Question Number 226291 by fantastic2 last updated on 25/Nov/25
Answered by mr W last updated on 25/Nov/25
$${m}\omega^{\mathrm{2}} \left({r}+{l}\:\mathrm{sin}\:\theta\right)={mg}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}\:\mathrm{tan}\:\theta}{{r}+{l}\:\mathrm{sin}\:\theta}} \\ $$
Commented by fantastic2 last updated on 25/Nov/25
$${how}\:{you}\:{got}\:{mg}\mathrm{tan}\:\theta\:{sir} \\ $$
Commented by mr W last updated on 25/Nov/25
Commented by fantastic2 last updated on 25/Nov/25
$${thanks}\:{sir} \\ $$