Question

A uniform soda can of mass$\mathbf{0}\mathbf{.}\mathbf{140}\mathbf{}\mathbf{kg}$is$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$tall and filled with$\mathbf{0}\mathbf{.}\mathbf{354}\mathbf{}\mathbf{kg}$of soda (Figure 9-41). Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. (a) What is the height hof the com of the can and contents initially and (b) After the can loses all the soda? (c) What happens to has the soda drains out? (d) If xis the height of the remaining soda at any given instant, find x when the com reaches its lowest point.

Short answer

a. Height of center of mass of can and content initially is $6\mathrm{cm}$.

b. Height of center of mass of can and content after it loses all the soda is $6\mathrm{cm}$.

c. h when soda drains out is $6\mathrm{cm}$.

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

Step-by-step solution

Given data

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

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

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

Understanding the concept of center of mass

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:

${\overrightarrow{r}}_{com}=\frac{1}{M}\sum _{i=1}^n{m}_i{r}_i$ … (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.

(a) Determination of the height of the center of mass of a can and content initially

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

(b) Determination of the height of the center of mass of empty can

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.

(c) Determination of the height when soda drains out

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

(d) Determination of the value of x

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$4.2\mathrm{cm}$