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Solving the Olympic Ring Puzzle: Calculate Max Mass
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SuperHero
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mgcosθ + Fn = mv2/R
mgh = (1/2)mv2
2mgh = mv2
mgcosθ + Fn = 2mgh/R
but for the height, i have been searching in my notes and even online, maybe its just a remember thing, but i can't seem to do anything that would be equal to h.
mgh = (1/2)mv2
2mgh = mv2
mgcosθ + Fn = 2mgh/R
but for the height, i have been searching in my notes and even online, maybe its just a remember thing, but i can't seem to do anything that would be equal to h.
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What height above the circle's centre does the bead start? How high is it above the circle's centre when it has moved angle theta around the circle? h is the difference of the two.
SuperHero
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haruspex said:
h = R - Rcosθ
Since R represents the radius
and Rcosθ is the side,
if we subtract the two the h is that right?
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Yes! Now plug that into find Fn. Next, determine the forces on the ring.SuperHero said:
SuperHero
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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!
mgcosθ + Fn = 2mg (R -Rcosθ) / R
Rmgcosθ + RFn = 2mgR - 2mgRcosθ
RFn = 2mgR - 3Rmgcosθ
RFn = (R) 2mg - 3mgcosθ
Fn = 2mg - 3mgcosθ
tada!
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SuperHero said:
Good. So the forces on the ring are? And when you have that, at what angle theta is the tension minimised?
SuperHero
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haruspex said:
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?
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SuperHero said:
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
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Alright so for the vertical 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
the two 2(Fn) towards the middle and the Fg as well as the tension so.
2(Fn = 2mg - 3mgcosθ) + T = Fg
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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
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TSny said:
ah, i see.
but doesn't me expression still go with your statement?
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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.)SuperHero said:
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
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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 vertical force. the equation would become T = Fg
ok so if the Fn is not a vertical force. the equation would become T = Fg
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No, Fn is neither vertical nor horizontal. It has a vertical component. What is the magnitude of that vertical component?SuperHero said:
SuperHero
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haruspex said:
it if Fncosθ
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Yes. So now write your equation with T, Fn, θ and Fg again.SuperHero said:
SuperHero
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haruspex said:
2(Fncosθ) + T = Fg
like this?
since you pointed out this equation was correct earlier.
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Yes, but substituting what you already worked out for Fn.SuperHero said:
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
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haruspex said:
So wouldn't the tension have to be the same as the gravity Fg force
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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.SuperHero said:
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
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haruspex said:
would it be that θ<π/2 would make, T < Fg, so couldn't it be 45°
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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
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haruspex said:
The Fg of the ring? which includes the ring and the beads
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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.SuperHero 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. 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
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haruspex 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?
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Theta has to be at least 1 what? Radian? How do you work that out? Never mind.SuperHero said:
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
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I have calculus next semester though
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Oh. I have no idea how you are supposed to solve this problem without differentiation to find a maximum value.SuperHero said:
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