91Ó°ÊÓ

Masses and coordinates of four particles are,

1. Mass of first particle${}^{{\mathrm{m}}_1=50\mathrm{g}}$, coordinate are ${}^{\mathrm{x}=2.0\mathrm{cm},\mathrm{y}=2.0\mathrm{cm}}$.
2. Mass of second particle ${}^{{\mathrm{m}}_2=25\mathrm{g}}$, coordinate are ${}^{\mathrm{x}=0\mathrm{cm},\mathrm{y}=4.0\mathrm{cm}}$.
3. Mass of third particle ${}^{{\mathrm{m}}_3=25\mathrm{g}}$, coordinate are${}^{\mathrm{x}=-3.0\mathrm{cm},\mathrm{y}=-3.0\mathrm{cm}}$.
4. Mass of fourth particle ${}^{{\mathrm{m}}_4=30\mathrm{g}}$,coordinate are ${}^{\mathrm{x}=-2.0\mathrm{cm},\mathrm{y}=4.0\mathrm{cm}}$.

By using the concept of moment of inertia you can calculate the moment of inertia about its axis. The moment of inertia of the system is the mass times perpendicular distance square at that point.

The formula for the moment of inertia is written as,

$l=\underset{}{\overset{}{∑}}{m}_i{r_i}^2$(i)

Substitute the values of mass and distances in equation (i) to calculate the moment of inertia about the x-axis.

$$\begin{gathered} l_x=\underset{}{\overset{}{∑}}{m}_i{y_i}^2 \\ =\left[50.0\mathrm{g}{(2.0\mathrm{cm})}^2+25.0\mathrm{g}{(4.0\mathrm{cm})}^2+25.0\mathrm{g}{(-3\mathrm{cm})}^2+30.0\mathrm{g}{(4.0\mathrm{cm})}^2\right] \\ {\text{=1305g.cm}}^2 \\ =1.3×{10}^3\mathrm{g}.{\mathrm{cm}}^2 \\ {l}_x=\left[50.0{(2.0)}^2+25.0{(4.0)}^2+25.0{(-3)}^2+30.0{(4.0)}^2\right] \\ {l}_x=200+400+225+480 \\ {l}_x=1305\mathrm{g}.{\mathrm{cm}}^2 \\ {l}_x=1.3×{10}^3\mathrm{g}.{\mathrm{cm}}^2 \end{gathered}$$

The rotational inertia about the x-axis is${}^{1.3×{10}^3\mathrm{g}.{\mathrm{cm}}^2}$.

Step 3: (b) Calculation for the rotational inertia about y axis

Substitute the values of mass and distances in the equation (i) calculate moment of inertia about y-axis.

$$\begin{gathered} l_y=\underset{}{\overset{}{∑}}{m}_i{y_i}^2 \\ =\left[50.0\mathrm{g}{(2.0\mathrm{cm})}^2+25.0\mathrm{g}{(0\mathrm{cm})}^2+25.0\mathrm{g}{(-3\mathrm{cm})}^2+30.0\mathrm{g}{(-2.0\mathrm{cm})}^2\right] \\ {\text{=545g.cm}}^2 \\ =5.5×{10}^2\mathrm{g}.{\mathrm{cm}}^2 \\ {l}_y=\left[50.0{(2.0)}^2+25.0{\left(0\right)}^2+25.0{(-3)}^2+30.0{(-2.0)}^2\right] \\ {l}_y=200+0+225+120 \\ {l}_y=545\mathrm{g}.{\mathrm{cm}}^2 \\ {l}_y=5.5×{10}^2\mathrm{g}.{\mathrm{cm}}^2 \end{gathered}$$

The rotational inertia about the y-axis is${}^{5.5×{10}^2\mathrm{g}.{\mathrm{cm}}^2}$.

Step 3: (c) Calculation for the rotational inertia about z axis

Using Pythagoras theorem, find the value of distance about z-axis and substitute in equation (i).

$l_z=\underset{}{\overset{}{∑}}{m}_i{y_i}^2$

where, $\sqrt[]{\left({x_i}^2+{y_i}^2\right)}$is the perpendicular distance of the particle from ${}^z$axis

$$\begin{gathered} l_{\mathrm{z}}=\underset{}{\overset{}{∑({\mathrm{m}}_{\mathrm{i}}}}{\mathrm{xi}}^2+{\mathrm{m}}_{\mathrm{i}}{\mathrm{yi}}^2) \\ =\left[50.0\mathrm{g}{(2.0\mathrm{cm})}^2+25.0\mathrm{g}{(4.0\mathrm{cm})}^{2}25.0\mathrm{g}{(-3\mathrm{cm})}^2+30.0\mathrm{g}{(4.0\mathrm{cm})}^2\right] \\ =\left[50.0\mathrm{g}{(2.0\mathrm{cm})}^2+25.0\mathrm{g}{(0\mathrm{cm})}^{2}25.0\mathrm{g}{(-3\mathrm{cm})}^2+30.0\mathrm{g}{(-2.0\mathrm{cm})}^2\right] \\ =1850\mathrm{g}.{\mathrm{cm}}^2 \\ =1.9×{10}^3\mathrm{g}.{\mathrm{cm}}^2 \end{gathered}$$

The rotational inertia about the z-axis is${}^{1.9×{10}^3\mathrm{g}.{\mathrm{cm}}^2}$

To calculate the rotational inertia in terms of $l_x$and $l_y$, use the equations for $l_x$and $l_y$ in the equation for $l_z$.

$$\begin{gathered} l_z=\underset{}{\overset{}{∑}}{m}_i({x_i}^2+{y_i}^2) \\ =\underset{}{\overset{}{∑}}{m}_i{x_i}^2+\underset{}{\overset{}{∑}}{m}_i{y_i}^2 \\ =l_y+l_x \\ =A+B{l}_z \\ {l}_z=\underset{}{\overset{}{∑(l_x}}+l_y) \\ {l}_z=A+B \end{gathered}$$

Therefore, the rotational inertia about z-axis is equal to the sum of rotational inertial about x and y axis.