Circular Motion - Hanging object swinging around

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Homework Statement


An object of mass m is rotating, hanging from a string of known length l. The string is attached to a pole, which rotates with a known angular velocity ω and forms a to-be-determined angle α with the string. Find α.

I think I have solved it (the numbers match the results of the book), but I could use a double-check from you veterans :D

Homework Equations


F = m*g for the weight force of an object.
Centripetal Force = m*ω2*r, with ω2*r = centripetal acceleration (a).

The Attempt at a Solution


I drew the free-body diagram (imagine the object rotating "into" the screen):
HR54Bg5.png


T = tension
F = centripetal force
W = weight

So on the X axis and Y axis we have:

X axis:
Tx + F = 0
T*sin(α) + m*a = 0

Y axis:
Ty - W = 0
T*cos(α) - m*g = 0
=> T = mg/cos(α)

Inputting the T found on the Y axis into the X equation:
mg*sin(α)/cos(α) + ma = 0
g*sin(α)/cos(α) + a = 0

Since a = ω2*r, and the radius is also the length of the rope multiplied by the sine of α (the rope is the hypotenuse of a triangle; in the picture the rope would be aligned with the vector T, and the radius would be aligned with the force F), then a = ω2*l*sin(α)

So the equation becomes:
g*sin(α)/cos(α) + ω2*l*sin(α) = 0
sin(α) * (g/cos(α) + ω2*l) = 0

1. sin(α) = 0 => α = 0
2. g/cos(α) + ω2*l = 0
=> g + ω2*l*cos(α) = 0
=> cos(α) = -g/(ω2*l)
(Then input the numbers and calculate the arc-cos)

Are there any inaccuracies/mistakes, or is this how it's supposed to be done?

Also, with these results (α = 0; the other angle α is impossible because it gives a cos(α) < -1), I'm not sure how to interpret the solution. With α = 0, then r = 0, so a = 0, so F = 0. But with a centripetal force = 0, how can there be circular motion at all? The premise of the problem is that we have an angular velocity, so the object IS rotating, but then it turns out that the angle is zero, so it's "spinning on the center of the circle (and the circle is a single point, which in this case would be the pole)"). Is this still considered circular motion, even if r = 0, or does this have a different name?
 

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  • #2
PeroK
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Homework Statement


An object of mass m is rotating, hanging from a string of known length l. The string is attached to a pole, which rotates with a known angular velocity ω and forms a to-be-determined angle α with the string. Find α.

I think I have solved it (the numbers match the results of the book), but I could use a double-check from you veterans :D

Homework Equations


F = m*g for the weight force of an object.
Centripetal Force = m*ω2*r, with ω2*r = centripetal acceleration (a).

The Attempt at a Solution


I drew the free-body diagram (imagine the object rotating "into" the screen):
View attachment 216679

T = tension
F = centripetal force
W = weight

So on the X axis and Y axis we have:

X axis:
Tx + F = 0
T*sin(α) + m*a = 0

Y axis:
Ty - W = 0
T*cos(α) - m*g = 0
=> T = mg/cos(α)

Inputting the T found on the Y axis into the X equation:
mg*sin(α)/cos(α) + ma = 0
g*sin(α)/cos(α) + a = 0

Since a = ω2*r, and the radius is also the length of the rope multiplied by the sine of α (the rope is the hypotenuse of a triangle; in the picture the rope would be aligned with the vector T, and the radius would be aligned with the force F), then a = ω2*l*sin(α)

So the equation becomes:
g*sin(α)/cos(α) + ω2*l*sin(α) = 0
sin(α) * (g/cos(α) + ω2*l) = 0

1. sin(α) = 0 => α = 0
2. g/cos(α) + ω2*l = 0
=> g + ω2*l*cos(α) = 0
=> cos(α) = -g/(ω2*l)
(Then input the numbers and calculate the arc-cos)

Are there any inaccuracies/mistakes, or is this how it's supposed to be done?

Also, with these results (α = 0; the other angle α is impossible because it gives a cos(α) < -1), I'm not sure how to interpret the solution. With α = 0, then r = 0, so a = 0, so F = 0. But with a centripetal force = 0, how can there be circular motion at all? The premise of the problem is that we have an angular velocity, so the object IS rotating, but then it turns out that the angle is zero, so it's "spinning on the center of the circle (and the circle is a single point, which in this case would be the pole)"). Is this still considered circular motion, even if r = 0, or does this have a different name?
The only things I'd say are: I don't like that minus sign that crept in; and, you made things a little more complicated than needed.

PS on your last question, you just note that ##\alpha =0## is a trivial static solution.
 
  • #3
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Mmh. Why not the minus sign though? I thought + and - depended on the chosen axis. So if I chose +Y upwards and +X leftwards, for Tension it should be positive & positive, for F it should be positive too, but Weight force goes down, so it should be -

If that - was a +, then the solution would be different, no (it would become a cosine > +1, which is still impossible but if the cosine were possible then the solution would be a different angle) ?

I also think you are right (I checked the result again, and the book does say the cosine is > +1, and not < -1), but I was sure you put + or - depending on whether the direction of the vector matches the direction of the +Y/+X axis you chose. Am I supposed to only use + signs in the equation, and then use g = -9.81 if I chose +Y upwards (so the - signs still ends up being in there, but it's more formally correct) ?
 
