Solving the Olympic Ring Puzzle: Calculate Max Mass

AI Thread Summary
The discussion revolves around solving a physics problem involving two identical beads sliding on a circular Olympic ring. Participants suggest using conservation of energy to relate the height and velocity of the beads, as well as analyzing the forces acting on the ring, including gravitational force and normal force. Key equations discussed include the relationship between centripetal force and the normal force exerted by the beads on the ring. The conversation emphasizes the importance of determining the angle θ to find the maximum tension in the rope, which occurs when the tension is zero. The overall goal is to calculate the maximum mass of the ring based on these dynamics.
  • #51
SuperHero said:
So wouldn't the tension have to be the same as the gravity Fg force
No! You have your equation: T = Fg - 2Fncos(θ) = Fg - 2mg(2-3cos(θ))cos(θ). Do you get that? It really bothers me that having come this far you would suggest T = Fg.

Let's look at some values here. When θ = 0, T = Fg+2mg (correct). When θ = π/2, T = Fg (correct). When θ > π/2, T > Fg. But perhaps somewhere between 0 and π/2, T < Fg. How might we find the most interesting value of θ?
 
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  • #52
haruspex said:
No! You have your equation: T = Fg - 2Fncos(θ) = Fg - 2mg(2-3cos(θ))cos(θ). Do you get that? It really bothers me that having come this far you would suggest T = Fg.

Let's look at some values here. When θ = 0, T = Fg+2mg (correct). When θ = π/2, T = Fg (correct). When θ > π/2, T > Fg. But perhaps somewhere between 0 and π/2, T < Fg. How might we find the most interesting value of θ?

would it be that θ<π/2 would make, T < Fg, so couldn't it be 45°
 
  • #53
It could be, or it could be something else. We want to know whether T can reach zero, right? And in particular we want to know the largest Fg for which T can reach zero. So of all the values of 2mg(2-3cos(θ))cos(θ), which one are we looking for?
 
  • #54
haruspex said:
It could be, or it could be something else. We want to know whether T can reach zero, right? And in particular we want to know the largest Fg for which T can reach zero. So of all the values of 2mg(2-3cos(θ))cos(θ), which one are we looking for?

The Fg of the ring? which includes the ring and the beads
 
  • #55
SuperHero said:
The Fg of the ring? which includes the ring and the beads
Not sure what you intended Fg to include, but based on the equations it is only the weight of the ring. That's how it should be. Any contribution from the beads comes in via Fn.

So, to recap, we have T = Fg - 2mg(2-3cos(θ))cos(θ), and we want to know the largest Fg for which T goes as low as 0 for some theta. We can turn that around and ask what's the lowest T for a given Fg? I.e. for a given Fg, what value of theta will make T go lowest?
 
  • #56
haruspex said:
Not sure what you intended Fg to include, but based on the equations it is only the weight of the ring. That's how it should be. Any contribution from the beads comes in via Fn.

So, to recap, we have T = Fg - 2mg(2-3cos(θ))cos(θ), and we want to know the largest Fg for which T goes as low as 0 for some theta. We can turn that around and ask what's the lowest T for a given Fg? I.e. for a given Fg, what value of theta will make T go lowest?

Well the Fg has to be a bit bigger than T, so the value of theta has to be atleast 1 to be allow the tension to be smaller than fg right?
 
  • #57
SuperHero said:
Well the Fg has to be a bit bigger than T, so the value of theta has to be atleast 1 to be allow the tension to be smaller than fg right?
Theta has to be at least 1 what? Radian? How do you work that out? Never mind.
Concentrate on this: given the equation T = Fg - 2mg(2-3cos(θ))cos(θ), where Fg is fixed, how do we find the value of θ that produces the smallest value for T? (Think calculus.)
 
  • #58
I have calculus next semester though
 
  • #59
SuperHero said:
I have calculus next semester though
Oh. I have no idea how you are supposed to solve this problem without differentiation to find a maximum value.
 
  • #60
haruspex said:
So, to recap, we have T = Fg - 2mg(2-3cos(θ))cos(θ), and we want to know the largest Fg for which T goes as low as 0 for some theta.

can't you just complete the square ? :smile:
 
  • #61
tiny-tim said:
can't you just complete the square ? :smile:

Yes, good point. Silly part is I've made exactly the same point in other threads in past!
SuperHero, the expression for T is a quadratic in cos(θ). Try to write it in the form (A cos(θ) + B)2 + C.
 
