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Question-135101




Question Number 135101 by BHOOPENDRA last updated on 10/Mar/21
Commented by BHOOPENDRA last updated on 10/Mar/21
find moment of inertia?
$${find}\:{moment}\:{of}\:{inertia}? \\ $$
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.
$${moi}\:{is}\:{always}\:{in}\:{relation}\:{to}\:{an}\:{axis}. \\ $$$${without}\:{saying}\:{about}\:{which}\:{axis}, \\ $$$${there}\:{is}\:{no}\:{answer}\:{possible}. \\ $$
Commented by BHOOPENDRA last updated on 16/Mar/21
about both axis X and Y
$${about}\:{both}\:{axis}\:{X}\:{and}\:{Y} \\ $$
Answered by mr W last updated on 17/Mar/21
Commented by mr W last updated on 17/Mar/21
s=((70×10×5+60×10×(10+30))/(70×10+60×10))     ≈19.538 mm  I_x =((70×10^3 )/(12))+70×10×(19.538−5)^2         +((10×60^3 )/(12))+10×60×(40−19.538)^2        ≈584997 mm^4   I_y =((10×70^3 )/(12))+((60×10^3 )/(12))      ≈290833 mm^4
$${s}=\frac{\mathrm{70}×\mathrm{10}×\mathrm{5}+\mathrm{60}×\mathrm{10}×\left(\mathrm{10}+\mathrm{30}\right)}{\mathrm{70}×\mathrm{10}+\mathrm{60}×\mathrm{10}} \\ $$$$\:\:\:\approx\mathrm{19}.\mathrm{538}\:{mm} \\ $$$${I}_{{x}} =\frac{\mathrm{70}×\mathrm{10}^{\mathrm{3}} }{\mathrm{12}}+\mathrm{70}×\mathrm{10}×\left(\mathrm{19}.\mathrm{538}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:+\frac{\mathrm{10}×\mathrm{60}^{\mathrm{3}} }{\mathrm{12}}+\mathrm{10}×\mathrm{60}×\left(\mathrm{40}−\mathrm{19}.\mathrm{538}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\approx\mathrm{584997}\:{mm}^{\mathrm{4}} \\ $$$${I}_{{y}} =\frac{\mathrm{10}×\mathrm{70}^{\mathrm{3}} }{\mathrm{12}}+\frac{\mathrm{60}×\mathrm{10}^{\mathrm{3}} }{\mathrm{12}} \\ $$$$\:\:\:\:\approx\mathrm{290833}\:{mm}^{\mathrm{4}} \\ $$

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