Question

Four holes of radius $R$ are cut from a thin square plate of side $4R$ and mass $M$.Determine inertia of the remaining portion about z-axis.

• A. $\left[\frac{10\pi }{16}-\frac{8}{3}\right]M{R}^2$
• B. $\left[\frac{8\pi }{3}-\frac{10}{16}\right]M{R}^2$
• C. $\left[\frac{10\pi }{3}-\frac{10}{16}\right]M{R}^2$
• D. $\left[\frac{10\pi }{16}-\frac{10}{3}\right]M{R}^2$

Answer 1

The correct option is A $\left[\frac{10\pi }{16}-\frac{8}{3}\right]M{R}^2$

We know that,

$\sigma =\frac{M}{16R^2}$ per unit area

A single disc

$m_d=\sigma \pi R^2=\frac{\pi }{16}M$

$I_{d}cm=\frac{1}{2}md{R}^2=\frac{\pi }{32}M{R}^2$

Using the parallel axis theorem,

$I_d:zz=T_d:cm+m_d{d}^2=\frac{\pi }{32}m{R}^2+\frac{\pi }{16}M(\sqrt{2}R{)}^2=\frac{5\pi }{32}M{R}^2$

So, the inertia of the sollowed.

$I_{net}:zz=I_p:zz-4I_d:zz=\frac{64-15\pi -15\pi }{24}M{R}^2$