- #1

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**Pendulum problem -- there's got to be an easier way!**

*What is the amplitude of a simple pendulum of length L = 0.60m if the bob has a speed v = 0.37 m/s when passing through the equilibrium position?*

I don't see why this is any different than figuring out the height a roller coaster will rise to, so... Work Energy theorum to get the max height of the pendulum. Then translate it into degrees.

[tex]

\Delta K=-\Delta U

\]

\[

\Delta K=\frac{1}{2}mv^2,

\quad

\Delta U=mg(y_f -y_i )

\]

\[

-mg(y_f -y_i )=\frac{1}{2}mv^2,

\quad

y_i =0

\]

\[

-mg(y_f )=\frac{1}{2}mv^2

\]

\[

(y_f )=\frac{\frac{1}{2}\rlap{--} {m}v^2}{-\rlap{--} {m}g}

\quad

\Rightarrow

\quad

y_f =\frac{\frac{1}{2}v^2}{-g}

\]

\[

h=y_f ,

\quad

h=\ell (1-\cos \theta )

\]

\[

\frac{\frac{1}{2}v^2}{-g}=\ell (1-\cos \theta )

\quad

\Rightarrow

\quad

\frac{\frac{1}{2}v^2}{-g}=\ell -\ell \cos \theta

\quad

\Rightarrow

\quad

\frac{\frac{1}{2}v^2}{-g}-\ell =-\ell \cos \theta

\quad

\Rightarrow

\]

\[

\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta

\quad

\Rightarrow

\quad

\theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell }

\right)}{-\ell }} \right)

\]

\[

\theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}\left( {0.37m/s}

\right)^2}{-9.8m/s^2}-0.6m} \right)}{-0.6m}} \right)

\quad

\Rightarrow

\quad

\theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}0.37^2\rlap{--}

{m}^{\rlap{--} {2}}/\rlap{--} {s}^{\rlap{--} {2}}}{-9.8\rlap{--}

{m}/\rlap{--} {s}^{\rlap{--} {2}}}-0.6\rlap{--} {m}} \right)}{-0.6\rlap{--}

{m}}} \right)

[/tex]

[tex]

\theta =8.75 deg

[/tex]

Same answer as the back of the book. But there's got to be an easier way. The book gives no examples of this type of problem.