Question Number 74274 by Learner-123 last updated on 21/Nov/19
Answered by mr W last updated on 21/Nov/19
$${ring}: \\ $$$${r}=\sqrt{{l}^{\mathrm{2}} −{h}^{\mathrm{2}} } \\ $$$${M}=\mathrm{2}\pi{r}\rho \\ $$$${I}_{{z}} ={Mr}^{\mathrm{2}} =\mathrm{2}\pi\rho\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$${each}\:{rod}: \\ $$$${M}=\rho{l} \\ $$$${I}_{{z}} =\frac{{Mr}^{\mathrm{2}} }{\mathrm{3}}=\frac{\rho{l}\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)}{\mathrm{3}} \\ $$$$ \\ $$$${total}: \\ $$$${I}_{{z}} =\mathrm{2}\pi\rho\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{3}×\frac{\rho{l}\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right)}{\mathrm{3}} \\ $$$$=\left(\mathrm{2}\pi\sqrt{{l}^{\mathrm{2}} −{h}^{\mathrm{2}} }+{l}\right)\rho\left({l}^{\mathrm{2}} −{h}^{\mathrm{2}} \right) \\ $$$$=\mathrm{2}\left(\mathrm{2}\pi\sqrt{\mathrm{0}.\mathrm{5}^{\mathrm{2}} −\mathrm{0}.\mathrm{4}^{\mathrm{2}} }+\mathrm{0}.\mathrm{5}\right)\left(\mathrm{0}.\mathrm{5}^{\mathrm{2}} −\mathrm{0}.\mathrm{4}^{\mathrm{2}} \right) \\ $$$$=\mathrm{0}.\mathrm{429}\:{kgm}^{\mathrm{2}} \\ $$
Commented by Learner-123 last updated on 21/Nov/19
thank you sir!