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Q.

Four holes of radius R are cut from a thin square plate of side 4R and mass M .The moment of inertia of the remaining portion about z-axis is $(\frac{x}{3} - \frac{10 π}{16}) {MR}^{2}$ where 'x' is____

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answer is 8.

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Detailed Solution

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Mass of each removed portion $= \frac{M}{16 R^{2}} \times {πR}^{2} = \frac{πM}{16}$

$I_{total} = I_{removed} + I_{remaining} \Rightarrow I_{remaining} = I_{total} - I_{removed}$

$I_{remaining} = \frac{M [{(4 R)}^{2} + {(4 R)}^{2}]}{12} - 4 [\frac{1}{2} . (\frac{πM}{16} {(R)}^{2}) + \frac{Mπ}{16} {(\sqrt{2} R)}^{2}]$

$I_{remaining} = (\frac{8}{3} - \frac{10 π}{16}) {MR}^{2}$

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