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\begin{center}
\large{\textbf{Indian Institute of Information Technology Allahabad}}
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\begin{center}
\large{\textbf{Quiz I}}
\end{center}
\hfill \footnotesize{Date: 19/09/2017}
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\noindent
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\begin{center}
\footnotesize{B.Tech. IT, Dual Degree - Semester I}
\end{center}
\vspace{-20pt}

\begin{center}
\footnotesize{Paper Code: SPHY132C}
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%\footnotesize{Paper Setter: Abdullah Bin Abu Baker \& Sumit Kumar Upadhyay}
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\noindent
\footnotesize{Max Marks: 20} \hfill \small{Duration: 1 hour}

\noindent
\footnotesize{Attempt \textbf{all} the questions. There is \textbf{no credit} for an answer if proper justification is not given, even if the answer is correct. Notations are standard. \textbf{Do not write} on question paper and cover pages except your details.}

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\begin{enumerate}

\item Assume a frictionless, massless pulley of radius $R$ and a massless rope of length $l$. Two masses $m_1$ and $m_2$ are attached at the two ends of the rope. Let at an instant, the coordinates of the masses are $(-x_1, y_1)$ and $(x_1,y_2)$ respectively (with respect to a Cartesian coordinate system erected at the centre of the pulley). What constraint (equation) does this system of masses satisfy? What type of constraint (holonomic or non-holonomic) is this? Is the constraint sceleronomous or rheonomous? What is the number of degrees of freedom of this system? Compute the Lagrangian of the system. Hence deduce the Hamiltonian of the system. Find out the equation of motion. \hfill[1+1+1+1+3+2+1]

\noindent

{\bf Ans.} Total length of the rope should be $=l$. So the constraint equation is 
$y_1+y_2+\pi R=l$ or $y_1 + y_2= c$. \rh{[1]}

Holonomic constraint. Sceleronomous. \rh{[1 +1]}

Number of degrees of freedom is $1$, since two coordinates $y_1, y_2$ are related by one equation. \rh{[1]}

Kinetic energy : $T=\frac{1}{2}m_1\dot{y_1}^2 + \frac{1}{2}m_2\dot{ y_2}^2=\frac{1}{2}(m_1+m_2)\dot{y_1}^2$. \rh{[1]}\\

Potential energy: $V=m_1gy_1 + m_2g (c-y_1)$. \rh{[1]}\\

Lagrangian: \[L=\frac{1}{2}(m_1+m_2)\dot{y_1}^2 - m_1gy_1 - m_2g (c-y_1) \]\rh{[1]}

Hamiltonian: \[p_{y_1}=\frac{\partial L}{\partial \dot{y_1}}=(m_1+m_2)\dot{y_1}\]\rh{[1]}
 \[H(y_1, p_{y_1})=\frac{p_{y_1}^2 }{2(m_1+m_2)}+V(y_1)\]\rh{[1]}
 
 Equation of motion: \[\ddot{y_1}=\frac{m_2-m_1}{m_1+m_2}g \] \rh{[1]}

\item A double plane pendulum consists of two simple pendulums, with one pendulum
suspended from the bob of the other. The ``upper'' pendulum has mass $m_1$ and length $l_1$,
the ``lower'' pendulum has mass $m_2$ and length $l_2$, and  both pendulums move in the
same vertical plane. Find the Lagrangian and deduce the Lagrange's equations of motion. 
\hfill[4+4]

{\bf Ans.} Let $(x_1,y_1)$ and $(x_2,y_2)$ be the coordinates of the masses $m_1$ and $m_2$ respectively. Let $\theta_1$ and $\theta_2$ be the angles that the upper and the lower pendulum blobs make with the vertical. Then,
\be x_1=l_1 Sin\theta_1,~ y_1=l_1Cos \theta_1; \qquad x_2=l_1 Sin\theta_1+l_2 Sin \theta_2,~ y_2=l_1 Cos\theta_1+ l_2Cos \theta_2\ee
%\rh{[1]}

