\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\begin{document}
\parindent=0pt


{\bf Question}

A particle of mass $m$ rests on a smooth plane.  The plane is
raised to an incline $\phi$ at a constant rate $\alpha$ ($\phi =
0$ at $t = 0$), causing the particle to move down the plane.
Express the Lagrangain in polar coordinates of the particle in a
coordinate system whose origin is at the foot of the plane.  Hence
determine the motion of the particle.

\vspace{.25in}

{\bf Answer}


$$$$

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\begin{picture}(5,1.5)

\put(0.5,.5){\line(1,0){2.5}}

\put(1,0.5){\line(2,1){2}}

\put(1,.7){\vector(2,1){1.5}}

\put(1,.7){\vector(-2,-1){0}}

\put(1,0.3){\makebox(0,0){O}}

\put(2.5,1.3){$\bullet$}

\put(2.6,.5){\vector(0,1){0.75}}

\put(2.6,.5){\vector(0,-1){0}}

\put(2.7,1){\makebox(0,0)[l]{$r\sin\phi$}}

\put(1.75,.7){\makebox(0,0){$\phi$}}

\put(1.75,1.25){\makebox(0,0){$r$}}

\put(5,1){\makebox(0,0)[l]{$\phi=\alpha t$}}

\end{picture}
\end{center}

$\ds L = K.E. - P.E. = \frac{1}{2}m(\dot r^2 - r^2 \dot \phi^2) -
mgr\sin\alpha t$

$\ds \frac{d}{dt} \left(\frac{\pl L}{\pl \dot r}\right) -
\frac{\pl L}{\pl r} = 0 \Rightarrow \ddot r - \alpha^2 r = -g \sin
\alpha t$

This has solution $\ds r(t) = Ae^{\alpha t}+ Be^{-\alpha t} +
\frac{g}{2\alpha^2} \sin \alpha t$

Initially $r(0) = r_0$, $\dot r(0) = 0$ whence we can find A and B

Thus $\ds r(t) = \frac{1}{2} \left[ r_0 - \frac{g}{2 \alpha^2}
\right]e^{\alpha t} + \frac{1}{2} \left[ r_0 + \frac{g}{2
\alpha^2}\right]e^{-\alpha t} + \frac{g}{2\alpha^2} \sin \alpha t$




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