(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A mass m is attached to one end of a light rod of length [itex]l[/itex]. The other end of the rod is pivoted so that the rod can swing in a plane. The pivot rotates in the same plane at angular velocity [itex]\omega[/itex] in a circle of radius R. Show that this "pendulum" behaves like a simple pendulum in a gravitational field [itex]g=\omega ^2 R[/itex] for all values of l and amplitudes of oscillations.

2. Relevant equations

[itex]L=T-V[/itex].

I used the trigonometric formula [itex]\sin \theta _1 \sin \theta _2 + \cos \theta _1 \cos \theta _2 =\cos (\theta _1 - \theta _2)[/itex] for a part.

3. The attempt at a solution

I place my coordinate system (Cartesian) in the center of the circle that the pivot describes. [itex]\vec r = x \hat i +y \hat j=(R \cos \theta _1 +l \cos \theta _2 ) \hat i +(R \sin \theta _1 + l \sin \theta _2 ) \hat j[/itex].

Since [itex]T=\frac{m}{2} ( \dot { \vec r} )^2[/itex], I calculated it to be worth [itex]\frac{m}{2}[\omega ^2 R^2+ \dot \theta _2 ^2 l^2+2Rl\omega \dot \theta _2 \cos (\theta _1 - \theta _2)]=L[/itex].

I've got the Euler-Lagrange equation to be (for [itex]\theta _2[/itex]) [itex]l^2 \ddot \theta _2 -Rl\omega \sin (\theta _1 - \theta _2 )(\omega - \dot \theta _2 )-Rl \omega \dot \theta _2 \sin (\theta _1 - \theta _2 )=0[/itex]. I don't think I can simplify this anymore.

However the answer should be [itex]ml^2 \ddot \theta =-mR\omega ^2 l \sin \theta[/itex] where their [itex]\theta[/itex] is worth my [itex]\theta _2[/itex].

I don't know what I did wrong.

The given answer starts with [itex]x=R\cos ( \omega t )+l \cos (\omega t + \theta )[/itex]. I don't really understand why the argument of the second cos term is this way though. Also I don't see why my "method" failed.

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# Homework Help: Lagrangian, weird pendulum

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