Question

28)Four small spheres, each of which you can regard as a point of mass of 0.200 kg, are arranged in a square of 0.400 m on a side and connected by extremely light rods (Fig. E9.28). Find the moment of inertia of the system about an axis

(a) through the center of the square, perpendicular to its plane (an axis through point O in the figure);

(b) bisecting two opposite sides of the square (an axis along the line AB in the figure);

(c) that passes through the centers of the upper left and lower right spheres and point O.

Short answer

(a) the center of the square, perpendicular to its plane is 0.0640km.m²

(b) Bisecting two opposite sides of the square is $0.032kg.m^2$.

(c) The center of the upper left and lower right sphere and through point o is $0.032kg.m^2$.

Step-by-step solution

Step:-1 explanation 

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

$I=\sum m_{i}r{i}_i^2$

Where, l = amoment of the inertia

$m_i=$mass

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

Step:-2 Concept

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)

Step:-3 calculation 

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,

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

Step:-4 explanation 

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.

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

Step:-5 calculation            

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

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

Step:-6 explanation  

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

Step:-7 concept

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,$I=2\times (m\times r^2)$

Step:-8 calculation 

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,

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