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Accelerated mathemathical pendulum
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[QUOTE="skrat, post: 4728358, member: 455703"] [h2]Homework Statement [/h2] Mathemathical pendulum with mass ##m## on a line ##L## is moving with constant acceleration in xz plane, ##\vec{a}=(a_x,a_z)##. Let's consider only the oscillations in xz plane. Find stationary value of ##\theta ## and find the frequency of small movements around the stationary angle if ##a_x=0##. [ATTACH=full]69024.vB[/ATTACH] [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] Ok, let's put one coordinate system somewhere :D and another one on the top of line ##L##. Let's say that these two coordinate systems are connected with vector R. Let's express everything in accelerated system: ##\vec{r}=(Lsin\theta , Lcos\theta )## and ##\vec{R}=(-\frac{1}{2}a_xt^2,-\frac{1}{2}a_yt^2)## Both together give me ##v=(\dot{\theta }Lcos\theta -a_xt,-\dot{\theta }Lsin\theta -a_yt)## and ##v^2=\dot{\theta }^2L^2+t^2(a_x^2+a_y^2)+2\dot{\theta }Lt(sin\theta a_y-cos\theta a_x## So ##L=T-V=\frac{1}{2}m(\dot{\theta }^2L^2+t^2(a_x^2+a_y^2)+2\dot{\theta }Lt(sin\theta a_y-cos\theta a_x)+mgLcos\theta ##. I won't even write the equations from here one because this brings me to wrong simple solution: ##\ddot{\theta }+\frac{g}{l}\theta =0## which can not be ok. It looks like I am missing something when calculating the velocity? But what? [/QUOTE]
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Accelerated mathemathical pendulum
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