As shown in the figure, a particle A of mass | vedclass

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Question

$I=\frac{m u \ell^{2}}{9}+\frac{2 m \ell^{2}}{9}$

$I=\frac{6 m \ell^{2}}{9}$

${	au _{{m{net}}}} = qE\ell \sin 	heta $

${	au _{{m{net}}

Vedclass · Accepted Answer

$qE+qE	heta_0^2$

Vedclass · Answer

$I=\frac{m u \ell^{2}}{9}+\frac{2 m \ell^{2}}{9}$

$I=\frac{6 m \ell^{2}}{9}$

${	au _{{m{net}}}} = qE\ell \sin 	heta $

${	au _{{m{net}}}} = I\alpha $

for small oscillation $\alpha=\omega^{2} 	heta$

${m{qE}}\ell 	heta  = \frac{{2{m{m}}{\ell ^2}}}{3}\alpha  \Rightarrow \frac{{3{m{qE}}	heta }}{{2{m{m}}\ell }} = \omega _{phase}^2$

${f_{{m{net}}}}$ toward centre

${m{T}} - {m{qE}} = {m{m}}\omega _{{m{particle }}}^2{m{R}}\quad \left[ {{{m{w}}_{{m{partice }}}} = {	heta _{\max }}{\omega _{{m{phase}}{m{. }}}}} ight]$

$\mathrm{T}=\mathrm{qE}+\mathrm{m} \frac{3 \mathrm{qE} 	heta_{0}^{2}}{2 \mathrm{m} \ell}\left(\frac{2 \ell}{3}ight) \quad\left[\mathrm{R}=\frac{2 \ell}{3}ight]$

$\mathrm{T}=\mathrm{q} \mathrm{E}+\mathrm{q} \mathrm{E} 	heta_{0}^{2}$

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