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Consider a simple pendulum of length $l$ moving in $(x,y)$ plane. Assume its point of attachment is being jiggled by external forces with prescribed acceleration $a(t)=(a_x(t),a_y(t))$. Let $\theta(t)$ be the angle of deflection from stable equilibrium. What are the kinetic and potential energies of the system?

P.S. If there is no external force then the potential energy is $P=mgl(1-\cos{\theta})$ and the kinetic energy is $K=\frac{ml^2}{2} \dot{\theta}^2$. How these change in the above case?

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  • $\begingroup$ The potential energy is not the same. The expression you have for $P$ does not account for lifting the entire rod vertically. Just as the kinetic energy does not account for $\theta(t=0)=0$ and lifting the rod in the air. It would have vertical momentum that is unaccounted for by your current expressions for the kinetic and potential energy. Someone else will have to provide a detailed answer but they might be guided by this very clear write up phys.lsu.edu/faculty/gonzalez/Teaching/Phys7221/… $\endgroup$ – LasersMatter Jul 29 '16 at 4:29
  • $\begingroup$ Thank you! The detailed answer appears in the linked file. $\endgroup$ – user16015 Jul 29 '16 at 5:16
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The problem is that the motion of your pendulum is no longer "simple harmonic motion". And the expression for potential energy (essentially height above instantaneous equilibrium position) and kinetic energy (due to velocity of rod) have to change - because the velocity of the bob is now the velocity due to angular motion PLUS the velocity of the support point. Similarly the definition of potential energy may need adjusting because the pivot is no longer at its original position.

The work done when accelerating the pivot point is a function of the angle of the pendulum at the time of the motion - for instance, moving the support point to the right while the pendulum has swung to the right will take less energy than doing so while the pendulum is to the left of the pivot (draw a force diagram to convince yourself)

There is not really a simple general formula. Are you sure you want the complicated one?

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    $\begingroup$ Thanks, I found the answer in the file given in the comment to the question. $\endgroup$ – user16015 Jul 29 '16 at 5:17
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Your potential and kinetic energies are both still the same. What changes are the functions $\theta$ and its time derivatives, which you have in your KE equation already.

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  • $\begingroup$ How do they change? Can we find explicitly? (I'm confused :)) $\endgroup$ – user16015 Jul 29 '16 at 3:49
  • $\begingroup$ So the equation of motion is still $l \ddot{\theta}+ g \sin{\theta}=0$? $\endgroup$ – user16015 Jul 29 '16 at 3:54
  • $\begingroup$ No. You have to add $(a_x+a_y)$ $\endgroup$ – hebetudinous Jul 29 '16 at 4:00
  • $\begingroup$ It's no longer a homogeneous differential equation with the addition of the driving force. $\endgroup$ – hebetudinous Jul 29 '16 at 4:00
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    $\begingroup$ I disagree. Motion of the pivot affects the expression for the kinetic energy. $\endgroup$ – Floris Jul 29 '16 at 4:43

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