  • #4
PeroK
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Mmh. Why not the minus sign though? I thought + and - depended on the chosen axis. So if I chose +Y upwards and +X leftwards, for Tension it should be positive & positive, for F it should be positive too, but Weight force goes down, so it should be -

If that - was a +, then the solution would be different, no (it would become a cosine > +1, which is still impossible but if the cosine were possible then the solution would be a different angle) ?

I also think you are right (I checked the result again, and the book does say the cosine is > +1, and not < -1), but I was sure you put + or - depending on whether the direction of the vector matches the direction of the +Y/+X axis you chose. Am I supposed to only use + signs in the equation, and then use g = -9.81 if I chose +Y upwards (so the - signs still ends up being in there, but it's more formally correct) ?
You need to use minus signs consistently. The cosine of an acute angle is positive, so something must have gone wrong.

Personally, I never use negative quantities unless I have to. But, that may be my bad habit!

That said, I think your problem goes back to having

##F + ma =0##

Instead of ##F = ma##
 
  • #5
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Ok ok ok I think I got it this time. When I do these exercises I need to put the sum of all the forces on the "centripetal force axis" equal to the centripetal force. So in this case I can do this:

X axis:
+Tx = F
(the only x component here is Tx, aside from F)

Y axis I can keep what I did before:
+Ty - W = 0

So it then becomes T*cos(α) - mg = 0 => T = mg/cos(α) on the Y axis.
=> (on the X equation) +mg*sin(α)/cos(α) = m*ω2*l*sin(α)

This "feels" more correct now.
PS on your last question, you just note that α=0 is a trivial static solution.
Meaning? :p
If I had to try and interpret this, I think you mean "it's a side-effect solution that doesn't really matter", but that's just a guess. Also, looking at the equation again, I could just as well simplify both terms by dividing by /sin(α) (after all I did that with the mass too), and that would "eliminate" this "first solution" entirely. So...sin(α) = 0, hence α=0, is just a "trivial" solution for this reason I take? And by "static" you mean that this solution implies that there is no circular motion (or that it "degenerates" into a non-motion, really; kinda like when you have those geometry problems and one of the sides of a figure becomes 0, so the figure (rectangle, etc.) becomes a segment or something) ? After all, if sin(α) = 0 then r = 0, so F = 0, so no centripetal force = no circular motion.

The only question I have about this is, again, if there is motion at all or we are in an actually-static situation. In other words, does this solution mean that the system is "idle" (completely static), or does this mean that "there is no circular motion [but there is still some other kind of motion]" ? If I think about it, I feel like the pole is still rotating on itself, with the string and object both lying on the pole and rotating while remaining attached to the pole. So maybe there is no circular motion, but there is still another form of motion, which is maybe "rotation" (around the pole's axis)...?
 
  • #6
PeroK
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The static solution is simply an object hanging freely. It should be obvious that it is of no real interest.
 
  • #7
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Ok, but does it hang at rest or does it hang while rotating along the pole? Given the numbers of this particular problem (one solution is zero, the other solution is impossible because of cosine > +1), my conclusion is that the only way the object can rotate is if it remains "along the side" of the pole. Also, aside from sin(α) = 0, the other solution is cos(α) = g/(ω2*l); to get a "lower" result we would need a bigger denominator, i.e. either a longer string or a faster rotation speed. If I think about it, it makes sense that the only acceptable solution is α = 0, because it basically means (at least how I'm interpreting it) that "with this relatively slow rotation speed (and/or relatively short string) the object can't "lift-off" the center of rotation (i.e. the pole), whereas if it were to gain more speed it would be able to do so. With the given data, however, the circular motion is just rotation around the pole's axis (α = 0)".

Is this a correct interpretation of the solution α = 0? If that's the case, I wouldn't really call it a "trivial" solution. The way I see it, it's one thing to say "cosine > +1 => this is just impossible", but it's another thing to say "either cosine > +1 => impossible, or you can have rotation "along the side" of the pole".

Also, now that I think about it again, I take back what I said earlier about dividing by /sin(α) at this point:
g*sin(α)/cos(α) + ω2*l*sin(α) = 0

After all x2+x = 0 is not the same equation as x+1 = 0. So I would think that mathematically we need to accept the solution sin(α) = 0, and physically this solution has a meaning too (the meaning being: the problem is not an impossible situation per-se; it's just that we don't have enough ω or rope length to have proper circular motion).

On the other hand, I don't know how I would interpret the solution if the cosine was a number between -1 and +1. In that case, the "real" solution would definitely be the one found with cosine, but then the "sine-solution" (= 0) would become difficult to interpret (it's not really a valid alternative I think; if you have enough velocity/string length then it's not like you can "either rotate at X° or 0° ", you would just rotate at X°). Mmh...
 
  • #8
haruspex
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I wouldn't really call it a "trivial" solution. The way I see it, it's one thing to say "cosine > +1 => this is just impossible", but it's another thing to say "either cosine > +1 => impossible, or you can have rotation "along the side" of the pole".
We're not concerned with an object of some dimension spinning on its axis; it is a point mass. So "revolving on a vertical axis" effectively means it is stationary.
Here's a question to consider: what if you were to hold it at some angle from the vertical and give it a slight sideways push, one insufficient to satisfy ω2≥g/l. What would happen?
 

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