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

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In summary, The problem involves finding the amplitude of a simple pendulum given its length and the speed of the bob at the equilibrium position. The solution involves using the Work-Energy Theorem to calculate the maximum height of the pendulum and then converting it into degrees. While there may be an easier way to solve this problem, the given solution is algebraically straightforward and gives the same result as the book. An alternative approach using the equation for the motion of a simple pendulum is also mentioned, but it may not necessarily involve less work.
  • #1
tony873004
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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.
 
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  • #2
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

[tex]\cos{\theta} = 1 - \frac{v^2}{2gl}[/tex]

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

--J
 
  • #3
Justin Lazear said:
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.

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

giving me a modified formula of:

[tex]\cos{\theta} = \frac{v^2-l}{2gl}[/tex]

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

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??
 
  • #4
tony873004 said:
giving me a modified formula of:

[tex]\cos{\theta} = \frac{v^2-l}{2gl}[/tex]

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

[tex]\cos{\theta} = 1 + \frac{v^2}{2gl}[/tex]

[tex]\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[/tex]

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. :biggrin:

--J

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

All better now.
 
Last edited:
  • #5
How did you modify the formula without taking the L.C.M.?
[tex]\frac{\left( {\frac{\frac{1}{2}v^2}{-g}-\ell } \right)}{-\ell }=\cos \theta [/tex]
It should be
[tex]\cos{\theta} =\frac{v^2 + 2gl}{2gl}[/tex]
which becomes
[tex]\cos{\theta} =\frac{v^2}{2gl}+ 1[/tex]
 
  • #6
So we're agreed.

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

--J
 
Last edited:
  • #7
Here's an alternative approach, though it probably doesn't involve any less work. The motion of a simple pendulum is described by:

[tex]\theta(t) = \theta_0 \sin (\sqrt{\frac{l}{g}}t)[/tex]

where [itex]\theta_0[/itex] 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.
 
  • #8
g/l, not l/g. The argument of cosine must be unitless.

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

--J

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

What is a pendulum problem and why is it considered difficult?

A pendulum problem refers to a physics problem that involves analyzing the motion of a pendulum, which is a weight suspended from a fixed point that swings back and forth. It is considered difficult because it requires knowledge of several concepts, including gravity, energy, and motion, and involves complex calculations.

What are the main factors that affect the motion of a pendulum?

The main factors that affect the motion of a pendulum are the length of the pendulum, the mass of the weight, and the amplitude (or angle of swing) of the pendulum. These factors determine the period (time for one full swing) and frequency (number of swings per unit of time) of the pendulum's motion.

Is there an easier way to solve pendulum problems?

Yes, there are easier ways to solve pendulum problems. One approach is to use simple harmonic motion equations, which describe the motion of an object back and forth in a regular pattern. Another approach is to use conservation of energy principles, which state that the total energy of a system remains constant. Both of these methods can simplify the calculations involved in solving pendulum problems.

What are some real-life applications of pendulum problems?

Pendulum problems have several real-life applications, including timekeeping devices such as grandfather clocks and metronomes, as well as seismometers used to measure earthquakes. They are also used in engineering to study the motion of structures and in sports to analyze the trajectory of a ball.

Can a pendulum ever reach a point where it stops swinging?

No, according to the law of conservation of energy, a pendulum will never stop swinging unless an external force is applied. This is because the potential energy of the pendulum is constantly being converted to kinetic energy, and vice versa, keeping the pendulum in motion.

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