91Ó°ÊÓ

If we want to solve our question. we will be applying an equation that determines the moment of inertia

$I=∑m_{i}r{i}_i^2$

Where, l = amoment of the inertia

$m_i=$mass

$r_i=$distance from the axis of the rotation.

The center of the square, perpendicular to its plane

Here, we will assume applying the Pythagorean theorem,

$r^2={\left(\frac{a}{2}+\frac{a}{2}\right)}^2$

First, we will calculate the distance from the axis.

$r^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2$____________(1)

Put the value in equation (1)

$$\begin{gathered} ={\left(\frac{0.400}{2}\right)}^2+{(0.400/2)}^2={(0.200m)}^2+{(0.200m)}^2=(0.04)(0.40) \\ =0.08m^2 \\ =0.0800m^2 \end{gathered}$$

Now we making an assumption that is four masses.

Now here we will calculate the moment of the inertia is,

$$\begin{gathered} I=4×(0.200kg×0.0800m^2) \\ =14×0.016 \\ =0.0640km.m^2 \end{gathered}$$

We will assume since of the rotation is placed at the AB line.

Here moment of inertia will be changed distance rotation squared will now.

$r^2={\left(\frac{a}{2}\right)}^2$

First, we will calculate distance,

$$\begin{gathered} r^2={\left(\frac{0.400m}{2}\right)}^2 \\ =\left(\frac{0.16}{4}\right) \\ =0.040m^2 \end{gathered}$$

Here we calculate the moment of inertia

$$\begin{gathered} I=4×(0.200kg×0.0400m^2) \\ =4×0.008 \\ =0.032kg.m^2 \end{gathered}$$

We will assume that since the place of the axis of the rotation is changed, the distance square will now be,

First, we will calculate that distance squared as,

$r^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2$

We also need to find the value of,$I=2×(m×r^2)$

Put a = 4.00

First, we will calculate that distance squared as,

$$\begin{gathered} r^2={\left(\frac{0.400}{2}\right)}^2+{\left(\frac{0.400}{2}\right)}^2 \\ =0.0800m^2 \end{gathered}$$

We will calculate for two masses as,

$$\begin{gathered} I=2×(0.200kg×0.0800m^2) \\ =2×0.016 \\ =0.0320kg.m^2 \end{gathered}$$