Circular Motion

  • #36
Cyrus
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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).
 
  • #37
Sirus
574
2
ZapperZ said:
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... :tongue:
 
  • #38
Cyrus
3,150
16
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.
 
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  • #39
Cyrus
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16
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?
 
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  • #40
Sirus
574
2
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?
 
  • #41
Cyrus
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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.
 
  • #42
ZapperZ
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cyrusabdollahi said:
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.
 
  • #43
Sirus
574
2
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 [itex]F_{g}[/itex] 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?
 
  • #44
Cyrus
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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.
 
  • #45
Tide
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If [itex]m_2[/itex] is the hanging mass and [itex]m_1[/itex] is the revolving mass then the tension on the string is

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

so that

[tex]v = \left(\frac {m_2^2 - m_1^2}{m_1^2}\right)^{1/4} \sqrt {gr}[/tex]
 
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  • #46
Cyrus
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tide you forgot to put a square inside the centripital force component. (quick edit while no ones looking!). :-)
 
  • #47
Cyrus
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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?
 
  • #48
Cyrus
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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.
 
  • #49
Sirus
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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. :smile:
 
  • #50
Cyrus
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16
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.
 
  • #51
fomenkoa
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Mass

Are you crazy? Clearly the masses are in a state of pure equilibrium with the net conservation of mass modeled by the equation F= GvC, where C is a constant! Chaos theory would predict that the flux of friction dissipation would be inversely proportional to its mass in a vaccuum!

Just kidding
 
  • #52
Cyrus
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I must be crazy what was i thinking. I forgot about the flux of friction dissipation! Tide, looks like we have some work to do! How could I have over looked that. Btw, I am working on a flux capacitor to put in my delorean. I think your Chaos theory equations might just be the model i need to make it work! :rofl:
 
  • #53
FZX Student
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0
Wow, thank you everyone who has gone through all this trouble to help me. Unfortunately, I feel the situation has gotten a little out of hand. So to prevent people from debating and hitting themselves with baseball bats (ie: zapperz), I'll tell you what my physics teacher explained to me. Basically, if the masses were doubled there should be no effect on velocity. I don't really want to go into equations, but I think Sirius got the gist of it.
2Fg = 2Fc
2mv^2/r = 2mg
*cancel 2m*
v^2/r = g
v^2 = gr (which is the same as the original weights)
How I got a completely different answer, is a complete mystery on its own.
 
  • #54
Cyrus
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16
FZX Student, in your case you have to remember that there was a simplifying assumption. You assumed that it spun on a horizontal plane. What tide and I did was a realistic case that includes the effects of it not being horizontal. In actuality its not really spinning around in a flat disk.
 

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