91Ó°ÊÓ

Mass of can, $\mathrm{M}=0.14\mathrm{kg}$

Mass of soda, $\mathrm{m}=0.354\mathrm{kg}$

Height of can,$\mathrm{H}=12\mathrm{cm}$.

For a system of particles, the whole mass of the system is concentrated at the center of mass of the system.

The expression for the coordinates of the center of mass are given as:

${\stackrel{→}{r}}_{com}=\frac{1}{M}\underset{i=1}{\overset{n}{∑}}{m}_i{r}_i$ … (i)

Here, $\mathrm{M}$is the total mass, ${\mathrm{m}}_1$is the individual mass of ith particle and ${\mathrm{r}}_{\mathrm{i}}$ is the coordinates of ithparticle.

As the can is made of uniform material then center of mass is at its geometrical center. So, it will be at$\frac{\mathrm{H}}{2}$

When the can is full of soda, height of soda is H and its center of mass will be at$\frac{\mathrm{H}}{2}$

Let M be the mass of the can, m be the mass of soda. When the can is full, the height of center of mass is,

$$\begin{gathered} \mathrm{h}=\frac{\left[\mathrm{M}\left(\frac{\mathrm{H}}{2}\right)+\mathrm{m}\left(\frac{\mathrm{H}}{2}\right)\right]}{\mathrm{M}+\mathrm{m}} \\ =\frac{\mathrm{H}}{2} \\ =\frac{12.0\mathrm{cm}}{2} \\ =6.0\mathrm{m} \end{gathered}$$

Thus, the center of mass is located at$6.0\mathrm{cm}$above of the base

When the can is empty, the center of mass will be at the same point, 6 cm above the vertical and along with the vertical axis symmetry.

As x decreases, the center of mass of the soda in the can at first drops and then rises to$\frac{\mathrm{H}}{2}=6.0\mathrm{cm}$

Let ${\mathrm{m}}_{\mathrm{p}}$is the mass of the soda in the can when its top surface is x above the base of the can. Then,

${\mathrm{m}}_{\mathrm{p}}=\mathrm{m}\left(\frac{\mathrm{X}}{\mathrm{H}}\right)$

The height of the center of mass of soda alone is$\frac{\mathrm{x}}{2}$above the base of the can.

Hence,

$$\begin{gathered} h=\frac{\left[\mathrm{M}\left({\displaystyle \frac{\mathrm{H}}{2}}\right)+{\mathrm{m}}_{\mathrm{p}}\left({\displaystyle \frac{\mathrm{x}}{2}}\right)\right]}{\mathrm{M}+{\mathrm{m}}_{\mathrm{p}}} \\ =\frac{\left[\mathrm{M}\left({\displaystyle \frac{\mathrm{H}}{2}}\right)+\mathrm{m}\left({\displaystyle \frac{\mathrm{x}}{\mathrm{H}}}\right)\left({\displaystyle \frac{\mathrm{x}}{2}}\right)\right]}{\mathrm{M}+\left({\displaystyle \frac{\mathrm{mx}}{\mathrm{H}}}\right)} \\ =\frac{{\mathrm{MH}}^2+{\mathrm{mx}}^2}{2(\mathrm{MH}+\mathrm{mx})} \end{gathered}$$

For the lowest position of the center of mass,

$$\begin{gathered} \frac{dh}{dx}=0 \\ \frac{2\mathrm{mx}}{2(\mathrm{MH}+\mathrm{mx})}-\frac{({\mathrm{MH}}^2+{\mathrm{mx}}^2)\mathrm{m}}{2{(\mathrm{MH}+\mathrm{mx})}^2}=0 \\ \frac{{\mathrm{m}}^2{\mathrm{x}}^2+2\mathrm{MmHx}-{\mathrm{MmH}}^2}{2{(\mathrm{MH}+\mathrm{mx})}^2}=0 \\ {\mathrm{m}}^2{\mathrm{x}}^2+2\mathrm{MmHx}-{\mathrm{MmH}}^2=0 \end{gathered}$$

The solution of localid="1654491409859" ${\mathrm{m}}_2{\mathrm{x}}_2+2\mathrm{MmHx}-{\mathrm{MmH}}^2=0$is,

$\mathrm{x}=\frac{\mathrm{MH}}{\mathrm{m}}\left(-1+\sqrt{1+\left(\frac{\mathrm{m}}{\mathrm{M}}\right)}\right)$

Since x must be positive therefore only positive root is taken.

$$\begin{gathered} \mathrm{h}=\frac{{\mathrm{MH}}^2+{\mathrm{mx}}^2}{2(\mathrm{MH}+\mathrm{mx})} \\ =\frac{\mathrm{HM}}{\mathrm{M}}\left(\sqrt{1+\left(\frac{\mathrm{m}}{\mathrm{M}}\right)}-1\right) \\ =\frac{\left(12\right)(0.14)}{0.354}\left(\sqrt{1+\left(\frac{0.354}{0.14}\right)}-1\right) \\ =4.2\mathrm{cm} \end{gathered}$$

Therefore, if x is the height of the remaining soda at any instant, then x when the center of mass reaches its lowest point is$4.2\mathrm{cm}$