anonymous
  • anonymous
Hi All, I am struggling to get this using the Lagrangian formula, would anyone be able to assist? http://s3.amazonaws.com/answer-board-image/628ae3fdb906a2b8b95fb53923eb34c7.jpg
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
experimentX
  • experimentX
|dw:1378492897098:dw|
anonymous
  • anonymous
Yes, that angle is constant.
anonymous
  • anonymous
any ideas?

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experimentX
  • experimentX
for the particular case, Newtonian approach seems better than Lagrangian approach. centripetal force + hooks force = 0 the coordinate is given by \[(x, y) = (r_1 \cos (\omega t) + r_2(t) \sin( \theta), r_1 \sin(\omega t) + r_2(t)\cos(\theta) ) \] differentiate it square it and put it below and solve the Euler-Lagrange eqn. \[L = \frac 1 2 m (\dot x^2 + \dot y^2) - \frac 1 2 k r_2(t)^2 \]
anonymous
  • anonymous
thanks so much experimentX, greatly appreciate that. can i ask a tad bit more of your time and ask if you get the following for the E-L equation:
experimentX
  • experimentX
\[ \frac{d}{dt}\left( \frac{\partial L }{\partial \dot{r_2(t)}} \right) - \frac{\partial L }{\partial r_2(t)} = 0\] also I think you also need initial condition to solve it.
anonymous
  • anonymous
sorry, my equation is not coming through for some reason....
experimentX
  • experimentX
did you find Lagrangian?
anonymous
  • anonymous
what I have does not really make sense. \[L = \frac{ 1 }{ 2 } mr _{1}^{2}w ^{2}(\sin ^{2}(wt) + \cos ^{2}(wt)) - \frac{ 1 }{ 2 }Kr _{2}t ^{2}\] Sorry, quite new at this and im finding it beyond challenging...
experimentX
  • experimentX
sin(theta) is missing
experimentX
  • experimentX
\[ (x, y) = (r_1 \cos (\omega t) + r_2(t) \sin( \theta), r_1 \sin(\omega t) + r_2(t)\cos(\theta) ) \] \[ \dot x = \omega r_1 \cos(\omega t ) + \dot {r}_2 \sin(\theta) \] square it ... and also for y
anonymous
  • anonymous
So, \[L = \frac{ 1 }{ 2 }m(r _{1}^{2}w ^{2}\cos ^{2}(wt) + r \prime_{2}^{2}\sin ^{2}(\theta)) + r _{1}^{2}w ^{2}\cos ^{2}(wt) + r \prime_{2}^{2}\cos ^{2}(\theta)) - \frac{ 1 }{ 2 }kr _{2}t ^{2})\]
experimentX
  • experimentX
where is the middle term 2ab of (a+b)^2
experimentX
  • experimentX
\[ m(r _{1}^{2}w ^{2}\cos ^{2}(wt) \] since it contains no r2 dot ... get wiped out by differentiation.
anonymous
  • anonymous
my mistake not typing middle term.. however, what happened to the \[\frac{ 1 }{ 2 }Kr _{2}t ^{2}\]
experimentX
  • experimentX
that thing is okay ... that is the potential of the spring.
experimentX
  • experimentX
that doesn't get involved with first diff of EL equation ...
anonymous
  • anonymous
Ok cool. So from there we can input into the d/dt(curl L.... = curl L / curl r2 t Then we can find the DE from there?
experimentX
  • experimentX
yes ... you should get ODE
anonymous
  • anonymous
thanks for your help experiemtnX, I appreciate it. Just to confirm \[L = m(r ^{2}_{1}w ^{2}\cos ^{2}(wt)) - \frac{ 1 }{ 2 }Kr _{2}t ^{2}\]
experimentX
  • experimentX
No ... you put what should have been ignored.
anonymous
  • anonymous
crap....
experimentX
  • experimentX
why this doesn't work `T = \frac {1}{ 2} m ( \dot{r}_2^2 + 2 \right \dot {r}_2 r_1 \omega \cos(\omega t) (\sin(\theta) + \cos(\theta)) )` http://www.codecogs.com/latex/eqneditor.php
anonymous
  • anonymous
sorry for the hassle eX.. understood.
experimentX
  • experimentX
moreover you an do this problem via Newtonian mechanics ... i suppose. just equate the centripetal force + hooks force = acceleration of mass m
experimentX
  • experimentX
no Coriolis force or angular acceleration is involed ... since it is assumed to be at constant angle and constant angular velocity.
anonymous
  • anonymous
indeed, however, we are covering Lagrangian mech so have to follow suite. It has been a struggle to be sure.
experimentX
  • experimentX
still ... i wouldn't be confident on my formulation ... also try asking here. to make sure ... also don't forget to show your working http://physics.stackexchange.com/
anonymous
  • anonymous
thanks kindly.
experimentX
  • experimentX
you are welcome

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