Solving the Olympic Ring Puzzle: Calculate Max Mass

[SuperHero]

So this is what i get:

mgcosθ + Fn = 2mg (R -Rcosθ) / R
Rmgcosθ + RFn = 2mgR - 2mgRcosθ
RFn = 2mgR - 3Rmgcosθ
RFn = (R) 2mg - 3mgcosθ
Fn = 2mg - 3mgcosθ

tada!

 

[haruspex]

So this is what i get:

mgcosθ + Fn = 2mg (R -Rcosθ) / R
Rmgcosθ + RFn = 2mgR - 2mgRcosθ
RFn = 2mgR - 3Rmgcosθ
RFn = (R) 2mg - 3mgcosθ
Fn = 2mg - 3mgcosθ

tada!

Good. So the forces on the ring are? And when you have that, at what angle theta is the tension minimised?

 

[SuperHero]

Good. So the forces on the ring are? And when you have that, at what angle theta is the tension minimised?

So the forces on the big ring are the 2 normal forces by the beads and the force of gravity. For the θ, r u supposed to know or calculate?

 

[haruspex]

So the forces on the big ring are the 2 normal forces by the beads and the force of gravity. For the θ, r u supposed to know or calculate?

We're not ready to determine theta yet. Just carry on using it as an unknown.
You missed out the tension. We're not ready to set that zero yet.
Write the equation for the vertical components of the forces.

 

[SuperHero]

Alright so for the verticle component of the ring we have

the two 2(Fn) towards the middle and the Fg as well as the tension so.

2(Fn = 2mg - 3mgcosθ) + T = Fg

 

[TSny]

Remember that the forces acting on the ring add together as vectors. So, you need to consider the vertical components separately from the horizontal components.

 

[SuperHero]

Remember that the forces acting on the ring add together as vectors. So, you need to consider the vertical components separately from the horizontal components.

ah, i see.
but doesn't me expression still go with your statement?

 

[haruspex]

Alright so for the verticle component of the ring we have

the two 2(Fn) towards the middle

Careful with the direction of Fn. As a force on the bead, you took the positive direction to be towards the centre of the circle. We're now dealing with the reaction force of the bead on the ring, so the direction is reversed. (But I see you have done that in the equation below.)

and the Fg as well as the tension so.

2(Fn = 2mg - 3mgcosθ) + T = Fg

We only want the vertical component of each force. Fn is not vertical.
When you have corrected that, the next thing is to find out for what values of Fg the tension, T, becomes 0 at some point during the movement of the beads.
How do you think we might do that?

 

[SuperHero]

would that mean that the fg should be less than T, in order for it to go lose or become slack?
ok so if the Fn is not a verticle force. the equation would become T = Fg

 

[haruspex]

ok so if the Fn is not a verticle force. the equation would become T = Fg

No, Fn is neither vertical nor horizontal. It has a vertical component. What is the magnitude of that vertical component?

 

[SuperHero]

No, Fn is neither vertical nor horizontal. It has a vertical component. What is the magnitude of that vertical component?

it if Fncosθ

 

[haruspex]

it if Fncosθ

Yes. So now write your equation with T, Fn, θ and Fg again.

 

[SuperHero]

Yes. So now write your equation with T, Fn, θ and Fg again.

2(Fncosθ) + T = Fg

like this?
since you pointed out this equation was correct earlier.

 

[haruspex]

2(Fncosθ) + T = Fg

Yes, but substituting what you already worked out for Fn.
The next thing is to find out for what values of Fg the tension, T, becomes 0 at some point during the movement of the beads.
How do you think we might do that?

 

[SuperHero]

Yes, but substituting what you already worked out for Fn.
The next thing is to find out for what values of Fg the tension, T, becomes 0 at some point during the movement of the beads.
How do you think we might do that?

So wouldn't the tension have to be the same as the gravity Fg force

 

[haruspex]

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?

 

[SuperHero]

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°

 

[haruspex]

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?

 

[SuperHero]

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

 

[haruspex]

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?

 

[SuperHero]

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?

 

[haruspex]

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.)

 

[SuperHero]

I have calculus next semester though

 

[haruspex]

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.

 

[tiny-tim]

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 ?

 

[haruspex]

can't you just complete the square ?

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.

 

[SuperHero]

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?

 

[TSny]

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.

 

[haruspex]

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.

 

[SuperHero]

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

 

[TSny]

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.

 

[SuperHero]

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

 

[haruspex]

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.

 

[TSny]

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?

 

[SuperHero]

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