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Geometry of problem

  • Thread starter Xyius
  • Start date
  • #1
508
4
This is a Lagrangian problem, I am posting it here in introductory physics because what I need help with isn't in Lagrangian mechanics, but rather geometry.

http://img97.imageshack.us/img97/7504/what3.png [Broken]

I am confused as how they got those relations for x and y. I have tried to make sense out of it but cannot figure it out. I am sure it is something really simple!
 
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Answers and Replies

  • #2
30
0
Let's start with y. You can probably understand where the (r+b)cos(θ) comes from I hope. Now imagine that you roll the cube around the sphere by an angle θ. The length of the line CB is then what? That's right! It's rθ. So obviously this adds a height rθsin(θ). Now you have

[tex] y = (r+b)cos(θ) + rθsin(θ) [/tex]
 
  • #3
473
13
Hm. I agree with Xyius; I don't think that, as stated, the formula necessarily holds. Consider a much smaller cube in the same place, at the same angle.

The unstated assumption is that the cube was originally placed square and centered on top of the cylinder and rolled to its current location, but that hasn't been specified.
 
  • #4
30
0
Hm. I agree with Xyius; I don't think that, as stated, the formula necessarily holds. Consider a much smaller cube in the same place, at the same angle.
A much smaller cube would not be able to get to as large of an angle, because the cube must roll without slipping. The formula is valid.
 
  • #5
508
4
Ah! I understand now! Thank you very much :]
 

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