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- Thread starter FZX Student
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[tex]F_{centripetal}=\frac{mv^2}{r}[/tex]

[tex]F_{g}=mg[/tex]

What is causing a centripetal force to exist?

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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?

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[tex]F_{g}=F_{centripetal}[/tex]

[tex]mg=\frac{mv^2}{r}[/tex]

[tex]v=\sqrt{gr}[/tex]

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?

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Sirus said:

[tex]F_{g}=F_{centripetal}[/tex]

[tex]mg=\frac{mv^2}{r}[/tex]

[tex]v=\sqrt{gr}[/tex]

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?

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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.

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In that case, I think it should come out to something pretty similar:

[tex]F_{c}=F_{T}\sin{\theta}[/tex]

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.

[tex]\frac{mv^2}{r}=mg\sin{\theta}[/tex]

[tex]v=\sqrt{\sin{\theta}gr}[/tex]

Does that look good?

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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.

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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.cyrusabdollahi said:I dont see how you can say the tension is the same though sirus.

Zz.

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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.

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Er... come again?cyrusabdollahi said:

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.

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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 [itex]F_{g}[/itex] on the swinging mass. I do not believe that gravity can be ignored here.ZapperZ said:All you do when you do a FBD on this swinging mass is the tension pointing inwards.

Other than that we are agreed.

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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.

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Once again... eh?Sirus said: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 [itex]F_{g}[/itex] on the swinging mass. I do not believe that gravity can be ignored here.

Other than that we are agreed.

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.

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This is getting absurd.cyrusabdollahi said:

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.

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.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.

Zz.

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ZapperZ, I am not very familiar with polar coordinates, but it seems to me that cartesian coordinates are efficient when dealing with force vectors in this problem, a concept I find to be key to understanding the motion of the mass.

I feel as if we are all talking about the same thing but in different languages, unless I am truly misunderstanding something.

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No. The mass swings in a flat circle, but the string providing the tension force is not flat. The tension force vector for the swinging mass points upward at an angle, towards the edge of the tube, but since the length of the string (from the mass to the tube's edge) is constant, the mass is kept at contant horizontal distance form the tube (and also at constant vertical distance from the edge of the tube) while it swings, causing it to follow a horizontally-flat circular path. Again, see my posts about vector components to explain the difference between the centripetal force and the tension force in this case, as I see it.cyrusabdollahi said:

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Someone, please smack me silly for sticking my nose into this! It has gotten to be utterly ridiculous.

This is such a STANDARD problem, they even make a JAVA applet for it, for heavens sake!

http://www.phy.ntnu.edu.tw/java/circularMotion/circular3D_e.html

Zz.

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