# Four holes of radius R are cut from a thin square plate of the side 4R and mass M in XY plane as shown. Then moment of inertia of the remaining portion about z-axis is=? Plz explain clearly!!

## Apr 12, 2018

$\frac{64 - 15 \pi}{24} M {R}^{2}$

#### Explanation: For the $4 R \times 4 R$ square plate, I get ${I}_{p : z z} = \frac{8}{3} M {R}^{2}$ using a standard formula for a rectangular plate:

• ${I}_{z z} = \frac{1}{12} m \left({h}^{2} + {w}^{2}\right)$

That's for rotation around an axis perpendicular to it's centre.

The plate has density $\sigma = \frac{M}{16 {R}^{2}}$ per unit area.

A single disc, therefore, has mass:

${m}_{d} = \sigma \pi {R}^{2} = \frac{\pi}{16} M$

A disc's inertia about it's centre is therefore:

${I}_{d : c m} = \frac{1}{2} {m}_{d} {R}^{2} = \frac{\pi}{32} M {R}^{2}$

Using the parallel axis theorem (see drawing):

${I}_{d : z z} = {I}_{d : c m} + {m}_{d} {d}^{2}$

$= \frac{\pi}{32} M {R}^{2} + \frac{\pi}{16} M {\left(\sqrt{2} R\right)}^{2} = \frac{5 \pi}{32} M {R}^{2}$

So the inertia of the hollowed out plate is:

${I}_{n e t : z z} = {I}_{p : z z} - 4 {I}_{d : z z} = \frac{64 - 15 \pi}{24} M {R}^{2}$