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I've got the equation [tex] T = mg(3-2cos\theta)[/tex] but when I solve that for [tex] \theta[/tex] with T = 48.9, it says it's wrong. any thoughts?

- Thread starter finlejb
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- #1

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I've got the equation [tex] T = mg(3-2cos\theta)[/tex] but when I solve that for [tex] \theta[/tex] with T = 48.9, it says it's wrong. any thoughts?

- #2

Doc Al

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Show how you arrived at that equation.finlejb said:I've got the equation [tex] T = mg(3-2cos\theta)[/tex]

- #3

Andrew Mason

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What is the acceleration of the ball as it moves in this path? What does it depend on? What provides this acceleration? Hint: it is not just gravity that provides the acceleration here.finlejb said:

I've got the equation [tex] T = mg(3-2cos\theta)[/tex] but when I solve that for [tex] \theta[/tex] with T = 48.9, it says it's wrong. any thoughts?

AM

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Andrew, I don't see any other source of acceleration. Explain.

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Andrew Mason

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The ball is falling (gravity) and it is moving in a circular arc.Andrew, I don't see any other source of acceleration. Explain.

AM

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Doc Al

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The ball is attached to a fishing line.Andrew, I don't see any other source of acceleration. Explain.

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