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Physics Question on System of Particles & Rotational Motion

Four holes of radius RR are cut from a thin square plate of side 4R4R and mass MM. The moment of inertia of the remaining portion about z-axis is

A

π12MR2\frac{\pi}{12}MR^{2}

B

(43−π4)MR2\left(\frac{4}{3}-\frac{\pi}{4}\right)MR^{2}

C

(43−π6)MR2\left(\frac{4}{3}-\frac{\pi}{6}\right)MR^{2}

D

(83−10π16)MR2\left(\frac{8}{3}-\frac{10\pi}{16}\right)MR^{2}

Answer

(83−10π16)MR2\left(\frac{8}{3}-\frac{10\pi}{16}\right)MR^{2}

Explanation

Solution

If M is mass of the square plate before cutting the holes, then mass of portion of each hole, m=M16R2×πR2=Mπ16m=\frac{M}{16R^{2}}\times\pi R^{2}=\frac{M \pi}{16} ∴ \therefore\, Moment of inertia of remaining portion about Z axis I=Isquare−4IholeI=I_{\text{square}}-4 I_{\text{hole}} =M12(16R2+16R2)−4[mR22+m(2R)2]=\frac{M}{12}\left(16 R^{2}+16 R^{2}\right)-4\left[\frac{mR^{2}}{2}+m\left(\sqrt{2}R\right)^{2}\right] =M12×32R2−10mR2=\frac{M}{12}\times32 R^{2}-10m R^{2} =83MR2−1016MR2π=\frac{8}{3} MR^{2}-\frac{10}{16}MR^{2} \pi I=(83−10π16)MR2.I=\left(\frac{8}{3}-\frac{10\pi}{16}\right) MR^{2}.