# How does the centrifugal force affect your weight on a scale?

1. Oct 11, 2009

### BobbyBear

Um, I suppose this is very basic but I just got kind of confused.

Okay suppose you are standing and swinging a rope with a mass attached to its end, so that the circular trajectory of the mass at the end of the rope lies in a vertical plane. And suppose you are stood on a weighing scale. Does the weight indicated by the scale vary as the position of the mass in its trajectory varies? When the mass is at the top, it's 'pulling' you upwards with a force equal to $$m w^2 R$$, where m is the mass at the end of the rope, w the angular velocity of its movement, and R the radius of the rope. So does that mean that at that instant the weight shown on the scale is less? And when the mass is at the bottom, it's more? I don't think so, but I'm not too sure now...

2. Oct 11, 2009

### Bob_for_short

Yes, the weight readings will be variable. You correctly described physics.

3. Oct 11, 2009

### BobbyBear

Oh no, I was hoping I was wrong! Shouldn't the scale show a constant weight (the weight of the person plus the weight of the mass at the end of the rope) because, the centrifugal force upon the person is not an external force... ?

4. Oct 11, 2009

### Bob_for_short

Stop rotating and it will show the sum of weights.

The problems are

1) in relative motion of two bodies and

2) measuring the weight of one of them rather than of the center of inertia of the system.

5. Oct 11, 2009

### BobbyBear

I don't understand :( The scale is measuring the weight of both of the bodies, no? So the internal forces (the ones transmitted by the rope) don't alter that weight! The external forces upon the system are the gravitational forces that act upon the person and the mass at the end of the rope, and the normal upward reaction of the scale, the sum of which is zero as the system as a whole is not accelerating. Therefore the scale (which exerts the normal reaction force) measures the constant value of the weight of the person and the weight of the mass at the end of the rope, no?

By centre of inertia you mean centre of mass?

6. Oct 12, 2009

### tramar

If the string is in your hand and you're swinging a mass in a circle, there will be centripetal force acting on the mass. Whenever the mass is at an angle between 0 and 180 with the horizontal then the centripetal force will have a y component aiming downwards. Add this to the gravitational force which is F = mg where m is the mass of everything on the scale, and yes the reading on the scale should change.

At least that's how I see it, someone tell me if I'm wrong.

7. Oct 13, 2009

### Pythagorean

The normal force acting up decreases when your "effective mass" decreases as the weight on the string is at the top, "pulling" you up.

The same kind of thing will happen if you bend your knees and then straighten them real quick, you'll impart an impulse onto the scale, and it will show heavier for a second as you straighten them, BUT when you first bend your knees to "charge" up, the scale will read lighter (well, it depends on how quickly the scale refreshes, but your weight is effectively smaller while you're bending your knees). Over time, if you average the two, they should work out to your nominal mass before as if you just stood there.

In the same way, as you rotate the weight around the mass, if you average over time, what you expect to be true will, you will get the nominal weight of you plus the mass, but you're doing the experiment slow enough that you get to see instantaneous frames of the experiment where the momentum of the weight is imparting a force by the impulse: F = dp/dt (the change in momentum over time).

The problem is statics vs. dynamics. You've likely been trained in statics, and this is more of a dynamic problem because of the constantly changing momentum of the mass (even if you twirl it a constant speed, the direction is changing, so velocity and thus momentum is actually changing)

8. Oct 13, 2009

### Bob_for_short

Yes. It is the center of mass which is not affected by the internal interaction.

Your weight is affected by the rotating body. For you it is an external variable force.
As soon as you are the sole body in contact with the scale, it will show a variable weight.

Instead of rope, you can use a spring with the same success.

Last edited: Oct 13, 2009
9. Oct 13, 2009

### BobbyBear

Thank you so much, everyone, for explaining! And yes, I understand what you're saying . . . but, on the other hand, I don't see how my other reasoning is wrong :(

I mean: the interaction between the object at the end of the rope, and the person, transmitted through the rope as tensile forces, are internal forces, right? (assuming the person and the object are the system being analysed). And according to Newton's second law, only external forces can vary the amount of movement of the system. So, if initially the person is standing still and is simply holding the rope without swinging it, the scale will simply show the sum of masses of the person and the object, right? Then, the person starts swinging it, and according to Newton's law, the centre of mass has to continue being stationary and in the same position it was before (as it was stationary before the person started swinging the rope). Is this true so far?

Bob_for_short: For you it is an external variable force. For me, the person, yes . . . but the system under study is the person and the object at the end of the rope. So it is an internal force, yes? The object pulls me, but I pull the object (reaction force), so they cancel out and have no effect whatever on the state of movement of the centre of mass of the system. Yes?

Pythagorean: The normal force acting up decreases when your "effective mass" decreases as the weight on the string is at the top, "pulling" you up. Yes. . . but, when the weight is at the top pulling me up, I am pulling it down with an equal but contrary force . . . so as far as the entire system goes, they cancel each other out and the scale should not be aware that the mass is pulling me up . . . at least that's my reasoning. Can you explain me why it's wrong?

tramar: If the string is in your hand and you're swinging a mass in a circle, there will be centripetal force acting on the mass. Yes yes, but! there is an equal but contrary force to the centripetal force acting on the mass, that is acting on the person!