Kinetic energy: \beq T&=&\frac{1}{2}m_1(\dot{x_1}^2+\dot{y_1}^2)+\frac{1}{2}m_2(\dot{x_2}^2+\dot{y_2}^2)\\
	                &=& \frac{1}{2}m_1l_1^2\dot{\theta_1}^2+ \frac{1}{2}m_2\left(l_1^2\dot{\theta_1}^2+2l_1l_2\dot{\theta_1}\dot{\theta_2}Cos(\theta_2-\theta_1)+l_2^2\dot{\theta_2}^2\right)\eeq \rh{[2]}
	                
Potential energy: $V=-m_1gl_1Cos \theta_1 -m_2g (l_1 Cos\theta_1+ l_2Cos \theta_2). $ 
\rh{[1]} (assuming the zero level of the potential is at the top).\\

Lagrangian: \[L=\frac{1}{2}m_1l_1^2\dot{\theta_1}^2+ \frac{1}{2}m_2\left(l_1^2\dot{\theta_1}^2+2l_1l_2\dot{\theta_1}\dot{\theta_2}Cos(\theta_2-\theta_1)+l_2^2\dot{\theta_2}^2\right)+m_1gl_1Cos \theta_1 +m_2g (l_1 Cos\theta_1+ l_2Cos \theta_2)\]\rh{[1]}

\newpage

\beq \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta_1}}\right)&=&(m_1+m_2)l_1^2\ddot{\theta_1}+m_2l_1l_2\ddot{\theta_2} Cos(\theta_2-\theta_1)-m_2l_1l_2\dot{\theta_2}(\dot{\theta_2}-\dot{\theta_1})Sin(\theta_2-\theta_1)\\
\frac{\partial L}{\partial \theta_1}&=&m_2l_1l_2\dot{\theta_1}\dot{\theta_2}Sin(\theta_2-\theta_1)-(m_1+m_2)gl_1Sin\theta_1
\eeq\rh{[2]}

\beq \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta_2}}\right)&=&m_2l_2^2\ddot{\theta_2}+m_2l_1l_2\ddot{\theta_1} Cos(\theta_2-\theta_1)-m_2l_1l_2\dot{\theta_1}(\dot{\theta_2}-\dot{\theta_1})Sin(\theta_2-\theta_1)\\
\frac{\partial L}{\partial \theta_2}&=&-m_2l_1l_2\dot{\theta_1}\dot{\theta_2}Sin(\theta_2-\theta_1)-m_2gl_2Sin\theta_2
\eeq\rh{[2]}

                
 \item Using Hamilton's principle show that Euler-Lagrange equation remains unchanged under the transformation 
 \[L(q, \dot{q}, t)\rightarrow L(q, \dot{q}, t) + \frac{d F(q, t)}{dt}.\]
  \hfill[2]
  
  {\bf Ans.} Hamilton's principle: \[ \delta \int_{t_1}^{t_2} dt L(q, \dot{q}, t)=0 .\]\rh{[1]}
  
  This gives the E-L equation. Let,
  \[ L^{'}=L +  \frac{d F(q, t)}{dt}.\]
  
  Now, \beq \delta \int_{t_1}^{t_2} dt \left(L(q, \dot{q}, t)+\frac{d F(q, t)}{dt}\right)&=&\delta \int_{t_1}^{t_2} dt L(q, \dot{q}, t) + \left[ \delta F(q,t)\right]_{t_1}^{t_2}.\eeq
  As boundary variation of $F$ is zero, we get, \be \delta \int_{t_1}^{t_2} dt L(q, \dot{q}, t) =0. \ee
  This is again the Hamilton's principle in terms of the old Lagrangian. Hence the equation of motion will remain unchanged. 
  
  \rh{[1]}
    
\end{enumerate}

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