Question

Four holes of radius R are cut from a thin square plate of side $4$R and mass M. The moment of inertia of the remaining portion about z=axis is?

• A. $\frac{64-15\pi }{24}M{R}^2$
• B. $\left(\frac{4}{3}-\frac{\pi }{4}\right)M{R}^2$
• C. $\left(\frac{8}{3}-\frac{10\pi }{16}\right)M{R}^2$
• D. $\left(\frac{4}{3}-\frac{\pi }{6}\right)M{R}^2$

Answer 1

The correct option is A $\frac{64-15\pi }{24}M{R}^2$

MI of square $:I_s=\frac{M{l}^2}{6}$

MI of square $:I_d=\frac{M{r}^2}{2}$

Mass of disk produced by holding the square plate $:M_d=\frac{\pi R^2}{{\left(4R\right)}^2}m$

Now, $I_d^0=I_d^{0^{\prime }}+M_d{\left({00}^{\prime }\right)}^2$

$\Rightarrow I_d^0=\frac{M_d{R}^2}{2}+M_d{\left(\sqrt{2}R\right)}^2$

$=\frac{5}{2}{M}_d{R}^2$

$=\frac{5\pi }{32}M{R}^2$

$\therefore$ Total M.I of $4$ disks :

$\Rightarrow I_4=4I_d^0=\frac{5\pi }{8}M{R}^2$

MI of square plate with $l=4R$

$\Rightarrow I_p=\frac{M{\left(4R\right)}^2}{6}=\frac{8}{3}M{R}^2$

M.I of remaining part,

$\Rightarrow I=I_p-I_4=\frac{8}{3}M{R}^2-\frac{5\pi }{8}M{R}^2$

$=\frac{16-15\pi M{R}^2}{24}$

Hence, the answer is $\frac{16-15\pi M{R}^2}{24}.$