91Ó°ÊÓ

Given in the question l =2.00 m.

M =4.00 kg and m =0.300 kg

M= mass of the bar

$I_{total}=I_{bar}+I_{leftball}+I_{rightball}$________(1)

$=\frac{1}{12}M{L}^2+m{r}^2+m{r}^2$

We find r value here

$$\begin{gathered} r=\frac{L}{2} \\ r=\frac{2.00}{2} \\ r=1.00m \end{gathered}$$

Put the value in the (1) equation

$$\begin{gathered} l_{total}=\frac{1}{2}(4.00)(2.00{)}^2+0.300(1.00{)}^2+0.300(1.00{)}^2=\frac{1}{12}×4.00×4+0.300×1+0.300×1 \\ =\frac{1}{12}×16+0.3+0.3 \\ =1.34+0.3+0.3 \\ =1.34+0.6 \\ =1.94kg{m}^2 \end{gathered}$$

Perpendicular to the bar through its center is $1.94kg{m}^2$.

in the given question L =2.00 m.

$$\begin{gathered} r_{left}=0 \\ {r}_{right}=2.00 \end{gathered}$$

We know that, the formula

$I_{total}=I_{bar}+I_{leftball}+I_{rightball}$

$=\frac{1}{3}M{L}^2+m{r_{left}}^2+m{r_{right}}^2$____________(1)

Put the value in equation (1). Then we get

$$\begin{gathered} l_{total}=\frac{1}{3}×\left(4.00\right)×{\left(2.00\right)}^2+0.300{\left(0\right)}^2+0.300{\left(2.00\right)}^2 \\ =\frac{1}{3}×(4.00)×4+0+0.300×4 \\ =\frac{16.00}{3}+1.2 \\ =5.34+1.2 \\ =6.53kg{m}^2 \end{gathered}$$

Perpendicular to the bar through one of the balls $6.53kg{m}^2$.

$$\begin{gathered} leftball=0 \\ rightball=0 \\ bar=0 \end{gathered}$$

Using,$I_{total}=I_{bar}+I_{leftball}+I_{rightball}$

as the bar is uniform, we can assume that the whole M mass is fixed

standing in the middle. as the bar is small and the moment of inertia

The bar is empty.

as the balls can be considered a pile of points and the rotating axis passes through them, their moment of hypocrisy is zero.

Put the value here,

$$\begin{gathered} I_{total}=0+0+0 \\ =0 \end{gathered}$$

in the given question,

d = 0.500 m, M =4.00 and m = 0.300 .

Here we solve our question,

$I_{total}=I_{totalCM}+M{d}^2+m{d}^2+m{d}^2$___________(1)

Now, put the value in equation (1) .then we get

$$\begin{gathered} =0+\left(4.00\right)(0.500{)}^2+(0.300\left)\right(0.500{)}^2+\left(0.300\right)(0.500{)}^2 \\ =0+4.00×0.25+0.300×0.25+0.300×0.25 \\ =1+0.075+0.075 \\ =1.15{\mathrm{kgm}}^2 \end{gathered}$$

Hence, Parallel to the bar and 0.500m form $1.15kg{m}^2$.