question_answer

A stone of mass m tied to the end of a string is whirled around in a horizontal circle (neglect force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then the tension in the string is given by  \[T=A{{r}^{n}}\], where A is a constant and r is instantaneous radius of the circle. Show that n = - 3.

Answer:

                For circular motion, tension in the string is \[T=\frac{m{{\upsilon }^{2}}}{r}\]                                           ...(i) Angular momentum, \[L=m\,\upsilon \,r\,=K\,=\operatorname{co}ns\tan t\]. \[\upsilon =\frac{K}{mr}\] Put in (i), \[T=\frac{m}{r}{{\left( \frac{K}{mr} \right)}^{2}}=\frac{{{K}^{2}}}{m}{{r}^{-3}}\] As       \[T=A{{r}^{n}}\]                    \[\therefore \] \[n=-3\], which was to be proved.