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I have to prove the following equations using the Lagrangian equations, The figure shows the image, I know how to do lagrangian. I just don't know how to solve the kinetic and potential energies of this pendulum. Can someone show me what the value are for the kinetic and potential energy? I have tried but I don't know how to solve them

This is the question https://i.stack.imgur.com/UJvfm.jpg

user2076774
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1 Answers1

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The horizontal speed of the mass $m$ is zero, and its vertical speed is $-\dot d$. Its kinetic energy is therefore $\frac12 m \dot d^2$. Its potential energy is obviously $\text{constant}-mgd$.

The horizontal speed of the mass $M$ is $\dot{(l\sin(\theta))} = l\cos(\theta)\dot\theta$. The vertical speed of the mass $M$ is $\dot{(-d - l\cos(\theta))} = -\dot d +l\sin(\theta)\dot\theta$. Its kinetic energy is therefore $\frac12M(\dot d^2 - 2\dot d l \sin(\theta) + l^2\dot\theta^2$). Its potential energy is $-Mg(d + l\cos(\theta))$.

apt1002
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  • I tried this method but when I calculate the torque equation from the lagrangian I am missing the Ml(d ̈)sinθ term. I think there is something missing in the kinetic energy equatino but I am not sure – user2076774 Dec 09 '13 at 08:03
  • I made a slip, didn't I? I did not correctly square the term $-\dot d + l\sin(\theta)\dot\theta$. I will fix that. – apt1002 Dec 09 '13 at 08:09
  • I tried this again, still not getting the correct answer. Adding the sine in the kinetic energy gives me a cosine in the final answer for Torque but I need two sines – user2076774 Dec 09 '13 at 18:46