  • #62
haruspex said:
Yes, good point. Silly part is I've made exactly the same point in other threads in past!
SuperHero, the expression for T is a quadratic in cos(θ). Try to write it in the form (A cos(θ) + B)2 + C.

T = Fg - 2mg(2-3cos(θ))cos(θ)
(A cos(θ) + B)2 + C.

T = -mg(-3cosθ +2)^2 + Fg
like this?
 
  • #63
SuperHero said:
T = Fg - 2mg(2-3cos(θ))cos(θ)
(A cos(θ) + B)2 + C.

T = -mg(-3cosθ +2)^2 + Fg
like this?

The last equation above is not correct. But before worrying anymore about that, consider the following. You are interested in the point where the tension goes to zero. So, let T = 0 in the first equation above and solve for Fg. Write Fg in terms of M (the mass of the ring) and solve for M.
 
  • #64
SuperHero said:
T = Fg - 2mg(2-3cos(θ))cos(θ)
(A cos(θ) + B)2 + C.

T = -mg(-3cosθ +2)^2 + Fg
like this?
No, it has to be the same equation as T = Fg - 2mg(2-3cos(θ))cos(θ), just written out in the form I said. Work backwards. Start with the (A cos(θ) + B)2 + C form and expand it. Match up the powers of cos(θ) (i.e. 0, 1, 2) between that and the original expression. That should tell you what to write for A, B and C.
 
  • #65
TSny said:
The last equation above is not correct. But before worrying anymore about that, consider the following. You are interested in the point where the tension goes to zero. So, let T = 0 in the first equation above and solve for Fg. Write Fg in terms of M (the mass of the ring) and solve for M.

but see i do not know the angle which is really hard
 
  • #66
Right now the angle is a variable that could have different values. If you pick a value of the angle and plug it into the formula after setting T = 0, then the equation will tell you the mass of the ring such that the tension would go to zero at that angle. Don't worry about a particular value of theta right now. Just find out how the mass M of the ring is related to the angle by solving the equation for M while leaving theta unspecified.
 
  • #67
TSny said:
Right now the angle is a variable that could have different values. If you pick a value of the angle and plug it into the formula after setting T = 0, then the equation will tell you the mass of the ring such that the tension would go to zero at that angle. Don't worry about a particular value of theta right now. Just find out how the mass M of the ring is related to the angle by solving the equation for M while leaving theta unspecified.

T = Fg - 2mg(2-3cos(θ))cos(θ)
0 = Fg - 2mg(2-3cos(θ))cos(θ)
Fg = 2mg(2-3cos(θ))cos(θ)
Mg = (4mg - 6cosθmg) cosθ
Mg = 4mgcosθ - 6(cosθ^2)mg
M = (4mgcosθ - 6(cosθ^2)mg)/g
M = 4mcosθ - 6(cosθ^2)m
 
  • #68
SuperHero said:
T = Fg - 2mg(2-3cos(θ))cos(θ)
0 = Fg - 2mg(2-3cos(θ))cos(θ)
Fg = 2mg(2-3cos(θ))cos(θ)
Mg = (4mg - 6cosθmg) cosθ
Mg = 4mgcosθ - 6(cosθ^2)mg
M = (4mgcosθ - 6(cosθ^2)mg)/g
M = 4mcosθ - 6(cosθ^2)m

... which gets us back to the necessary step of writing the RHS in the form A(cosθ + B)2 + C.
 
  • #69
OK, great. Since you let T = 0, the equation you have derived is going to give you the mass of the ring that would make the tension go to zero when the beads reach the angle theta.

You are trying to find the maximum value that M can have and still have the tension go to zero. That means that you need to find the angle theta that would make the right hand side of the equation take on its maximum possible value.

Usually finding the maximum value of a function requires calculus. But, here it turns out we can get it without calculus. For convenience, suppose we let the symbol x stand for cosθ. Since theta is a variable, so is x. What does your equation for M look like in terms of x?
 
  • #70
TSny said:
OK, great. Since you let T = 0, the equation you have derived is going to give you the mass of the ring that would make the tension go to zero when the beads reach the angle theta.

You are trying to find the maximum value that M can have and still have the tension go to zero. That means that you need to find the angle theta that would make the right hand side of the equation take on its maximum possible value.

