# Circular Motion

1. Nov 27, 2004

### FZX Student

I have a physics lab on centripetal force and I'm stuck on an anaylsis question. The lab requires me to have a piece of string with masses on opposite ends threaded through a tube. The mass at the bottom should hang, while the mass on the top are swung in circular motion. The question asks, if the masses on the top and bottom were to be doubled, what effect would it have on velocity? I predicted that there would be no effect because the masses are still proportionate to one another, but in the experiment I got something completely different. Can you help me?

2. Nov 27, 2004

### Sirus

Consider that:
$$F_{centripetal}=\frac{mv^2}{r}$$
$$F_{g}=mg$$
What is causing a centripetal force to exist?

3. Nov 28, 2004

### FZX Student

Still confused

Thanks for your help but I'm still confused,
So what you are saying is that gravity is causing the centripetal force to exist and that Fc = Fg?
If so then if the gravitational force was doubled then the centripetal force would be doubled as well as velocity? How would I show that using the equation mv2/r? If I equated 2Fc = 2Fg wouldn't the 2 cancel?

4. Nov 28, 2004

### mgiddy911

It seems to me that the mass would have no effect with the equation sirius presented, mg=(mv^2)/2 the m's canel out since they are on both sides of the equation, I could be wrong however...

5. Nov 29, 2004

### Sirus

FZX, changing both masses should have no effect because
$$F_{g}=F_{centripetal}$$
$$mg=\frac{mv^2}{r}$$
$$v=\sqrt{gr}$$
A couple things to think about: how different was you experimental data? What types of error were present in your lab and how significant do you think their effect was?

6. Nov 29, 2004

### Cyrus

I dont see how you equated the force due to gravity and the centripital acceleration. They are in different axis. Gravity acts down and the centripital acceleration acts side ways. They are compeletly different from one another. Based on what FZX said, the mass did have an affect when he did the experiment. Maybe I am mistaken in how he did the experiment though.

7. Nov 29, 2004

### Sirus

I am not equating them vectorally, because yes, they point in different directions. But their magnitudes should be the same. Assuming no friction, the gravitational force is the only force keeping the revolving mass from flying off along with the string.

Remember that I am not talking about two forces acting on the same mass. Perhaps I should have made this more clear in my equations, but it is the magnitude of the gravitational force on the mass in the tube that equals the magnitude of the centripetal force on the revolving mass.

8. Nov 30, 2004

### Cyrus

I dont see how you can say the tension is the same though sirus. I would agree with you if the mass swinging was on some sort of a board or something. Because when you equate the magnitudes like that, you assume that the mass is spinning along a flat plane with no angle. In this case the tension is equal to the centripital acceleration. But its impossible to spin it with zero angle, because that would require infinte tension. You would have it spinning in the air horiziontally with zero tension in the rope. So there has to be some angle to it. And the component of force in the x direction should equal the centripital acceleration. But not the total magnitude of force on the string. Hmmm, Im trying to figure out how to go around this problem but am not having any luck.

9. Nov 30, 2004

### Sirus

I agree. I was assuming that the tension force on the swinging mass had no vertical component. In reality, the tension force will be the resultant vector of the gravitational force on the swinging mass and the centripetal force.

In that case, I think it should come out to something pretty similar:
$$F_{c}=F_{T}\sin{\theta}$$
Since the tube basically acts like a pully (no friction), the tension should be constant throughout the string, and since the mass in the tube is in equilibrium, its tension force must be equal to its gravitational force.
$$\frac{mv^2}{r}=mg\sin{\theta}$$
$$v=\sqrt{\sin{\theta}gr}$$
Does that look good?

10. Nov 30, 2004

### Cyrus

Hmm, im not sure what im doing wrong but im not getting it to work out for some reason. Here is my reasoning: The tension in the rope is equal to the tension in the segment of rope that has the hanging mass. Thats just equal to M(hanging)*g. But since its the same rope, the tension in the portion of wire that spinning must also be equal to Mg. Yet at the same time, the Y component of tension of the spinning mass has to be equal to the weight of the spinning mass, otherwise it would not stay up, it would fall and hit the tube. But if the two tensions are equal, this implies that the tension in the x direction has to be equal to zero. Other wise when you find the tension due to the X component and the Y component, they will be greater than the tension in the part that has a mass hanging on it. This implies that when the two masses are equal it spins on a flat plane, but thats wrong. The only way around this problem seems to be if the tensions in the two portions are not equal. But that does not seem right either. Its an enigma to me. :-(

11. Nov 30, 2004

### rayjohn01

I believe that ALL forces irrispective of how generated , gravity or centrifugal, are proportional to the mass , so from an ideal viewpoint no change should be required.
There are some hidden things which may change that according to setup , friction , air drag, mass errors , and so on . ( the string tension WILL be uniform unless it has non-negligeable mass ) if it was not then you have disturbed the mass distribution !!!!!!!
and the above is NOT true.

