Rotational Equilibrium and Dynamics

[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.

 

[nrqed]

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.

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

 

[Jasonn]

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

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.

 

[nrqed]

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.

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?

 

[Jasonn]

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?

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:

 

[nrqed]

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:

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)

 

[Jasonn]

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)

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?

 

[nrqed]

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?

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...

 

[Jasonn]

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...

I made a mistake...

[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.

 

[nrqed]

I made a mistake...

[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.

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)

 

[Jasonn]

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)

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.)

 

[nrqed]

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.)

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?

 

[Jasonn]

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?

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?

 

[nrqed]

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?

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!)

 

[Jasonn]

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!)

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)

 

[nrqed]

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)

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)

 

[Jasonn]

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)

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)?

 

[nrqed]

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)?

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.

 

[Jasonn]

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.

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.

 

[nrqed]

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.

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

 

[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.)

 

[nrqed]

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.

I know that

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.)

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!

 

[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?

 

[nrqed]

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?

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??

 

[Jasonn]

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??

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?

 

[nrqed]

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?

Ok.. that's the whole point... If the net torque is zero, what does it tell you? That the object won't rotate. So if the center of the stick is to the right of your finger, on what side do you think you will have to place th emass in order to have equilibrium? Obviously, on the left of your finger. Now, do you think you will have the choice in placing the mass wherever you want?? what do you think will happen as you place the mass at different spots?

 

[Jasonn]

Ok.. that's the whole point... If the net torque is zero, what does it tell you? That the object won't rotate. So if the center of the stick is to the right of your finger, on what side do you think you will have to place th emass in order to have equilibrium? Obviously, on the left of your finger. Now, do you think you will have the choice in placing the mass wherever you want?? what do you think will happen as you place the mass at different spots?

K, I understand that much. You obviously need the mass to the left of the axis (finger) if the center of the stick is to the right. However, I don't understand WHERE to place the mass to the left exactly. I do see how the torque changes depending on where you place it from the axis.

 

[nrqed]

K, I understand that much. You obviously need the mass to the left of the axis (finger) if the center of the stick is to the right. However, I don't understand WHERE to place the mass to the left exactly. I do see how the torque changes depending on where you place it from the axis.

Ok.. that's the key point of the entire thing. When you were writing your equations, you were setting the total torque equal to zero. In that case the system won't rotate.

Now, forget about equations for a second...Just use your intuition (or use a ruler right now and use an eraser to play the role of the mass...). You put your ruler on your finger and put the eraser at different points...What will happen? Is there a place for which the system will not rotate and stay at equilibrium? Is there more than one place for the eraser for which the system will stay at equilibrium? (while keeping your finger at the same position under the ruler)

Now, if you repeat after placing the ruler at a different position...will you have to place the eraser at the same spot as before (that is, at the same distance from your finger)??

Do you see what I am getting at?

 

[Jasonn]

I see where I made a mistake in the equations. Rather, the two torques should be equal, correct?

So I need to find a distance from the axis, which I can place the mass and still remain in equilibrium?

 

[nrqed]

I see where I made a mistake in the equations. Rather, the two torques should be equal, correct?

So I need to find a distance from the axis, which I can place the mass and still remain in equilibrium?

Yes, the two torques must add up to zero, so they must be equal in magnitude. The point is to find the spot (the unique spot) where the mass can be placed and the whole system (meter stick and mass) stays in equilibrium. you can see why this is impossible if your finger is under the middle point of the meter stick, by the way.

If you place the meter stick at a different position, the mass will have to be placed somewhere else...the two torques will be different from the values you had in th eprevious position (but still equal to each other) and of course, when you solve, you will get the same mass for the meter stick.

The farther your finger is from the middle of the stick, the farther the mass will have to be placed from your finger. If you put the center of th emiddle stick too far from yoru finger, it will be impossible to find a spot where the mass will keep the system in equilibrium.

Does this make sense?

 

[Jasonn]

Yes, thank you for your help. :)

 

[nrqed]

Yes, thank you for your help. :)

You're welcome.

Hopefully, you now see the answer to your question "if I use 70 cm instead of 50 cm, I will get a different mass"...

The answer is that you don't get to pick the distance of the mass...The experiment will tell you that only one distance (for a given mass and a given position of the ruler on your finger) will keep the system at equilibrium.

godo luck!