Question Number 24873 by Tinkutara last updated on 28/Nov/17
$$\mathrm{A}\:\mathrm{rigid}\:\mathrm{body}\:\mathrm{is}\:\mathrm{made}\:\mathrm{of}\:\mathrm{three}\:\mathrm{identical} \\ $$$$\mathrm{thin}\:\mathrm{rods},\:\mathrm{each}\:\mathrm{of}\:\mathrm{length}\:{L},\:\mathrm{fastened} \\ $$$$\mathrm{together}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\mathrm{letter}\:{H}.\:\mathrm{The} \\ $$$$\mathrm{body}\:\mathrm{is}\:\mathrm{free}\:\mathrm{to}\:\mathrm{rotate}\:\mathrm{about}\:\mathrm{a}\:\mathrm{horizontal} \\ $$$$\mathrm{axis}\:\mathrm{that}\:\mathrm{runs}\:\mathrm{along}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{one}\:\mathrm{of} \\ $$$$\mathrm{legs}\:\mathrm{of}\:{H}.\:\mathrm{The}\:\mathrm{body}\:\mathrm{is}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{fall} \\ $$$$\mathrm{from}\:\mathrm{rest}\:\mathrm{from}\:\mathrm{a}\:\mathrm{position}\:\mathrm{in}\:\mathrm{which}\:\mathrm{plane} \\ $$$$\mathrm{of}\:{H}\:\mathrm{is}\:\mathrm{horizontal}.\:\mathrm{The}\:\mathrm{angular}\:\mathrm{speed} \\ $$$$\mathrm{of}\:\mathrm{body}\:\mathrm{when}\:\mathrm{plane}\:\mathrm{of}\:{H}\:\mathrm{is}\:\mathrm{vertical}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 28/Nov/17
Commented by mrW1 last updated on 28/Nov/17
$${let}'{s}\:{say}\:{each}\:{leg}\:{has}\:{a}\:{mass}\:{m}. \\ $$$${I}=\frac{{mL}^{\mathrm{2}} }{\mathrm{3}}+{mL}^{\mathrm{2}} =\frac{\mathrm{4}{mL}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${I}\alpha={mg}\left(\frac{{L}}{\mathrm{2}}+{L}\right)\mathrm{cos}\:\theta=\frac{\mathrm{3}{mgL}}{\mathrm{2}}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\frac{\mathrm{4}{mL}^{\mathrm{2}} }{\mathrm{3}}\alpha=\frac{\mathrm{3}{mgL}}{\mathrm{2}}\mathrm{cos}\:\theta \\ $$$$\Rightarrow\alpha=\frac{\mathrm{9}{g}\mathrm{cos}\:\theta}{\mathrm{8}{L}}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} ={mg}\left(\frac{{L}}{\mathrm{2}}+{L}\right)\mathrm{sin}\:\theta=\frac{\mathrm{3}{mgL}}{\mathrm{2}}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}{mL}^{\mathrm{2}} }{\mathrm{3}}\omega^{\mathrm{2}} =\frac{\mathrm{3}{mgL}}{\mathrm{2}}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\omega^{\mathrm{2}} =\frac{\mathrm{9}{g}}{\mathrm{4}{L}}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\omega=\left(\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\frac{{g}}{{L}}}\right)\sqrt{\mathrm{sin}\:\theta}\:\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${at}\:\theta=\frac{\pi}{\mathrm{2}}: \\ $$$$\alpha=\mathrm{0} \\ $$$$\omega=\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\frac{{g}}{{L}}} \\ $$
Commented by Tinkutara last updated on 28/Nov/17
$${What}\:{is}\:\theta? \\ $$
Commented by mrW1 last updated on 28/Nov/17
$${that}\:{is}\:{the}\:{position}\:{of}\:{the}\:{plane}: \\ $$$$\theta=\mathrm{0}\:{means}\:{horizontal}\:{position} \\ $$$$\theta=\frac{\pi}{\mathrm{2}}\:{means}\:{vertical}\:{position} \\ $$
Commented by mrW1 last updated on 28/Nov/17
Commented by Tinkutara last updated on 28/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$