Question

The masses and coordinates of four particles are as follows:${}^{\mathbf{50}\mathbf{g}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}}$;${}^{\mathbf{25}\mathbf{}\mathbf{g}\mathbf{,}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}}$;${}^{\mathbf{25}\mathbf{}\mathbf{g}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{c}\mathbf{m}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{c}\mathbf{m}\mathbf{;}}$${}^{\mathbf{30}\mathbf{}\mathbf{g}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}}$.

What are the rotational inertias of this collection about the (a) x, (b) y, and (c) z axes? (d) Suppose that we symbolize the answers to (a) and (b) as A and B, respectively. Then what is the answer to (c) in terms of A and B.

Short answer

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

b. The rotational inertia about y-axis is${}^{5.5\times {10}^{2{\mathrm{gcm}}^2}}$.

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

d. If answer a) is A and b) is B) then answer c) in terms of A and B is${}^{\mathrm{A}+\mathrm{B}}$

Step-by-step solution

Understanding the given information

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 ${}^{\mathrm{x}=-2.0\mathrm{cm},\mathrm{y}=4.0\mathrm{cm}}$.

Concept and Formula used for the given question    

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,

$l=\sum _{}^{}{m}_i{r_i}^2$(i)

(a) Calculation for the rotational inertia of the x-axis

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

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

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

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.

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

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

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).

$l_z=\sum _{}^{}{m}_i{y_i}^2$

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

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

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

(d) The answer to (c) in terms of A and B

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 $l_z$.

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

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