Consider a simple plane pendulum with a bob of mass \(m\) attached to a string of length \(l\). After the
pendulum is set in motion the string is shortened by a constant rate \(\frac {dl}{dt}=-\alpha \). Formulate Hamilton’s
principle and determine the equation of motion. Compare the Hamiltonian to the total energy. Is
energy conserved?
Solution:
Assume initial string length is \(l\), and assume \(t\left ( 0\right ) =0\), then at time \(t\) we have
This is a second order nonlinear differential equation. Notice that when \(l=\alpha t\) it will mean that the
string has been pulled all the way back to the pivot and \(r\left ( t\right ) =0\). So when running the solution it needs to
run from \(t=0\) up to \(t=\frac {l}{\alpha }\).
A small simulation was done for the above solution which can be run for different parameters to
see the effect more easily. Here is a screen shot.
Now we need to determine the Hamiltonian of the system.
Now we are asked to compare \(H\) to the total energy. The total instantaneous energy of the system is
\(\left ( T+V\right ) \), hence we need to determine if \(H=T+V\) or not.
To make comparing (5) and (6) easier, I need to either replace \(p\) by \(mr^{2}\dot {\theta }\) in (5) or replace \(\dot {\theta }\) by \(\frac {p}{mr^{2}}\). Lets do
the later, hence \((6)\) becomes
Hence \(H\) does not represents the total energy, and the energy of the system is not conserved.\(\footnote {Reading a reference on Noether's theorem, total energy is written as $-\left ( T+V\right ) $ not $\left ( T+V\right ) $, this would not make a difference in showing that $H\neq $ total energy, just different calculations results as shown below, but the same conclusion \par \begin {equation} -\left ( T+V\right ) =-\left ( \frac {1}{2}m\left ( \alpha ^{2}+r^{2}\dot {\theta }^{2}\right ) +mg\left ( l-r\cos \theta \right ) \right ) \tag {6}\end {equation} \par To make comparing (5) and (6) easier, I\ need to either replace $p$ by $mr^{2}\dot {\theta }$ in (5) or replace $\dot {\theta }$ by $\frac {p}{mr^{2}}$, let do the later, hence $(6)$ becomes \par \begin {align} -\left ( T+V\right ) & =-\left ( \frac {1}{2}m\left ( \alpha ^{2}+r^{2}\left ( \frac {p}{mr^{2}}\right ) ^{2}\right ) +mg\left ( l-r\cos \theta \right ) \right ) \nonumber \\ & =-\left ( \frac {1}{2}m\left ( \alpha ^{2}+\frac {p^{2}}{m^{2}r^{2}}\right ) +mg\left ( l-r\cos \theta \right ) \right ) \nonumber \\ & =-\left ( \frac {1}{2}m\alpha ^{2}+\frac {1}{2}\frac {p^{2}}{mr^{2}}+mg\left ( l-r\cos \theta \right ) \right ) \tag {7}\end {align} \par If $H$ is the total energy, then (7)-(6) should come out to be zero, lets find out \par \begin {align*} H-\left ( -\left ( T+V\right ) \right ) & =\left ( \frac {1}{2}\frac {p^{2}}{mr^{2}}+mg\left ( l-r\cos \theta \right ) -\frac {1}{2}m\alpha ^{2}\right ) +\left ( \frac {1}{2}m\alpha ^{2}+\frac {1}{2}\frac {p^{2}}{mr^{2}}+mg\left ( l-r\cos \theta \right ) \right ) \\ & =\frac {p^{2}}{mr^{2}}+2mg\left ( l-r\cos \theta \right ) \end {align*} \par Now we ask, can the above be zero? Since $\frac {p^{2}}{mr^{2}}$ is always $\geq 0$, and since $\left ( l-r\cos \theta \right ) $ represents the remaining length of the string, hence it is a positive quantity (until such time the string has been pulled all the way in), Therefore RHS\ above $>0$. Hence $H$ does not represent the total energy of the system. Hence the energy is not conserved.}\)
3 Problem 1 (section 3.5,#9, page 197)
problem: A bead of mass \(m\) slides down the rim of a circular hoop of radius \(R\). The hoop stands
vertically and rotates about its diameter with angular velocity \(\omega \). Determine the equation of motion
of the bead.
Answer:
Kinetic energy \(T\) is made up of 2 components, one due to motion of the bead along the hoop itself
with speed \(R\dot {\theta }\), and another due to motion with angular speed \(\omega \) which has speed given by
\(R\sin \theta \omega \)