question_answer

A ball of mass \[{{10}^{-2}}kg\] and having charge + \[3\times {{10}^{-6}}C\]is tied at one end of a 1 m long thread. The other end of the thread is fixed and a charge - \[3\times {{10}^{-6}}C\]is placed at this end. The ball can move in the circular orbit of radius 1 m in the vertical plane. Initially, the ball is at the bottom. Find the minimum initial horizontal velocity of the bally so that it will be able to complete the full circle.

Answer:

                Let F be the eictric force and T be the tension in the string, both towards O, Fig. 5(HT).6. The centripetal force required = \[m{{\upsilon }^{2}}/r\] As is clear from Fig. 5(HT).6, \[T+F-mg\,\cos \text{ }\!\!\theta\!\!\text{ =}\frac{m{{\upsilon }^{2}}}{r}\]                          or                            \[T\text{=}\frac{m{{\upsilon }^{2}}}{r}-F+mg\,\cos \text{ }\!\!\theta\!\!\text{ }\]                                            ... (i) T will be minimum, when \[\cos \text{ }\!\!\theta\!\!\text{ }\,\text{=}\,\text{min}\text{.}\,\text{=}\,\text{-1}\] i.e. 0 = 180° i.e. at highest point H of the vertical circle, where velocity is \[{{\upsilon }_{H}}\]. From (i), \[{{T}_{\min }}=\frac{m\upsilon _{H}^{2}}{r}-F-mg\]                                                                                                                     ..(ii) For looping the loop, \[{{T}_{\min }}\ge 0\] From (ii), \[\left[ m\frac{\upsilon _{H}^{2}}{r}-F-mg \right]\ge 0\]                            or            \[\frac{\upsilon _{H}^{2}}{r}\ge (F+mg)\]                                              ... (iii) or            \[\upsilon _{H}^{2}\ge \frac{r}{m}\left[ F+mg \right]\] or            \[\upsilon _{H}^{2}\ge \left[ \frac{F}{m}+g \right]\]                                                                                                        \[\left( \because \,\,\,r=1m \right)\]      . (iv) According to Coulomb's law in electrostatics, \[F=\frac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] Where \[k=9\times {{10}^{9}}N{{m}^{2}}{{c}^{-2}}\] \[F=\frac{9\times {{10}^{9}}\left( 3\times {{10}^{-6}} \right)\left( 3\times {{10}^{-6}} \right)}{{{1}^{2}}}=81\times {{10}^{-3}}N\] From (iv), \[{{\left[ \upsilon _{H}^{2} \right]}_{\min }}=\left[ \frac{81\times {{10}^{-3}}}{{{10}^{-2}}}+10 \right]=18.1\] If \[{{\upsilon }_{L}}\] is minimum velocity of projection at lowest point, then applying the principle of conservation of energy, total energy at L = total energy' at H \[\frac{1}{2}m\upsilon _{L}^{2}=\frac{1}{2}m\upsilon _{H}^{2}+mg\left( 2r \right)\] \[\upsilon _{L}^{2}=\upsilon _{H}^{2}+4gr\] \[{{\upsilon }_{L}}=\sqrt{18.1+4\times 10\times 1}=\sqrt{58.1}=7.6\,m\,{{s}^{-1}}\]