Applications of Derivatives Set 13


Example - 29

Consider a particle free to move on the inner side of a smooth vertical circular ring of radius r. It is projected from the lowest point with a velocity just sufficient to carry it to the highest point. Show that  the reaction between the particle and the ring is zero after time t given by

\[t = \sqrt {\frac{r}{g}} \;\log \left( {\sqrt 5  + \sqrt 6 } \right)\]

Solution:         Let P be the lowest point and Q be the point at which the reaction becomes zero. Let \({v_i}\)and \({v_f}\) be the respective velocities of the particle t = 0 and later instant.

By energy conservation, we have

\[\frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2 = mgr(1 - \cos \theta )\]

At \(\theta  = \pi ,\;{v_f} = 0\)

\[ \Rightarrow   \frac{1}{2}mv_i^2 = 2\;mgr\]

At the point    \(Q,\;v_f^2 = 2gr(1 + \cos {\theta _Q})\)

       \[R = 0 = \frac{{mv_{f\theta }^2}}{r} + mg\cos {\theta _Q}   \Rightarrow  \cos {\theta _Q} =  - \frac{2}{3}\]


\[\begin{align}&{v_f} = \frac{{rd\theta }}{{dt}} = \sqrt {2gr(1 + \cos \theta )}  = 2\sqrt {gr} \cos \frac{\theta }{2}\\\\ &\Rightarrow \quad dt = \frac{1}{2}\sqrt {\frac{r}{g}} \sec \frac{\theta }{2}d\theta    \Rightarrow  {t_Q} = \frac{1}{2}\sqrt {\frac{r}{g}} \;\int\limits_0^{{\theta _Q}} {\sec \frac{\theta }{2}\;d\theta } \\\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{r}{g}} \log \;\sec \frac{{{\theta _Q}}}{2} + \tan \frac{{{\theta _Q}}}{2}\\\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{r}{g}} \log \left( {\sqrt 6  + \sqrt 5 } \right)\end{align}\]


Example - 30

Consider the parabola \({x^2} = 4ay,\;a > 0\). A rod of length l (greater than the semi-latus-rectum of the parabola) rests on the parabola as shown.

Let \(\theta\) be the inclination of the rod. Find the value(s) of  \(\theta\) for which the height of the center of the rod is minimum.

Solution: From the geometry of the figure,

\[P \equiv \left( {2at,\;a{t^2}} \right),\;\theta  \equiv \left( {2at + 2l\cos \theta ,\;a{t^2} + 2l\sin \theta } \right)\]

From the fact that Q must satisfy \({x^2}4ay,\)we obtain

\[t = \tan \theta  - \frac{l}{{2a}}\cos \theta \]

Therefore, the height is

\[h = a\,{t^2} + l\sin \theta  = a{\tan ^2}\theta  + \frac{{{l^2}}}{{4{a^2}}}{\cos ^2}\theta \]

\(h(\theta )\)has extreme (minima) points for \(\begin{align}{\cos ^2}\theta  = \frac{{2a}}{l}\;{\rm{or}}\;\sin \theta  = 0\end{align}\)

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