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A spherical hailstone is falling under gravity in still air and due to condensation its radius increases at a rate governed by $\frac{dr} {dt} = kr;$ where $k$ is a constant. Neglecting air resistance show that the hailstone approaches a limiting speed of $ \frac{g}{3k} $.

So, my initial thoughts, given there's no air resistance yet must be an upwards force to balance $ F=mg $ as it has a terminal speed, that there is a force due to the still air upon collision with the hailstone. Now I'm unsure how to find this force. I then went on to think that when the mass of the hailstone is the same as the mass of the still water hit every second and we can then say the same for volume. Then it should be at terminal velocity when volumes are equal. Then when the radius is equal to the area (Which is never) it should be at top speed. Is this true? How could I approach this. Any tips since my logic isn't working?

Leyla Alkan
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Dan D'silva
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3 Answers3

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You have $r=Ae^{kt}$ for the radius of the hailstone as a function of time. You are expected to add the new water at zero velocity and use all the force of gravity to accelerate the new water up to speed. The mass of the hailstone is $\frac 43\pi r^3\rho$ where $\rho$ is the density of water. In a small time $\Delta t$ the impulse delivered by gravity is $mg\Delta t$ and the impulse delivered to the new water is $v \Delta m$ You want to equate these, giving $$v \frac {dm}{dt}=mg\\v\rho4\pi r^2\frac {dr}{dt}=\frac 43 \pi r^3 \rho g\\v=\frac r {3\frac {dr}{dt}}g\\v=\frac g{3k}$$

Ross Millikan
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Newton's second law states that: $$ \frac{d(mv)}{dt}=F$$ here $F=mg$ so: $$m \frac{d v}{dt}+v \frac{d m }{dt} =mg$$.

So if there is a terminal velocity $v_\infty$ you must have: $$ \frac{d m}{dt} v_\infty = mg$$

But denoting $\rho$ the density you have: $$m =\frac{4}{3} \pi \rho r^3$$ so: $$\frac{dm}{dt}=4 \pi \rho r^2 \frac{dr}{dt}=3 k m$$ Finally you obtain: $$3 k m v_\infty=mg$$.

Delta-u
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The mass of the spherical hailstone is $$m=\frac 43\pi r^3\rho$$ where$\rho$ is the density. So $$\frac{dm}{dt}=4\pi r^2\rho \frac{dr}{dt}=3mk$$

Meanwhile the equation of motion is $$\frac{d}{dt}(mv)=mg$$ $$\implies m\frac{dv}{dt}+v\frac{dm}{dt}=mg$$

For the limiting speed we require $\frac{dv}{dt}\rightarrow0$

In which case, $$v(3mk)\rightarrow mg$$

So the terminal velocity is $\frac{g}{3k}$

David Quinn
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