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

[tony873004]

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.

<br /> \Delta K=-\Delta U<br /> \]<br /> \[<br /> \Delta K=\frac{1}{2}mv^2,<br /> \quad<br /> \Delta U=mg(y_f -y_i )<br /> \]<br /> \[<br /> -mg(y_f -y_i )=\frac{1}{2}mv^2,<br /> \quad<br /> y_i =0<br /> \]<br /> \[<br /> -mg(y_f )=\frac{1}{2}mv^2<br /> \]<br /> \[<br /> (y_f )=\frac{\frac{1}{2}\rlap{--} {m}v^2}{-\rlap{--} {m}g}<br /> \quad<br /> \Rightarrow <br /> \quad<br /> y_f =\frac{\frac{1}{2}v^2}{-g}<br /> \]<br /> \[<br /> h=y_f ,<br /> \quad<br /> h=\ell (1-\cos \theta )<br /> \]<br /> \[<br /> \frac{\frac{1}{2}v^2}{-g}=\ell (1-\cos \theta )<br /> \quad<br /> \Rightarrow <br /> \quad<br /> \frac{\frac{1}{2}v^2}{-g}=\ell -\ell \cos \theta <br /> \quad<br /> \Rightarrow <br /> \quad<br /> \frac{\frac{1}{2}v^2}{-g}-\ell =-\ell \cos \theta <br /> \quad<br /> \Rightarrow <br /> \]<br /> \[<br /> \frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta <br /> \quad<br /> \Rightarrow <br /> \quad<br /> \theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } <br /> \right)}{-\ell }} \right)<br /> \]<br /> \[<br /> \theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}\left( {0.37m/s} <br /> \right)^2}{-9.8m/s^2}-0.6m} \right)}{-0.6m}} \right)<br /> \quad<br /> \Rightarrow <br /> \quad<br /> \theta =\cos ^{-1}\left( {\frac{\left( {\frac{\frac{1}{2}0.37^2\rlap{--} <br /> {m}^{\rlap{--} {2}}/\rlap{--} {s}^{\rlap{--} {2}}}{-9.8\rlap{--} <br /> {m}/\rlap{--} {s}^{\rlap{--} {2}}}-0.6\rlap{--} {m}} \right)}{-0.6\rlap{--} <br /> {m}}} \right)<br />

<br /> <br /> \theta =8.75 deg<br />

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.

 

[Justin Lazear]

Good job on solving the problem, but this problem isn't difficult, and the algebra is even pretty clean. You've made it a bit uglier than it needs be by not simplifying your expressions completely. Notice that your final expression can be written

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

Be happy that the problem is this straight-forward and get used to algebra. It only gets worse.

--J

 

[tony873004]

Be happy that the problem is this straight-forward and get used to algebra. It only gets worse.
--J

Thanks for the encouraging words

if the -L didn't exist in the numerator, then
I could see shoving the 1/2 downstairs as a 2 and
I could see shoving the /-g downstairs as simply g, and neutralizing its minus sign with L's minus sign.

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

giving me a modified formula of:

\cos{\theta} = \frac{v^2-l}{2gl}

But your forumla
\cos{\theta} = 1 - \frac{v^2}{2gl}

got rid of the -L in the numerator and introduced a "1-". Plugging in numbers, your forumla works, my original formula works, but my modified formula doesn't.

Also, I would have thought there would be some pendulum formula to take care of this whole thing in 2 lines. Is this the right way to do this problem??

 

[Justin Lazear]

giving me a modified formula of:

\cos{\theta} = \frac{v^2-l}{2gl}

Well, that's interesting. It looks like we both messed up the algebra! According to what you've written, the formula should simplify to

\cos{\theta} = 1 + \frac{v^2}{2gl}

\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }= \frac{\frac{\frac{1}{2}v^2}{-g}}{-l}-\frac{l}{-l} = \frac{v^2}{2gl} + 1

Your -g should have actually been a +g. You got lucky with your original formula because cosine is even. A happy accident where you screw up, but your profs don't mark you off.

--J

Geh... LaTeX didn't work out so well. Too many nested \frac{}{}s. Gimme a sec.

All better now.

 

[physics4ever]

How did you modify the formula without taking the L.C.M.?
\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta
It should be
\cos{\theta} =\frac{v^2 + 2gl}{2gl}
which becomes
\cos{\theta} =\frac{v^2}{2gl}+ 1

 

[Justin Lazear]

So we're agreed.

Oh, and I just distributed the denominator to each term in the numerator right away.

--J

 

[James R]

Here's an alternative approach, though it probably doesn't involve any less work. The motion of a simple pendulum is described by:

\theta(t) = \theta_0 \sin (\sqrt{\frac{l}{g}}t)

where \theta_0 is the (angular) amplitude.

The angular speed at any time is given by the time derivative of this equation, and from that you can work out an expression for the linear speed at any time. Use that formula to calculate the amplitude in your problem.

 

[Justin Lazear]

g/l, not l/g. The argument of cosine must be unitless.

And that equation assumes that \theta is sufficiently small that \sin{\theta} \approx \theta, which isn't necessary for this problem. The original solution was better.

--J

Edit: Been too long since I last used TeX...