar 14, 2006

Jasonn

Alright, the stick will be balanced on my finger, .5m from both ends. The mass will be suspended from a point (the axis?) on the stick.

Using T=F*d - F*d ... = 0, Fw of the stick should be solveable

Distances would be how far each force is from the axis, which can change depending on where you place the axis.

T- -Fw1 * .25 (assume the force of weight (center) is .25m from the axis) - Fw2 ) * (L) (the distance of this obviously depends on where the axis is. should the axis be placed where the mass is fixated? If so, the distance is 0.)

12. Mar 14, 2006

nrqed

I am not sure what you mean by the axis....The axis will be where you place your finger!! Are you placing the ruler on something else???

13. Mar 14, 2006

Jasonn

No no you are right...wasn't thinking straight at all.

If the axis is where my finger is (.5m) from both ends, how would the equation fit

T= -Fw1*Distance -Fw2*distance = 0

should the distance for Fw1 be 0 because it is at the axis or .5m?

14. Mar 14, 2006

nrqed

If your finger is at the center, then it will be impossible for you to get equilibrium!!! (unless you don't put the mass on the ruler at all in which case you can't measure the mass of the ruler!)

15. Mar 14, 2006

Jasonn

Alright, then just place the finger slighly off center (say .25m from the center), and solve for distances accordingly?

T = -Fw1 * .25m - (-Fw2 * D) = 0 (where D is the distance from the force of weight of the mass to the axis)..let's say .5m

T = -Fw1 * .25m +.981N * .5m = 0
T = -Fw1 * .25m + .736 = 0
T = -Fw1 * .25m = -.736
Fw = 2.944 N (.30kg)

16. Mar 14, 2006

nrqed

we are getting there, but I would like you to describe in words what *manipulations* you will do. Then I will really see if you understand the trick. Tell me what manipulations you will do and what you will measure. (no numbers, just words)

17. Mar 14, 2006

Jasonn

No numbers...hmm

We know that F = F(D) (perpendicular)

Torquenet = F*d - F*d ... = 0

F*d...force has to be perpendicular to the distance involved (which in this case it is (Force of weight of the meterstick is perpendicular to the distance to the axis).

Manipulations? I don't think I quite understand.
I understand the concept of the problem, just not the steps needed for the solution. To keep everything in equilibrium, the axis can not be on the very end or at the center. Check. The distance from the weight of the mass to the axis can be manipulated...however I do see a problem with that. If I were to move the mass, say from .5m from the axis to .7m, I will get a completed different answer.

Shouldn't there be a way to manipulate it so that, regardless of where the mass is in terms of distance to the axis, you will get the same answer (Fw of the meterstick)?

18. Mar 14, 2006

nrqed

Of course the mass of the meter stick does not depend on where you put the axis...

You see, this is why I am not sure you understand the concept behind the maths. This is why I am asking about the manipulation.

I give you a ruler and a mass of 0.1 kg. I'd like to know exactly what you will do with them. That's what I mean by manipulation. If you can answer that and understand the *physics*, then all your questions will have obvious answers.

19. Mar 14, 2006

Jasonn

I've already explained this though. I've explained that I would place an axis somewhere on the stick, calculate the forces perpendicular to the axis for both the Fw of the stick and the mass. You've told me that these equations/answers are either way off or getting there. Aside from what I've said in the past 8-9 posts, I have no clue.

20. Mar 14, 2006

nrqed

Ok, but I am not sure what you mean by "placing an axis" somewhere.
From your initial question, I assume that you would have to measure yourself the mass of the meter stick. So you have to know exactly what to do. Maybe you do but then you can tell me what you will do.

Ok, let's pretend that you talk to a friend on the phone...someone who knows *nothing* about physics... Absolutely nothing. Your friend has the meter stick and the 0.1 kg. You have to tell him/her what to do to get the measurements you need. If you tell him "place an axis somewhere", he will go "uh????". He will say "Dude, what do you mean by placing an axis? I have a meter stick in my hands and a mass, what do I do now?"

That's the kind of instructions that I have in mind. If you can answer this, then I will answer your question about why using, say 50 cm instead of 70 cm seems to give a different mass for the meter stick in your equations (if you understand the physics, then it will be clear why you would always get the same mass for the meter stick).

But if you don't know what to do with the mass and meter stick, then plugging numbers in an equation is not useful.

Pat

21. Mar 14, 2006

Jasonn

We will not be able to measure the mass of the meter stick. Only the mass of the weight (approx. .1kg). We are trying to FIND the mass of the meter stick.

Alright, first, you will draw a diagram of everything involved/labeled. The meterstick, the forces (force of weight of the stick and mass, found by mass * gravity). Also, an axis will be placed on the diagram. The axis is simply a position on the meter stick relative to the rotational motion.)

22. Mar 14, 2006

nrqed

I know that

if I understand your first post correctly, this is for a LAB (right??)! So before you do any calculation, diagram, etc, you will have to do *measurements*. Do you know what you will have to do with the mass and the meter stick??

By *manipulations*, I mean what you will actually be doing with the stick and the mass!!! That's before using any pencil to write or draw anything or to calculate anything.

If I give you a meter stick right now and a 0.1 kg mass, do you know what you will do with them??? You won't sit down and start drawing a diagram and do calculations until you have some measurements to work with. My question is about what you will actually do with the meter stick and the mass! That's the first step. If that is not clear, you will get stuck as soon as the prof puts in your hands the stick and the mass!

23. Mar 14, 2006

Jasonn

Alright. I have a .1kg mass and a 1m uniform stick.

The length of the stick is given. The axis (my finger) will be placed .25m from one of the ends. Measuring the center of the stick to this will yield a .25m distance. The mass will be placed .25m from the OTHER end. Measuring this place on the stick to the axis will give .5m.

Next, the mass needs to be weighed (for accuracy). We've been told ahead of time the approximate value will be .1k, but values need to be as exact as possible, of course.

Correct?

24. Mar 14, 2006

nrqed

Now we are getting to the core of the problem.

All right.... The first question is what makes you decide to put the mass .25 m from the end of the meter stick??? Why not somewhere else??

This is the *key* point that you are missing.... The idea is that you *cannot* place the mass wherever you want.... Do you see why??

25. Mar 14, 2006

Jasonn

If the mass is hanging from a point .25m from the end of one stick, then I can see why you can't do that (the stick wouldn't stay on your finger for one :tongue: ). Does that mean the weight should be placed in the center?