Thanks for all your answers, but I hope that my dilemma is clearer now :P

10. Oct 13, 2009

### BobbyBear

So . . . if you guys are right, then does that mean that, if I were to swing the mass fast enough, I could effectively be lifted off the ground when the mass is at the top?

11. Oct 13, 2009

### BobbyBear

Hmm, I was just thinking, the momentum of the centre of mass of the system (person plus weight at the end of the rope) is constantly changing, as you've said, right? But how can this be if the rope has been put into motion purely by internal forces? Newton's second law says that only external forces can change the state of motion of the centre of mass. Where am I going wrong?

12. Oct 13, 2009

### Bob_for_short

No, it is not so.

Consider two interacting classical particles, for simplicity, bound with an elastic force:

m1a1 = F(r1-r2)

m2a2 = -F(r1-r2)

These two equations are essentially coupled and are hard to solve. The dynamics of particles is rather complicated. Each equation contains a force so the force is not zero.

Then make the following variable changes: introduce R = (r1m1 + r2m2)/Mtot and r = r1 - r2. You obtain:

MtotA = 0,

µa = F(r)

The first one follows from the original Newton equations for r1 and r2: you add them and the forces cancel. So it is the center of masses that is not sensible to the internal forces, not each separate particle.

If there is an external force like mg, the CM equation will be MtotA = Mtotg

13. Oct 15, 2009

### BobbyBear

Um, what are µ and a in µa = F(r) ? Are you simply representing the dynamic equation of any one of the individual particles? I understood everything else, so thank you :)

14. Oct 15, 2009

### Bob_for_short

Sorry, I thought you knew the expression for the reduced mass µ: µ = m1m2/(m1+m2).

a is d2r/dt2, the acceleration of the relative distance r.

A is d2R/dt2, the acceleration of the center of mass R.

You know, the center of mass coordinate R and the relative distance r obey some equations. These coordinates do not correspond to any individual particles. Rather, it is "collective" variables. But their equations are Newton-like or particle-like. So the center of mass and the relative distance are called quasi-particles.

Last edited: Oct 15, 2009
15. Oct 15, 2009

### willem2

both gravity and the normal force are external forces on the rope/person system, and they are responsible for the motion of the centre of mass of the system.

16. Oct 15, 2009

### Franco_v

If you were suspended in mid-air with no gravity and were swinging the rope & mass around in a vertical plane, the center of mass of the system would not move, since there are only internal forces acting, and no other forces are acting on the system. This means, however, that your body would be moving (up and down) such that the position of the center of mass of the system (which consists of your body + rope & mass) does not change.

But, if you were to stand on a scale (with gravity present), then you are introducing external forces which are: 1) gravity, and 2) the upward normal force exerted on you by the scale. The presence of the normal force results from the fact that the position of your body is fixed in space (on the scale). Therefore, as you swing the rope & mass around, the center of mass of the system must change position with time. Hence (since gravity is constant), the normal force (scale reading) must change since the center of mass of the system is changing position (accelerating) with time.

Last edited: Oct 15, 2009
17. Oct 15, 2009

### Pythagorean

Willem answered your other question to me (gravity is not an internal force).

Yes, the weight IS pulling you up and you are pulling it down, but the scale isn't measuring the force between you and the string. They do cancel each other out at the interface between your hand and the string (which is why there is no net motion in the radial direction of the boundary between your hand and the string).

This means nothing to the scale, who is only measuring the force down on it (and pushing with an equal force up, as you'd presume). But that force that it's measuring is all the forces on you, including the force from the string has on you, but not the force that you exert back on the string.

When you draw a free-body diagram, you generally isolate the body, so you would draw the force up on you from the string, but you wouldn't draw the equal and opposite force that acts on the string acting on you, it doesn't act on you, you act on it. So it's not only opposite in direction, you also have to pay attention to the subscripts:

$$F_{12} = -F_{21}$$

so that in addition to the negative sign that implies opposite direction, the subscripts flip so that you recognize the force from 2 acting on 1 does not sum with the force from 1 acting on 2... at least, not on the same body.

18. Oct 31, 2009

### BobbyBear

Sorry for not thanking you all earlier, some things came up and I couldn't.

Thank you Bob_for_short, I read up a bit on two mass-systems and I understand now what the reduced mass is about =)

Yes, thank you! I forgot about gravity lol and it just didn't occur to me that the upward normal force exerted by the scale is an external force :P silly mee..
As Bob_for_Short and Franco_v have described, if there was no gravity and both bodies were suspended in Space, then the centre of mass wouldn't accelerate :)

Okay . . . just one thing: you said, when you draw a free body driagram, you isolate the body . . . so why are we isolating only the person? is it because the person IS in static equilibrium, and their centre of mass IS still, so that what we are applying are the equations of static equilirbrium upon the person? Because the system I was isolating was the person and the mass at the end of the string, but I understand that this system is not in static equilibrium, as everyone has pointed out, so I can't just balance forces in the latter case: I'd have to consider the accelertion of the centre of mass.

Thanks all for your great insights xD