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\large{\textbf{Indian Institute of Information Technology Allahabad}}
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\large{\textbf{Review Test- Classical Mechanics}}
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\noindent
\footnotesize{Max Marks: 20} \hfill \small{Duration: 1 hour}

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\begin{enumerate}

 \item A freely moving particle of mass $m$ is moving with velocity $\bf{v}=\dot{r}$ with respect to an inertial reference frame $K$. Let $K'$ be another inertial frame moving with a constant velocity $\bf{V}$ w.r.t. frame $K$. Write down the Lagrangian w.r.t. the frame $K'$ and show that the Euler-Lagrange equation remains the same.  \hfill[3]\\
 \noindent{\bf Ans.} $L_K=\frac{1}{2}m \textbf{v}^2$.\\
  $L_{K'}=\frac{1}{2}m (\textbf{v}-\textbf{V})^2$. \hfill[1]
  \\
  {\bf Method-1} Lagrange's eqn (in x direction) : $\frac{\partial L_{K'}}{\partial \dot{x}}=m( \dot{x}- V_x)\implies m\ddot{x}=0$. \\
  Similar expressions for $y$ and $z$ directions. These are the eqn. of motion for $L_K$. \hfill[1+1]

{\bf Method-2} $L_{K'}= L_K \,+\, \frac{d}{dt}(-m\textbf{r}.\textbf{V}+\frac{1}{2}mV^2t)$. Hence the Lagrange's eqn will be invariant. \hfill[2]

  \item A mass $m$ is suspended from the ceiling by a string of length $l$. The mass is free to move in all directions. This arrangement is called spherical pendulum. Write down the equation of constraint in terms of cartesian coordinates for this system. Is this constraint holonomic?  Obtain the Lagrangian and find out equations of motion for the mass. Identify the cyclic coordinate and corresponding conserved quantity. Find out generalized momenta for the system. Obtain the Hamilton's equations of motion. \hfill[1+1+1+1+2+1+1+2+2+2]\\
   \noindent{\bf Ans.} $x^2+y^2+z^2=l^2.$ \hfill[1]\\
  Yes. Holonomic, as it is of the form $f(x,y,z)=0$. \hfill[1]\\
  $T=\frac{1}{2}m l^2(\dot{\theta}^2\,+\,\sin^2 \theta \dot{\phi}^2)$. \hfill[1]\\
  $V=-mgl\cos \theta$ \hfill[1]\\
  $\frac{d}{dt}(ml^2\dot{\theta})-ml^2\sin \theta\cos \theta \dot{\phi}^2 + mgl\sin \theta=0.$\hfill[1]\\
  and,\\
  $\frac{d}{dt}(ml^2 \sin^2 \theta \dot{\phi})=0.$ \hfill[1]
  
  $\phi$ is the cyclic coordinate. \hfill[1]\\
  Angular momentum or conjugate momentum to $\phi$ , $p_{\phi}=ml^2 \sin^2 \theta \dot{\phi}$ is conserved. \hfill[1]\\
  $p_{\theta}=ml^2\dot{\theta}$; $p_{\phi}=ml^2 \sin^2 \theta \dot{\phi}$ \hfill[2]\\
  $ H=p_{\theta}\dot{\theta}+p_{\phi}\dot{\phi}-L=\frac{1}{2ml^{2}}\Big(p_{\theta}^{2}+\frac{p_{\phi}^{2}}{\sin^{2}\theta}\Big)-mgl\cos\theta $\hfill[2]\\
  EOM: momentum eqns as above. $\partial_{\theta}H=-\dot{p_{\theta}}\implies-ml^{2}\ddot{\theta}=-\frac{2p_{\phi}^{2} cot \theta \,cosec^{2}\theta }{2 m l^2}+ mgl\sin \theta.$\\
  $\partial_{\phi} H = -\dot{p}_{\phi}=0$
   \hfill[2]

  
\item Using Hamilton's principle directly show that equation of motion for a $1$ dimensional simple harmonic oscillator of mass $m$ is \[\ddot{x}=- {k\over m} x,\] where $k$ is Force constant.\hfill[3]\\
  \noindent{\bf Ans.} $L=\frac{m \dot{x}^2}{2}-\frac{1}{2}k x^2$. \\
    Hamilton's principle: $\delta S=\delta \int^{t_2}_{t_1} L dt =0$ \hfill[1]
   \begin{eqnarray*}
    \implies\int (m \dot{x} \delta \dot{x} -k x \delta x)dt=0&&\\
   \implies \int \left(\frac{d}{dt}(m \dot{x} \delta{x})\right)dt - (\int (m\ddot{x}+k x))\delta x dt=0&&[1]\\
   \implies (\int (m\ddot{x}+k x))\delta x dt=0,\quad \text{(since first integral is a boundary term which is zero)} \,[1]
   \end{eqnarray*}
   
   Hence the condition for exrtremum becomes the EOM: $m\ddot{x}+k x=0$
\end{enumerate}
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