Question Number 163053 by ajfour last updated on 03/Jan/22
Commented by ajfour last updated on 03/Jan/22
$${Q}.\mathrm{162949}\:\:\left({revisit}\right) \\ $$
Answered by ajfour last updated on 04/Jan/22
$$\:\:{e}=\frac{\mathrm{2}{R}}{\pi}=\frac{\mathrm{4}{a}}{\pi}\:\:\:\:\:\:\:\:\left(\mathrm{2}{a}={R}\right) \\ $$$$\:\:{r}^{\mathrm{2}} =\left(\mathrm{2}{a}\mathrm{cos}\:\phi\right)^{\mathrm{2}} +\left(\mathrm{3}{a}+\mathrm{2}{a}\mathrm{sin}\:\phi\right)^{\mathrm{2}} \\ $$$${dm}_{{sc}} =\mathrm{2}\lambda{ad}\phi \\ $$$${I}_{\mathrm{0}} =\frac{\mathrm{2}{a}\lambda\left(\mathrm{6}{a}\right)^{\mathrm{2}} }{\mathrm{3}}+\int{r}^{\mathrm{2}} {dm}_{{sc}} \\ $$$$\:\:=\mathrm{24}\lambda{a}^{\mathrm{3}} +\mathrm{2}\lambda{a}^{\mathrm{3}} \int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{13}+\mathrm{12sin}\:\phi\right){d}\phi \\ $$$$\:\:=\mathrm{24}\lambda{a}^{\mathrm{3}} +\mathrm{2}\lambda{a}^{\mathrm{3}} \left(\mathrm{13}\pi+\mathrm{24}\right) \\ $$$$\:{say}\:\:{m}_{{rod}} ={m}=\mathrm{6}\lambda{a} \\ $$$$\Rightarrow\:\:{I}_{\mathrm{0}} ={ma}^{\mathrm{2}} \left(\mathrm{12}+\frac{\mathrm{13}\pi}{\mathrm{3}}\right)=\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){ma}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${M}=\frac{\left(\pi+\mathrm{3}\right){m}}{\mathrm{3}} \\ $$$$\Rightarrow\:{I}_{\mathrm{0}} =\frac{\left(\mathrm{36}+\mathrm{13}\pi\right){MR}^{\mathrm{2}} }{\mathrm{4}\left(\pi+\mathrm{3}\right)} \\ $$$$ \\ $$$$\: \\ $$$$\:\:\:\checkmark \\ $$$${for}\:{center}\:{of}\:{mass}: \\ $$$$\mathrm{2}\pi{a}\lambda\left(\frac{\mathrm{4}{a}}{\pi}\right)=\left(\mathrm{2}\pi{a}\lambda+\mathrm{6}{a}\lambda\right){s} \\ $$$${s}=\frac{\mathrm{4}{a}}{\pi+\mathrm{3}} \\ $$$${p}^{\mathrm{2}} ={s}^{\mathrm{2}} +\mathrm{9}{a}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\beta=\frac{{s}}{\mathrm{3}{a}}=\frac{\mathrm{4}}{\mathrm{3}\left(\pi+\mathrm{3}\right)} \\ $$$${Mgp}\mathrm{cos}\:\left(\theta+\beta\right)={I}_{\mathrm{0}} \left(\frac{\omega{d}\omega}{{d}\theta}\right) \\ $$$$\omega^{\mathrm{2}} =\mathrm{2}{Mgp}\left\{\mathrm{sin}\:\left(\theta+\beta\right)−\mathrm{sin}\:\beta\right\} \\ $$$${T}=\sqrt{\frac{{I}_{\mathrm{0}} }{\mathrm{2}{Mgp}}}\int_{\mathrm{0}} ^{\:\:\frac{\pi}{\mathrm{2}}−\varphi} \frac{{d}\theta}{\:\sqrt{\mathrm{sin}\:\left(\theta+\beta\right)−\mathrm{sin}\:\beta}} \\ $$$${where}\:\:\varphi=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}} \\ $$$${T}=\sqrt{\frac{{a}\left(\mathrm{36}+\mathrm{13}\pi\right)}{\mathrm{4}\left(\mathrm{3}+\pi\right){g}\sqrt{\left(\frac{\mathrm{4}}{\pi+\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{9}}}}×\int\bigstar \\ $$$$\int\bigstar=\int_{\mathrm{0}} ^{\:\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}} \frac{{d}\theta}{\:\sqrt{\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{3}\left(\pi+\mathrm{3}\right)}\right)−\frac{\mathrm{4}}{\:\sqrt{\mathrm{16}+\mathrm{9}\left(\pi+\mathrm{3}\right)^{\mathrm{2}} }}}} \\ $$$${Sir},\:{please}\:\:{check}\:\:{the}\:{answer}\:{this} \\ $$$${yields}\:\:{with}\:{yours}..{i}\:{will}\:{try}\:{too}.. \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 04/Jan/22
$${i}\:{got}\:{with}\:{your}\:{formula}\:{T}\approx\mathrm{1}.\mathrm{3626}\sqrt{\frac{{R}}{{g}}} \\ $$$${that}\:{differs}\:{from}\:{my}\:{result}. \\ $$