Question Number 25013 by Tinkutara last updated on 01/Dec/17
$$\mathrm{With}\:\mathrm{reference}\:\mathrm{to}\:\mathrm{figure}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{of} \\ $$$$\mathrm{edge}\:{a}\:\mathrm{and}\:\mathrm{mass}\:{m},\:\mathrm{state}\:\mathrm{whether}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{are}\:\mathrm{true}\:\mathrm{or}\:\mathrm{false}.\:\left(\mathrm{O}\:\mathrm{is}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}.\right) \\ $$$$\left(\mathrm{1}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\ $$$$\mathrm{about}\:{z}-\mathrm{axis}\:\mathrm{is},\:{I}_{{z}} \:=\:{I}_{{x}} \:+\:{I}_{{y}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\ $$$$\mathrm{about}\:{z}'\:\mathrm{is},\:{I}_{{z}'} \:=\:{I}_{{z}} \:+\:\frac{{ma}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{cube} \\ $$$$\mathrm{about}\:{z}''\:\mathrm{is},\:{I}_{{z}'} \:=\:{I}_{{z}} \:+\:\frac{{ma}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:{I}_{{x}} \:=\:{I}_{{y}} \\ $$
Commented by Tinkutara last updated on 01/Dec/17
Commented by ajfour last updated on 01/Dec/17
$${I}_{{Z}} =\frac{{ma}^{\mathrm{2}} }{\mathrm{6}}\:,\:\:\:{I}_{{Z}\:'} =\frac{{ma}^{\mathrm{2}} }{\mathrm{6}}+\frac{{ma}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}{ma}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${I}_{{X}} ={I}_{{Y}} =\frac{{ma}^{\mathrm{2}} }{\mathrm{6}}+\frac{{ma}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{5}{ma}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${I}_{{Z}\:''} =\:?\:..... \\ $$
Commented by Tinkutara last updated on 02/Dec/17
$${How}\:{I}_{{z}} =\frac{{ma}^{\mathrm{2}} }{\mathrm{6}}? \\ $$