Mechanics limiting speed with variable radius.

Question

Okay so I'm trying to solve this problem and have ran into some difficulties.

Using impulse change of momentum principles I managed to figure out that the equation of motion for the hailstone is

$$\displaystyle \frac{dv}{dt}+\frac{v}{m}\frac{dm}{dt}=g$$ however I don't know how to work out $\displaystyle \frac{dm}{dt}$ I thought it must have something to do with the differential equation given about the radius in the question but I couldn't figure it out and got stuck.

Presumably when I figure out $$\displaystyle \frac{dm}{dt}$$ I can just solve to find $\displaystyle v(t)$and take the limit as $\displaystyle t \rightarrow \infty$ and the answer should drop out.

Any help?

QUESTION:

Answer 1

Observe that you can solve the equation for the increase of the radius by $r(t) = r_0 e^{kt}$ where $r_0$ is some initial radius (this won't be important later on). Assuming that the hailstone has a uniform mass density $\rho$ we can then give that

$$ m(t) \;\; =\;\; \rho \frac{4\pi}{3} r^3(t) \;\; =\;\; \rho\frac{4\pi}{3} r_0^3 e^{3kt}. $$

Observe that

$$ \frac{dm}{dt} \;\; =\;\; \rho \frac{4\pi}{3} r_0^3 \frac{d}{dt}e^{3kt} \;\; =\;\; 3k \left (\rho \frac{4\pi}{3} r_0^3 e^{3kt} \right ) \;\; =\;\; 3k m(t). $$

Therefore the term

$$ \frac{v}{m} \frac{dm}{dt} \;\; =\;\; 3kv $$

and the differential equation becomes

$$ \frac{dv}{dt} + 3kv \;\; =\;\; g. $$

We can solve this using an integrating factor. The rest should be straightforward from there.

Answer 2

Assuming spherical hailstones with density $\rho_h$, you can write the mass in terms of the radius as: $$ m(t) = (\text{density})\times(\text{volume})=\frac{4}{3}\pi\rho_h r^3(t), $$ so, differentiating both sides and using the chain rule you get that $$ \frac{dm}{dt} = 4\pi\rho_h r^2(t)\frac{dr}{dt}=4\pi\rho_h k r^3(t) = 3km(t). $$