Solving the Olympic Ring Puzzle: Calculate Max Mass

[TSny]

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?

 

[SuperHero]

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

 

[TSny]

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...?

 

[SuperHero]

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..

 

[TSny]

Good. Does it open "up" like a U or "down" like an n?

 

[SuperHero]

it is a parabola..

since i found the x = 2/3
can i do cos θ = 2/3

 

[SuperHero]

since i found the x = 2/3
can i do cos θ = 2/3

down :)

 

[TSny]

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?

 

[TSny]

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?

 

[SuperHero]

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

 

[TSny]

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]

 

[SuperHero]

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

 

[TSny]

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?

 

[SuperHero]

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 ?

 

[TSny]

but isn't y the mass ?

Not at all. Remember, M = 2m(2x-3x²) = 2my

 

[SuperHero]

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

 

[TSny]

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.

 

[SuperHero]

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.