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Four solid spheres each of mass ' m ' and radius ' r ', are placed with their centers at the corners of a square of side ' d '. Then the moment of inertia of the system about the diagonal of the square will be:-

Option: 1

\frac{m}{8}\left[5 r^2+8 d^2\right]


Option: 2

\frac{m}{5}\left[5 r^2+8 d^2\right]


Option: 3

\frac{m}{5}\left[8 r^2+5 d^2\right]


Option: 4

\frac{m}{8}\left[8 r^2+5 d^2\right]


Answers (1)

best_answer

\begin{aligned} I_P=I_R & =\frac{2}{5} m r^2+m\left(\frac{d}{\sqrt{2}}\right)^2 \\ & =\frac{2}{5} m r^2+\frac{m d^2}{2} \\ I_Q=I_S & =\frac{2}{5} m r^2 \\ I_T & =I_P+I_Q+I_{R+} I_S \\ & =2\left[\frac{2}{5} m r^2+\frac{m d^2}{2}\right]+2\left[\frac{2}{5} m r^2\right] \\ & =\frac{8}{5} m r^2+m d^2 \\ & =\frac{m}{5}\left[8 r^2+5 d^2\right] \end{aligned}

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