Boundry conditions on a string with a hoop at one end

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The discussion centers on understanding the tension in a string with a hoop at one end, particularly how Tsinθ relates to -T∂y/∂x. The small angle approximation simplifies sinθ to θ, leading to the conclusion that sinθ is approximately equal to tan(θ), which is the ratio of vertical to horizontal tension. This approximation is crucial for analyzing the forces acting on the hoop, which primarily moves vertically. The conversation emphasizes the importance of recognizing the instantaneous slope of the string in this context. Overall, the small angle approximation aids in clarifying the relationship between tension and the angle of the string.
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Homework Statement



problem 4.4

The question:
http://ocw.mit.edu/courses/physics/...ions-and-waves-fall-2004/assignments/ps4a.pdf

The solution:
http://ocw.mit.edu/courses/physics/...ons-and-waves-fall-2004/assignments/sol4a.pdf




Homework Equations





The Attempt at a Solution


For part a I am having trouble understanding why Tsinθ becomes -T∂y/∂x

If we were dealing with small angles would it not be the case that

sinθ ≈θ


and so we would have
Tsinθ≈Tθ
 
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\frac{\partial y}{\partial x} is essentially the ratio of vertical tension to horizontal tension, much like tan(\theta). It's a little confusing because obviously you're thinking about the hoop, which only moves vertically, but in this context you can think of it as representing the direction the hoop would go if it suddenly flew off of the rod, at which point the only force acting on the hoop would be the tension at the contact point (ignoring gravity), thus it would fly in that direction. Another way to think of it is simply as the instantaneous "slope" of the string.

The small angle approximation comes in because for small angles tan(\theta) \simeq sin(\theta). (or equivalently, T_{x} \simeq T_{tot}.)

Thus we can put it all together: sin(\theta) \simeq tan(\theta) = \frac{\partial y}{\partial x}
 
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Thank you very much bossman27!
 
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