Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

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

$${s}=\frac{\mathrm{70}×\mathrm{10}×\mathrm{5}+\mathrm{60}×\mathrm{10}×\left(\mathrm{10}+\mathrm{30}\right)}{\mathrm{70}×\mathrm{10}+\mathrm{60}×\mathrm{10}} \\ $$$$\:\:\:\approx\mathrm{19}.\mathrm{538}\:{mm} \\ $$$${I}_{{x}} =\frac{\mathrm{70}×\mathrm{10}^{\mathrm{3}} }{\mathrm{12}}+\mathrm{70}×\mathrm{10}×\left(\mathrm{19}.\mathrm{538}−\mathrm{5}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:+\frac{\mathrm{10}×\mathrm{60}^{\mathrm{3}} }{\mathrm{12}}+\mathrm{10}×\mathrm{60}×\left(\mathrm{40}−\mathrm{19}.\mathrm{538}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\approx\mathrm{584997}\:{mm}^{\mathrm{4}} \\ $$$${I}_{{y}} =\frac{\mathrm{10}×\mathrm{70}^{\mathrm{3}} }{\mathrm{12}}+\frac{\mathrm{60}×\mathrm{10}^{\mathrm{3}} }{\mathrm{12}} \\ $$$$\:\:\:\:\approx\mathrm{290833}\:{mm}^{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com