Tetherball Physics Problem: Understanding the Flaw in the dWork Formula

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Discussion Overview

The discussion revolves around a physics problem involving a tetherball apparatus, specifically focusing on the formula for work done as the ball wraps around a pole. Participants explore the implications of the formula and the dynamics of the ball's motion, including its height and velocity changes during the process.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of the formula dWork = -(mv^2/lsinX)*dxsinX + mgdlsinX, seeking clarification on its flaws.
  • Another participant suggests that there is a missing "l" in the first term of the formula, proposing a revised version.
  • A subsequent post corrects the formula, indicating a typo and providing an alternative expression for work done.
  • One participant expresses confusion about the problem statement and requests further elucidation.
  • A later post clarifies the scenario, stating that as the ball wraps around the pole, its height increases due to upward acceleration.
  • Another participant challenges the assertion that the height of the ball increases, suggesting it would decrease instead.
  • One participant argues that the tension in the string must counteract gravity, implying that a shortening string would lead to an increase in height.
  • Another participant inquires about the assumptions behind the original formula for work done, seeking insight into its derivation.

Areas of Agreement / Disagreement

Participants express differing views on the dynamics of the ball's motion, particularly regarding whether the height increases or decreases as it wraps around the pole. There is no consensus on the correctness of the original formula or the interpretations of the motion involved.

Contextual Notes

Participants have not fully resolved the assumptions underlying the problem, including the effects of tension and gravity on the ball's motion and the implications of the formula for work done.

doriang101
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i'm driving myself insane here...ok, you have a ball wrapping around a pole (was initialy spinning with a horizontal velocity). Ok, Can someone tell me what is wrong with the formula dWork = -(mv^2/lsinX)*dxsinX + mgdlsinX where X is the angle the rope makes with the pole, v is current velocity and l is length of string? I am making the assumption that the pole is small so the velocity is assumed to be horizontal.

I don't want a solution to the problem, just a reason why that formula is wrong.

Thanks
 
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I think there's an "l" missing in the first term : should be -(mv^2/lsinX)*lsinX*dX
 
Typo

typo

should read:
dW = -(mv^2/lsinX)*sinXdl + mgdlcosX

sorry
 
I'm confused as to the question. Can you state the problem?
 
Elucidation of the problem

Ok, you have a tetherball apparatus. Initially (before the rope wraps around the pole), the ball is moving with a horizontal velocity of Vo m/s (causing an angle Xi with the pole). Now, as the ball begins to wrap around, the height of the ball will increase (since it is accelerating upwards). I need to find what the final height of the ball is versus the initial height when the ball finishes wrapping around the pole. So far I have y = Integral(cosxdl) and tanx^2=v^2/(lgsinX) by F=ma. I tried dW = -(mv^2/lsinX)*sinXdl - mgcosXdl
(ie. dW = -Tx*dr - Ty*dy). But this formula isn't producing the right results. I then set dW=d(1/2*mv^2) = mvdv and so I have vdv=-(v^2/l)*dl - gcosXdl.
This however doesn't seem to work.
 
how will the height of the ball increase, wouldent it decrease??
 
Ok, you have a tetherball apparatus. Initially (before the rope wraps around the pole), the ball is moving with a horizontal velocity of Vo m/s (causing an angle Xi with the pole). Now, as the ball begins to wrap around, the height of the ball will increase (since it is accelerating upwards).

When you say horizontal velocity, I assume you are referring to the ball's tangential velocity relative to the pole. And maybe I'm off base here (or your description of the problem is), but I'm pretty sure that there is no vertical acceleration of the ball since all its acceleration takes place via the tension force in the centripetal direction, and the only vertical force acting on the ball countering the force of gravity is the vertical component of of the tension force, making the total force acting vertically zero, thereby making the acceleration vertically zero.
 
Last edited:
tension!

clearly the tension in the string would be >= to the force of gravity. Hence if the length of the string shortens, it will move up! I don't want to argue about that pointer further, because I'm sure about that at least.
 
Read my appended post above. And I was just curious of how you obtained
dWork = -(mv^2/lsinX)*dxsinX + mgdlsinX ? I don't need a detailed derivation, just your assumptions regaurding the problem.
 

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