Solving the Olympic Ring Puzzle: Calculate Max Mass

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SUMMARY

The discussion focuses on solving the Olympic Ring Puzzle, which involves calculating the maximum mass of a large circular ring supporting two identical beads, each with a mass of 30 kg. Participants suggest using conservation of energy to relate angular speed and height, and emphasize the importance of understanding the forces acting on the system, including gravitational force and normal force. The key equations discussed include mgh = 1/2mv² and mgcosθ + Fn = mv²/R, which are essential for determining the conditions under which the tension in the rope becomes zero.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with centripetal force and normal force concepts
  • Knowledge of angular motion and its equations
  • Ability to analyze forces in a dynamic system
NEXT STEPS
  • Study the derivation of centripetal acceleration in circular motion
  • Learn how to apply conservation of momentum in dynamic systems
  • Explore the relationship between angular displacement and height in circular motion
  • Investigate the effects of varying mass and angle on tension in a system
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in dynamics and force analysis in circular motion will benefit from this discussion.

  • #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.
 
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  • #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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