- #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
\]
\[
\Delta K=\frac{1}{2}mv^2,
\quad
\Delta U=mg(y_f -y_i )
\]
\[
-mg(y_f -y_i )=\frac{1}{2}mv^2,
\quad
y_i =0
\]
\[
-mg(y_f )=\frac{1}{2}mv^2
\]
\[
(y_f )=\frac{\frac{1}{2}\rlap{--} {m}v^2}{-\rlap{--} {m}g}
\quad
\Rightarrow
\quad
y_f =\frac{\frac{1}{2}v^2}{-g}
\]
\[
h=y_f ,
\quad
h=\ell (1-\cos \theta )
\]
\[
\frac{\frac{1}{2}v^2}{-g}=\ell (1-\cos \theta )
\quad
\Rightarrow
\quad
\frac{\frac{1}{2}v^2}{-g}=\ell -\ell \cos \theta
\quad
\Rightarrow
\quad
\frac{\frac{1}{2}v^2}{-g}-\ell =-\ell \cos \theta
\quad
\Rightarrow
\]
\[
\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta
\quad
\Rightarrow
\quad
\theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell }
\right)}{-\ell }} \right)
\]
\[
\theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}\left( {0.37m/s}
\right)^2}{-9.8m/s^2}-0.6m} \right)}{-0.6m}} \right)
\quad
\Rightarrow
\quad
\theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}0.37^2\rlap{--}
{m}^{\rlap{--} {2}}/\rlap{--} {s}^{\rlap{--} {2}}}{-9.8\rlap{--}
{m}/\rlap{--} {s}^{\rlap{--} {2}}}-0.6\rlap{--} {m}} \right)}{-0.6\rlap{--}
{m}}} \right)
[/tex]
[tex]
\theta =8.75 deg
[/tex]
Same answer as the back of the book. But there's got to be an easier way. The book gives no examples of this type of problem.
What is the amplitude of a simple pendulum of length L = 0.60m if the bob has a speed v = 0.37 m/s when passing through the equilibrium position?
I don't see why this is any different than figuring out the height a roller coaster will rise to, so... Work Energy theorum to get the max height of the pendulum. Then translate it into degrees.
[tex]
\Delta K=-\Delta U
\]
\[
\Delta K=\frac{1}{2}mv^2,
\quad
\Delta U=mg(y_f -y_i )
\]
\[
-mg(y_f -y_i )=\frac{1}{2}mv^2,
\quad
y_i =0
\]
\[
-mg(y_f )=\frac{1}{2}mv^2
\]
\[
(y_f )=\frac{\frac{1}{2}\rlap{--} {m}v^2}{-\rlap{--} {m}g}
\quad
\Rightarrow
\quad
y_f =\frac{\frac{1}{2}v^2}{-g}
\]
\[
h=y_f ,
\quad
h=\ell (1-\cos \theta )
\]
\[
\frac{\frac{1}{2}v^2}{-g}=\ell (1-\cos \theta )
\quad
\Rightarrow
\quad
\frac{\frac{1}{2}v^2}{-g}=\ell -\ell \cos \theta
\quad
\Rightarrow
\quad
\frac{\frac{1}{2}v^2}{-g}-\ell =-\ell \cos \theta
\quad
\Rightarrow
\]
\[
\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta
\quad
\Rightarrow
\quad
\theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell }
\right)}{-\ell }} \right)
\]
\[
\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.