Usually finding the maximum value of a function requires calculus. But, here it turns out we can get it without calculus. For convenience, suppose we let the symbol x stand for cosθ. Since theta is a variable, so is x. What does your equation for M look like in terms of x?

M = 4mcosθ - 6(cosθ^2)m
M = 4mx - 6(x^2)m
 
  • #71
OK. If you factor out 2m on the right, you have M = 2m(2x-3x2).

If you can find the maximum value that the factor 2x-3x2 can have, that will help you find the maximum value of M. Think of 2x-3x2 as some function y. So, y = 2x-3x2. This is a quadratic equation (or function). If you were to graph y vs. x, what would be the shape of the graph?
 
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  • #72
TSny said:
OK. If you factor out 2m on the right, you have M = 2m(2x-3x2).

If you can find the maximum value that the factor 2x-3x2 can have, that will help you find the maximum value of M. Think of 2x-3x2 as some function y. So, y = 2x-3x2. This is a quadratic equation. If you were to graph y vs. x, what would be the shape of the graph?

so i should use the quadratic formula
y = -3x^2 + 2x
x = -(2) +/- √ ((2^2)-4(-3)(0)) / 2(-3)

x = -(2) + 2 / -6
x = 0
or
x = -(2) - 2 / -6
x = -4/-6
x = 2/3
 
  • #73
Well, what you just did is find the values of x that make y= 0. That would be the two values of x where the graph of y vs. x crosses the x axis. I think you can use that. Did you learn in algebra the shape of the graph of a quadratic function? It's not a straight line...it's not a circle...?
 
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  • #74
TSny said:
Well, what you just did is find the values of x that make y= 0. That would be the two values of x where the graph of y vs. x would cross the x axis. I think you can use that. Did you learn in algebra the shape of the graph of a quadratic function? It's not a straight line...it's not a circle...?

it is a parabola..
 
  • #75
Good. Does it open "up" like a U or "down" like an n?
 
  • #76
SuperHero said:
it is a parabola..

since i found the x = 2/3
can i do cos θ = 2/3
 
  • #77
SuperHero said:
since i found the x = 2/3
can i do cos θ = 2/3

down :)
 
  • #78
SuperHero said:
since i found the x = 2/3
can i do cos θ = 2/3
No, x = 2/3 is one of the places where the parabola crosses the x axis. That does not correspond to the maximum value of the parabola. Likewise, x = 0 is another point on the x-axis where the parabola crosses the x-axis. Does the parabola open up or down?
 
  • #79
SuperHero said:
down :)

Great. You have a parabola that opens down and goes through x-axis at x = 0 and x = 2/3. Can you visualize what the graph must look like, and can you see what value of x corresponds to making y as large as possible?
 
  • #80
TSny said:
Great. You have a parabola that opens down and goes through x-axis at x = 0 and x = 2/3. Can you visualize what the graph must look like, and can you see what value of x corresponds to making y as large as possible?

x = 1/3 since the greatest y is at the maximum
 
  • #81
Fantastic! What is the value of y at x = 1/3? [EDIT: Don't really need to do this, just use x = 1/3 to find the maximum value of M]
 
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  • #82
TSny said:
Fantastic! What is the value of y at x = 1/3?

y = -3x^2 + 2x
y = -3(1/3)^2 + 2(1/3)
y = -3/9 + 2/3
y = 0.3333
 
  • #83
Very good. So, y = 1/3 when x = 1/3. Can you now get the maximum value for the mass of the ring that will still allow the tension to go to zero?
 
  • #84
TSny said:
Very good. So, y = 1/3 when x = 1/3. Can you now get the maximum value for the mass of the ring that will still allow the tension to go to zero?

but isn't y the mass ?
 
  • #85
SuperHero said:
but isn't y the mass ?

Not at all. Remember, M = 2m(2x-3x2) = 2my
 
  • #86
TSny said:
Not at all. Remember, M = 2m(2x-3x2) = 2my

M = 2m(2x-3x2) = 2my
M = 2(30kg)(1/3)
M = 20kg

YES THE MASS IS 20KG! :D
 
  • #87
Great! It would be a good idea to review the whole problem in a day or two to make sure that you understand the whole solution. Good work.
 
  • #88
TSny said:
Great! It would be a good idea to review the whole problem in a day or two to make sure that you understand the whole solution. Good work.

thx for your help,
i will look over.
 

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