91Ó°ÊÓ

Given that a machine part consists a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m.

The longer bar has a small but dense 2.00-kg ball at one end.

The vertical bar is pivoted counterclockwise through to make the entire part horizontal.

Let object 1 is horizontal bar, object 2 is vertical bar and object 3 be the ball. Let $\left({\mathrm{x}}_{\mathrm{i}},{\mathrm{y}}_{\mathrm{i}}\right)$and $\left({\mathrm{x}}_{\mathrm{f}},{\mathrm{y}}_{\mathrm{f}}\right)$be the coordinates of the bar before and after the vertical bar is pivoted.

$$\begin{gathered} {\mathrm{m}}_1=4\mathrm{kg} \\ {\mathrm{m}}_2=3\mathrm{kg} \\ {\mathrm{m}}_3=2\mathrm{kg} \end{gathered}$$

Center of mass is

${\mathbf{x}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{3}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}}$

Where${\mathbf{m}}_{\mathbf{i}}{\mathbf{,}}_{\mathbf{s}}$ are masses and${\mathbf{x}}_{\mathbf{i}}{\mathbf{,}}_{\mathbf{s}}$ are positions.

Find the coordinates$\left({\mathrm{x}}_{\mathrm{i}},{\mathrm{y}}_{\mathrm{i}}\right)$and$\left({\mathrm{x}}_{\mathrm{f}},{\mathrm{y}}_{\mathrm{f}}\right)$using center of mass

$$\begin{gathered} {\mathrm{x}}_{\mathrm{i}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3} \\ =\frac{\left(4\mathrm{kg}\right)\left(0.750\mathrm{m}\right)+0+0}{4\mathrm{kg}+3\mathrm{kg}+2\mathrm{kg}} \\ =0.333\mathrm{m} \end{gathered}$$

$$\begin{gathered} {\mathrm{y}}_{\mathrm{i}}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{m}}_3{\mathrm{y}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3} \\ =\frac{0+\left(3\mathrm{kg}\right)\left(0.9\mathrm{m}\right)+\left(2\mathrm{kg}\right)\left(1.8\mathrm{m}\right)}{4\mathrm{kg}+3\mathrm{kg}+2\mathrm{kg}} \\ =0.7\mathrm{m} \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}_{\mathrm{f}}=\frac{\left(4\mathrm{kg}\right)\left(0.750\mathrm{m}\right)+\left(3\mathrm{kg}\right)\left(-0.9\right)+2\mathrm{kg}\left(-1.8\mathrm{m}\right)}{4\mathrm{kg}+3\mathrm{kg}+2\mathrm{kg}} \\ =-0.366\mathrm{m} \\ {\mathrm{y}}_{\mathrm{f}}=0 \end{gathered}$$

Now,${\mathrm{x}}_{\mathrm{f}}-{\mathrm{x}}_{\mathrm{i}}=-0.7\mathrm{m}$

And ${\mathrm{y}}_{\mathrm{f}}-{\mathrm{y}}_{\mathrm{i}}=-0.7\mathrm{m}$

Hence, the center of mass moves 0.7 m to the right and 0.7 m upward.