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I'm working on the exact same problem described in this 3 year old thread:
https://www.physicsforums.com/showthread.php?t=433104
except the length of my rope is 3.8 m and my velocity is 5.6 m/s.
A tetherball is on a rope 3.8 m long and moves in a circle with velocity 5.6 m/s. Solve for the angle the rope makes with the pole.
I follow the explanation in the old thread up to this point.
[tex] \begin{array}{l}<br /> \left( {\frac{{mv^2 }}{{r\sin \theta }}} \right)\cos \theta = mg \\ <br /> \\ <br /> \frac{{\cos \theta }}{{\sin ^2 \theta }} = \frac{{rg}}{{v^2 }} \\ <br /> \end{array}[/tex]
When I rearrange the top expression into the bottom expression, I don't get sin^2. I just get sin. It becomes very simple to solve for θ at that point. How did sin end up getting squared?
My full attempt is this:
[tex] \begin{array}{l}<br /> T\cos \theta = mg \\ <br /> T\sin \theta = \frac{{mv^2 }}{r}\,\,\,\, \Rightarrow \,\,\,\,T = \frac{{mv^2 }}{{r\sin \theta }} \\ <br /> \frac{{mv^2 }}{{r\sin \theta }}\cos \theta = mg \\ <br /> \frac{{\sin \theta }}{{\cos \theta }} = \frac{{v^2 }}{{gr}} \\ <br /> \tan \theta = \frac{{v^2 }}{{gr}}\,\,\,\, \Rightarrow \,\,\,\,\theta = \tan ^{ - 1} \frac{{\left( {5.6\,{\rm{m/s}}} \right)^2 }}{{\left( {9.8\,{\rm{m/s}}^{\rm{2}} } \right)\left( {3.8\,{\rm{m}}} \right)}} = 40^\circ \\ <br /> \end{array}[/tex]
The back of the book says 48.4 degrees.
When I continue the problem with the sin^2, I get the correct answer. The logic makes sense. But where did that extra sin come from?
https://www.physicsforums.com/showthread.php?t=433104
except the length of my rope is 3.8 m and my velocity is 5.6 m/s.
A tetherball is on a rope 3.8 m long and moves in a circle with velocity 5.6 m/s. Solve for the angle the rope makes with the pole.
I follow the explanation in the old thread up to this point.
[tex] \begin{array}{l}<br /> \left( {\frac{{mv^2 }}{{r\sin \theta }}} \right)\cos \theta = mg \\ <br /> \\ <br /> \frac{{\cos \theta }}{{\sin ^2 \theta }} = \frac{{rg}}{{v^2 }} \\ <br /> \end{array}[/tex]
When I rearrange the top expression into the bottom expression, I don't get sin^2. I just get sin. It becomes very simple to solve for θ at that point. How did sin end up getting squared?
My full attempt is this:
[tex] \begin{array}{l}<br /> T\cos \theta = mg \\ <br /> T\sin \theta = \frac{{mv^2 }}{r}\,\,\,\, \Rightarrow \,\,\,\,T = \frac{{mv^2 }}{{r\sin \theta }} \\ <br /> \frac{{mv^2 }}{{r\sin \theta }}\cos \theta = mg \\ <br /> \frac{{\sin \theta }}{{\cos \theta }} = \frac{{v^2 }}{{gr}} \\ <br /> \tan \theta = \frac{{v^2 }}{{gr}}\,\,\,\, \Rightarrow \,\,\,\,\theta = \tan ^{ - 1} \frac{{\left( {5.6\,{\rm{m/s}}} \right)^2 }}{{\left( {9.8\,{\rm{m/s}}^{\rm{2}} } \right)\left( {3.8\,{\rm{m}}} \right)}} = 40^\circ \\ <br /> \end{array}[/tex]
The back of the book says 48.4 degrees.
When I continue the problem with the sin^2, I get the correct answer. The logic makes sense. But where did that extra sin come from?