Question Number 135101 by BHOOPENDRA last updated on 10/Mar/21
Commented by BHOOPENDRA last updated on 10/Mar/21
$${find}\:{moment}\:{of}\:{inertia}? \\ $$
Commented by mr W last updated on 15/Mar/21
$${moi}\:{is}\:{always}\:{in}\:{relation}\:{to}\:{an}\:{axis}. \\ $$$${without}\:{saying}\:{about}\:{which}\:{axis}, \\ $$$${there}\:{is}\:{no}\:{answer}\:{possible}. \\ $$
Commented by BHOOPENDRA last updated on 16/Mar/21
$${about}\:{both}\:{axis}\:{X}\:{and}\:{Y} \\ $$
Answered by mr W last updated on 17/Mar/21
Commented by mr W last updated on 17/Mar/21
$${s}=\frac{\mathrm{70}×\mathrm{10}×\mathrm{5}+\mathrm{60}×\mathrm{10}×\left(\mathrm{10}+\mathrm{30}\right)}{\mathrm{70}×\mathrm{10}+\mathrm{60}×\mathrm{10}} \\ $$$$\:\:\:\approx\mathrm{19}.\mathrm{538}\:{mm} \\ $$$${I}_{{x}} =\frac{\mathrm{70}×\mathrm{10}^{\mathrm{3}} }{\mathrm{12}}+\mathrm{70}×\mathrm{10}×\left(\mathrm{19}.\mathrm{538}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:+\frac{\mathrm{10}×\mathrm{60}^{\mathrm{3}} }{\mathrm{12}}+\mathrm{10}×\mathrm{60}×\left(\mathrm{40}−\mathrm{19}.\mathrm{538}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\approx\mathrm{584997}\:{mm}^{\mathrm{4}} \\ $$$${I}_{{y}} =\frac{\mathrm{10}×\mathrm{70}^{\mathrm{3}} }{\mathrm{12}}+\frac{\mathrm{60}×\mathrm{10}^{\mathrm{3}} }{\mathrm{12}} \\ $$$$\:\:\:\:\approx\mathrm{290833}\:{mm}^{\mathrm{4}} \\ $$