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Question Number 25013 by Tinkutara last updated on 01/Dec/17

$$\mathrm{With}\mathrm{reference}\mathrm{to}\mathrm{figure}\mathrm{of}\mathrm{a}\mathrm{cube}\mathrm{of}$$$$\mathrm{edge}a\mathrm{and}\mathrm{mass}m,\mathrm{state}\mathrm{whether}\mathrm{the}$$$$\mathrm{following}\mathrm{are}\mathrm{true}\mathrm{or}\mathrm{false}.(\mathrm{O}\mathrm{is}\mathrm{the}$$$$\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{cube}.)$$$$\left(1\right)\mathrm{The}\mathrm{moment}\mathrm{of}\mathrm{inertia}\mathrm{of}\mathrm{cube}$$$$\mathrm{about}z-\mathrm{axis}\mathrm{is},I_z=I_x+I_y$$$$\left(2\right)\mathrm{The}\mathrm{moment}\mathrm{of}\mathrm{inertia}\mathrm{of}\mathrm{cube}$$$$\mathrm{about}{z}^{\prime }\mathrm{is},I_{z^{\prime }}=I_z+\frac{{ma}^2}{2}$$$$\left(3\right)\mathrm{The}\mathrm{moment}\mathrm{of}\mathrm{inertia}\mathrm{of}\mathrm{cube}$$$$\mathrm{about}{z}^{″}\mathrm{is},I_{z^{\prime }}=I_z+\frac{{ma}^2}{2}$$$$\left(4\right)I_x=I_y$$

Commented by Tinkutara last updated on 01/Dec/17

Commented by ajfour last updated on 01/Dec/17

$$I_Z=\frac{{ma}^2}{6},I_{Z{}^{\prime }}=\frac{{ma}^2}{6}+\frac{{ma}^2}{2}=\frac{2{ma}^2}{3}$$$$I_X=I_Y=\frac{{ma}^2}{6}+\frac{{ma}^2}{4}=\frac{5{ma}^2}{12}$$$$I_{Z{}^{″}}=?.....$$

Commented by Tinkutara last updated on 02/Dec/17

$$How{I}_z=\frac{{ma}^2}{6}?$$

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