Question

A droplet of perspiration falls off a worker who is standing on a platform 40 m above the
ground. Assume throughout that the drop is spherical (volume of a sphere is equal to 4
3
πr3
while its surface area is equal to 4πr2
). Your objective is to investigate what happens to
the drop as it falls down.
(a) Suppose that as the drop falls, the droplet evaporates at a constant rate. If V is the
volume of the drop, this means that
dV
dt = −c,
where c is a positive constant. Find an expression for dr/dh, that is, the rate of change
of the droplet’s radius with respect to the height h above the ground. (3 marks)
(b) A more realistic model is to suppose instead that the rate of evaporation is proportional
to the surface area of the drop. Find dr/dh now. (3 marks)
(c) Is it possible, in either of these models, that the drop disappears via evaporation just
before hitting the ground?

Answer 1

Answer on Question #61883-Physics-Mechanics-Relativity

A droplet of perspiration falls off a worker who is standing on a platform $40\mathrm{m}$ above the ground. Assume throughout that the drop is spherical (volume of a sphere is equal to $\frac{4}{3}\pi r^3$ while its surface area is equal to $4\pi r^2$). Your objective is to investigate what happens to the drop as it falls down.

(a) Suppose that as the drop falls, the droplet evaporates at a constant rate. If $V$ is the volume of the drop, this means that $\frac{dV}{dt} = -c$, where $c$ is a positive constant. Find an expression for $\frac{dr}{dh'}$ that is, the rate of change of the droplet's radius with respect to the height $h$ above the ground. (3 marks)

(b) A more realistic model is to suppose instead that the rate of evaporation is proportional to the surface area of the drop. Find $\frac{dr}{dh}$ now. (3 marks)

(c) Is it possible, in either of these models, that the drop disappears via evaporation just before hitting the ground?

Solution

(a) The equation of vertical motion in this situation is

$$h = h_0 - \frac{gt^2}{2}.$$ $$\frac{dh}{dt} = -gt \rightarrow \frac{dt}{dh} = -\frac{1}{gt}.$$ $$\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt} \rightarrow -c = 4\pi r^2 \frac{dr}{dt}$$ $$\frac{dr}{dt} = -\frac{c}{4\pi r^2}.$$ $$\int dV = \int -cdt$$ $$V = V_0 - ct$$ $$\frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 - ct$$ $$r = \sqrt[3]{R^3 - \frac{3ct}{4\pi}}.$$ $$\frac{dr}{dt} = -\frac{c}{4\pi \left(R^3 - \frac{3ct}{4\pi}\right)^{\frac{2}{3}}}.$$ $$\frac{dr}{dh} = \frac{dr}{dt} \frac{dt}{dh} = -\frac{c}{4\pi \left(R^3 - \frac{3ct}{4\pi}\right)^{\frac{2}{3}}} \left(-\frac{1}{gt}\right) = \frac{c}{4\pi gt \left(R^3 - \frac{3ct}{4\pi}\right)^{\frac{2}{3}}}.$$

Using $t = \sqrt{\frac{2(h_0 - h)}{g}}$

$$\frac {d r}{d h} = \frac {c}{4 \pi g \sqrt {\frac {2 (h _ {0} - h)}{g}} \left(R ^ {3} - \frac {3 c}{4 \pi} \sqrt {\frac {2 (h _ {0} - h)}{g}}\right) ^ {\frac {2}{3}}}.$$

(b)

$$\frac {d V}{d t} = \frac {d V}{d r} \frac {d r}{d t} \rightarrow - k 4 \pi r ^ {2} = 4 \pi r ^ {2} \frac {d r}{d t}$$ $$\frac {d r}{d t} = - k.$$ $$\frac {d r}{d h} = \frac {d r}{d t} \frac {d t}{d h} = - k \left(- \frac {1}{g t}\right) = \frac {k}{g t}.$$

Using $t = \sqrt{\frac{2(h_0 - h)}{g}}$

$$\frac {d r}{d h} = \frac {k}{g \sqrt {\frac {2 (h _ {0} - h)}{g}}} = \frac {k}{\sqrt {2 g (h _ {0} - h)}}$$

(c) Yes, if the initial radius of a droplet is small enough.

1. $r = \sqrt[3]{R^3 - \frac{3ct}{4\pi}} = 0$, when $t = \sqrt{\frac{2h_0}{g}}$

$$R ^ {3} = \frac {3 c}{4 \pi} \sqrt {\frac {2 h _ {0}}{g}}$$ $$R = \sqrt[3]{\frac {3 c}{4 \pi} \sqrt {\frac {2 h _ {0}}{g}}}$$

2. $r = R - kt = 0$, when $t = \sqrt{\frac{2h_0}{g}}$

$$R = k \sqrt {\frac {2 h _ {0}}{g}}.$$

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