Solving a Centripetal Force Lab: Doubling the Masses

[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?

 

[Sirus]

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

 

[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?

 

[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...

 

[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?

 

[Cyrus]

I don't see how you equated the force due to gravity and the centripetal acceleration. They are in different axis. Gravity acts down and the centripetal 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.

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?

 

[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.

 

[Cyrus]

I don't 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 centripetal acceleration. But its impossible to spin it with zero angle, because that would require infinite 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 centripetal acceleration. But not the total magnitude of force on the string. Hmmm, I am trying to figure out how to go around this problem but am not having any luck.

 

[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?

 

[Cyrus]

Hmm, I am not sure what I am doing wrong but I am 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 that's 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. :-(

 

[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.

 

[ZapperZ]

I don't see how you can say the tension is the same though sirus.

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.

 

[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, that's 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.

 

[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.

 

[ZapperZ]

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, that's 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.

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.

 

[Sirus]

All you do when you do a FBD on this swinging mass is the tension pointing inwards.

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.

 

[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. Let's step through this together and see where my problem is occurring.

 

[ZapperZ]

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.

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.

 

[Cyrus]

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

 

[ZapperZ]

"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. Let's step through this together and see where my problem is occurring.

This is getting absurd.

Look at the ORIGINAL QUESTION:

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.

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.

 

[Cyrus]

Are you implying that the mass swings in a flat circle then? With absolute no sag, so that all of the centripetal force is along the string directly?

 

[Sirus]

I actually do believe the mass is moving in a horizontal plane. Its y-coordinate relative to the tube is not changing. It traces a horizontal circle around the tube with radius perpendicular to the vertical tube. Regarding your question about what holds it in mid air, please review my previous posts about the components of the tension force vector.

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.

 

[Cyrus]

The mass that is spinning has to make some angle with the horizontal... It cannot be in a plane. The X component of force can be treated as a circle moving in a plane. And that is the X component of tension in the string. But there also has to be a Y component of tension in the string to keep it from falling DOWN.

 

[Sirus]

Are you implying that the mass swings in a flat circle then? With absolute no sag, so that all of the centripetal force is along the string directly?

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.

 

[ZapperZ]

I GIVE UP!

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.

 

[Cyrus]

Read tt Zappper! LoL it says its on a table! the table holds it up! take away the table and it falls! We don't have a table to support the Y component of force. By the way, its a very nice little simulator i saw it too, thanks for the link. there is another one i will send you so you can see what I am talking about.

 

[Cyrus]

http://www.phys.unt.edu/~klittler/demo_room/mech_demos/1d50_10.html Look at the picture, the ball SAGS! Its NOT a plane circle!

 

[Sirus]

What can I say? This is hilarious. It's been a pleasure working with you, gentlemen.

Sirus

 

[Cyrus]

Lol Hugs For All Of You! This is too much. I love you guys.

 

[Sirus]

Nope, have to chime back in. Abdollahi, although in the picture the person is not holding the tube vertically, but at an angle, the ball would, as you say, "sag" even if she did. The ball would be orbiting below the edge of the tube, not on the same vertical level as it. Nevertheless, my previous statements regarding this problem still hold. Now see post #28.