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 \[\frac{1}{2}m\upsilon _{L}^{2}=\frac{1}{2}m\upsilon _{H}^{2}+mg\left( 2r \right)\] \[\upsilon _{L}^{2}=\upsilon _{H}^{2}+4gr\] \[{{\upsilon }_{L}}=\sqrt{18.1+4\times 10\times 1}=\sqrt{58.1}=7.6\,m\,{{s}^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec