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Question Number 74274 by Learner-123 last updated on 21/Nov/19

Answered by mr W last updated on 21/Nov/19

$$ring:$$$$r=\sqrt{l^2-h^2}$$$$M=2\pi r\rho$$$$I_z={Mr}^2=2\pi \rho {(l^2-h^2)}^{\frac{3}{2}}$$$$$$$$eachrod:$$$$M=\rho l$$$$I_z=\frac{{Mr}^2}{3}=\frac{\rho l(l^2-h^2)}{3}$$$$$$$$total:$$$$I_z=2\pi \rho {(l^2-h^2)}^{\frac{3}{2}}+3\times \frac{\rho l(l^2-h^2)}{3}$$$$=(2\pi \sqrt{l^2-h^2}+l)\rho (l^2-h^2)$$$$=2(2\pi \sqrt{0.5^2-0.4^2}+0.5)(0.5^2-0.4^2)$$$$=0.429{kgm}^2$$

Commented by Learner-123 last updated on 21/Nov/19

thank you sir!

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