Question

Six identical square plates are welded to form a hollow cubical box of dimension 10cm. The total mass of the box is 1kg.The box is rotated about an axis passing through its geometrical center and perpendicular to two of the faces. Calculate the moment of inertia.

Answer 1

Answer:   (2/9)*10^-2 kgm^2

Explanation: Dear student

four plates are equidistant to each other from geometrical center,and remaining two plates are parallel.

See the attachment for further process.

Answer 2

Answer:

The ‘moment of inertia’ of the square plates is $\bold{\frac{2}{9} \times 10^{-2} k g m^{2}}$

Given:

Hollow ‘cubical box of dimension’ is 10cm.

The total mass of the box is 1kg.

Explanation:

Step 1:

Plate $6_{1\ and\ 2}$ are parallel.

Hence, moment of Inertia.

$\begin{array}{l}{I_{6}+I_{2}=\frac{m l^{2}}{6}+\frac{m l^{2}}{6}} \\ \\{=\frac{m l^{2}}{3}}\end{array}$

Step 2:

Plate1,3,4,5 are equidistance from geometrical center.

$I_{1}+I_{3}+I_{4}+I_{5}=4 m d^{2}$

So, net moment of inertia $=2 \frac{m l^{2}}{6}+4 \frac{m l^{2}}{6}$

Total mass (M) = 1kg

So, $m=\frac{M}{6 A} \times A=\frac{M}{6}=\frac{1}{6} k g$

Step 3:

$\begin{array}{l}{I_{n e t}=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_{6}} \\ \\{=m\left(\frac{l^{2}}{3}+4 d^{2}\right)} \\ \\{=\frac{1}{6}\left(\frac{16^{2}}{3}+4 \times 5^{2}\right)} \\ \\{=\frac{1}{6}\left(\frac{100}{3}+100\right)}\end{array}$

$\begin{array}{l}{I_{n e t}=\frac{400}{3 \times 6}=\frac{200}{9} \mathrm{kg} \mathrm{cm}^{2}} \\ \\{=\frac{200}{9} \times \frac{1}{10^{4}} \mathrm{kg} \mathrm{cm}^{2}} \\ \\{I_{\text {net}}=\frac{2}{9} \times 10^{-2} \mathrm{kg} \mathrm{cm}^{2}}\end{array}$