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 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.

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 centripital 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 centripital 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 centripital 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 centripital 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.

 

[Cyrus]

My head is starting to hurt now lol. Can we walk through this step by step zapperz?

 

[ZapperZ]

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.

Sigh... maybe we ALL should wait till the original poster comes back and explain if this is all in a horizontal flat plane, or if *I* was the one who simply interpreted the question wrong...

Zz. <smacks himself silly with a baseball bat>

Edit: PS: I apologize for going off the bend with you two. I should have waited for the explanation.

 

[Cyrus]

LOL! Hey, leave some of that bat for me too!

 

[Cyrus]

Ok zapperz, let's just do my question then is that ok? let's see if we can make the mass in the tube remain static if we provide a centripital acceleration to the spinning mass, of equal value to the stationary one.

 

[Cyrus]

I will go through step by step and you can stop me if u find a problem. (like the ball and tube comibation in the picture i sent you)

 

[Cyrus]

Part a.) the hanging mass, The mass in the tube is hanging, and the force is just the weight Mg. So the tension in the segment of string in the tube holding up the mass will have to equal Mg if it is to be static. Because its the same piece of string, we all agree that that the tension in the segment of string connected to the spinning mass is also equal to Mg, (of the hanging mass).

 

[Sirus]

Sigh... maybe we ALL should wait till the original poster comes back and explain if this is all in a horizontal flat plane, or if *I* was the one who simply interpreted the question wrong...

Zz. <smacks himself silly with a baseball bat>

Edit: PS: I apologize for going off the bend with you two. I should have waited for the explanation.

I agree that the ball is moving in a flat horizontal plane. See my previous posts.

The original poster will come back to two pages of argument and regret what he started...

 

[Cyrus]

So the Magnitude of the tension in the string, (for the spinning mass), must equal Mg of the hanging string.


When the mass is spinning, there must be a component of tension in the Y direction to support the spinning mass from falling down. And it must equal the weight of the spinning mass.

Are we ok so far? Or do you disagree.

 

[Cyrus]

Now because the mass is spinning it has a ceneripital force along the plane of the circle. But this will ONLY contribute to the X component of force. Because as you stated, it acts along the radial direction. And the radius of the circle it traces is NOT along the string, but it is horizonal along the axis of the rod to the mass. So this force must be balanced by a tension in the X-direction of the string, for the part that is spinning.

But the two masses are equal. Any force in the x direction will make the magnitude of the tension in the spinning string more than the stationary one. So we can conclude that the stationary mass has to be heavier than the spinning one in order to allow for a component of tension in the x direction.

Did I goof off somewhere?

 

[Sirus]

I see what you are saying. Perhaps an additional force must be introduced into the system to keep the mass in circular motion? When you actually do this in the lab, you have to keep moving your arm around to keep the system the way you want it (outer mass circling, mass inside the tube stationary to keep radius constant). This could, however, only be necessary to overcome non-conservative forces such as friction, forces not present in the theoretical model of the problem we are concerned with. What do you guys think?

 

[Cyrus]

Hmm, that's a good point. I think your right though, when you flick your wrist all you do is keep up the motion as its lost to friction, also you support the string from falling as it supports itself on the rim of the tube.

 

[ZapperZ]

Now because the mass is spinning it has a ceneripital force along the plane of the circle. But this will ONLY contribute to the X component of force. Because as you stated, it acts along the radial direction. And the radius of the circle it traces is NOT along the string, but it is horizonal along the axis of the rod to the mass. So this force must be balanced by a tension in the X-direction of the string, for the part that is spinning.

But the two masses are equal. Any force in the x direction will make the magnitude of the tension in the spinning string more than the stationary one. So we can conclude that the stationary mass has to be heavier than the spinning one in order to allow for a component of tension in the x direction.

Did I goof off somewhere?

So obviously we want to jump the gun and run with this scenario.

Well, OK, if you want to do that... What you THEN have, assuming your picture is the correct one, is that you have a conical pendulum at the top. However, now the calculation of the tension in the string isn't as straightforward, since now, BOTH masses contribute to the tension since the swinging mass now will have a component of its weight along the string.

If this is true, then we cannot solve this problem till we first solve the correct angle that the conical pendulum makes with the vertical for the given mass.

Zz.

 

[Sirus]

What you were saying got me thinking that if the tension force acting on the swinging mass is the vector resultant of the centripetal force and the vertical gravitational force, it cannot, at the same time, be equal to the vertical gravitational force acting on the other mass, since this is a right triangle in which the tension force is the hypotenus, and F_{g} is the vertical component. This seems to point towards the presence of a non-conservative force here. However, the overcoming of frictional forces may explain the laboratory experience. Would the mass continue in circular motion indefinitely given an isolated system? ZapperZ?

 

[Cyrus]

http://physics.bu.edu/~duffy/java/Circular.html look look this is exactly what i mean. The component of force in the y direction is simply the weight of the spinning mass. the remainder of the tension in the x direction goes into the centripital force. so you don't need to know the angle. you can calculate the angle based on the weights of the two masses.

 

[Tide]

If m_2 is the hanging mass and m_1 is the revolving mass then the tension on the string is

T = m_2 g = \sqrt { \left(m_1 g\right)^2 + \left( \frac {m_1 v^2}{r^2}\right)^2}

so that

v = \left(\frac {m_2^2 - m_1^2}{m_1^2}\right)^{1/4} \sqrt {gr}

 

[Cyrus]

tide you forgot to put a square inside the centripital force component. (quick edit while no ones looking!). :-)

 

[Cyrus]

I have a somewhat, disturbing problem. I rigged up a homemade rod and string with equal masses on both ends. According to the equation, they should not be able to stay at equilibrium; however, they DO. I can't figure out for the life of me WHY!? They were light masses on each end. Maybe friction was large enough in this case? I don't know what's your take?

 

[Cyrus]

Is your name Sirus, or is that just the nickname you choose for this. I was wondering if you pronounce your name sea-ros, but spell it like sirus the cloud.

 

[Sirus]

I am still inclined to say that if the masses are equal, due to my earlier reasoning, a non-conservative force must be present, most likely one like that which you apply to keep the mass revolving when you actually do the experiment.

Cyrus, Sirus is the nickname I use on PF, and I pronounce it like sigh-rus. Good to see another sirus/cyrus on the forums.

 

[Cyrus]

Now, now. Look at the equation tide provided sirus. It shuold become evidently clear that the masses cannot be equal. The nonconservative forces go into play to overcome the friction between the rope and the rod, that's all.