12. Nov 30, 2004

### ZapperZ

Staff Emeritus
The tension HAS to be the same in magnitude because: (i) the string is inextensible and (ii) there is no movement either up and down for the hanging mass and no movement radially for the "orbiting" mass. Thus, there has to be a balance of the force along the string. If the tension is not uniform throughout the string, then there has to be a slack somewhere, or movement of either/both masses.

Zz.

13. Nov 30, 2004

### Cyrus

But Zapperz, what the centriptal force acts long the axis of rotation. But the swinging mass is not going to swing along a horizontal plane, thats impossible. So there must be a component of tension in the x and y directions. And the y component has to equal the swinging mass force. And the x direction has to be the centriptial acceleration.

14. Nov 30, 2004

### rayjohn01

non -horizontal

That is correct, but it makes no difference forces are still proportional to m
The string DOES make a difference if it's mass is not negligeable, the tension will then be non-uniform just like a hanging chain.
At this point there is not enough info in the question to resolve the issue
the questioner should explain the physical set up.
Ray.

15. Nov 30, 2004

### ZapperZ

Staff Emeritus
Er... come again?

The "swinging" mass has a radial tension that is keeping it in a circular motion. Do we agree on this?

Now what is providing this radial tension? Hint: if you cut the string connected to the hanging mass, the swinging mass will fly off in a straight line.

The issue here is that you TWO SEPARATE systems: one for the swinging mass, the other for the hanging mass. The only thing that connects these two systems is the tension on the rope. The swinging mass couldn't care less what is providing the centripetal force: be it a mass hanging at the end of the rope, or someone's hand pulling on it, or if it is simply attached to a hook at the end of a bottomless hole! It really doesn't care! All you do when you do a FBD on this swinging mass is the tension pointing inwards.

Now do the same thing with the hanging mass. The FBD that you sketch will only have TWO forces: one for the weight acting downwards, the other for the tension acting upwards, and they balance out under static equilibrium. The hanging mass also couldn't care less what is providing that tension: it could be a hand holding the other end of the rope, a mass "swinging" in a plane, or the string tied to a hook. All it cares about is that the tension is balancing out its weight. Period!

The ONLY thing that connects those two is the tension on the rope. If the rope is inextensible, then however you bend or twist the rope, if there is slack and you are at static equilibrium, the tension will be the same! I'm puzzle why this is so astounding. The same thing is done on a mass on a horizontal table being attached to a rope over a pulley and at the other end, another mass hanging vertically. Again, the tension on each one is of the same magnitude even when the tension act horizontally on one mass and acts vertically on the other. Why is this causing a problem?

Zz.

16. Nov 30, 2004

### Sirus

That is not entirely true. As we discussed earlier, the centripetal force points inward in a direction perpendicular to the vertical tube. It is the x-component of the tension force, who's vertical component is the gravitational force $F_{g}$ on the swinging mass. I do not believe that gravity can be ignored here.

Other than that we are agreed.

17. Nov 30, 2004

### Cyrus

"All you do when you do a FBD on this swinging mass is the tension pointing inwards"
NO! The tension acts along the rope and is NOT pointing inwards. There a Y component and an X component. Lets step through this together and see where my problem is occuring.

18. Nov 30, 2004

### ZapperZ

Staff Emeritus
Once again... eh?

Why are we using cartesian coordinates for the mass that's moving in a circle? What is wrong with plane polar coordinates for that?

They are two separate system connected by the tension of the string. You can use any coordinate system you wish for each system. Obviously, it appears that we'd rather use not the straightforward one, but rather make the problem more difficult than it is by using cartesian coordinates for something moving in a circle.

Zz.

19. Nov 30, 2004

### Cyrus

Because It Cant Move In A Plane! DO A FBD moving in a plane is imposibe it requires infinte force! If its moving an a plane, what component of tension in the Y directions going to hold it in mid air!?!?

20. Nov 30, 2004

### ZapperZ

Staff Emeritus
This is getting absurd.

Look at the ORIGINAL QUESTION:

My question is, why in hell are you using "x and y components" for the mass that's moving in a circular motion when a plane polar coordinate would do? This makes NO sense! The centripetal force provided via the tension in the string acts RADIALLY (as in the radial coordinate of a plane polar coordinate system), and it HAS to act inwards for it to be a "centripetal force" that is causing a circular motion! I can't believe this is still a mystery.

Zz.