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# phisics problem:a drop with primary radius r0 falls from height and gets moisture from air dm/dt=ks?

drop gets moisture according to its surface dm/ds=ks

s is surface of drop.

find the acceleration of drop ?

find radiuse of drop according to time?

Relevance

In this following, I'll assume that the drop maintains a spherical shape. Otherwise, the math will get horrendous, and one probably can't get an analytical solution.

The surface area of a sphere is given by:

S = 4*pi*r^2

The mass of a sphere of density D is simply the volume times the density. The volume of a sphere with radius r is given by:

V = 4/3 * pi * r^3,

so the mass is given by:

M = 4/3 * pi * r^3 * D.

We know that the drop gains mass according to:

dM/dt = k*S = k*4*pi*r^2

Substitute the expression for the mass as a function of radius into this differential equation:

4/3*pi*D*d(r^3)/dt = k*4*pi*r^2

Simplifying, we get:

d(r^3)/dt = 3*k*r^2/D

let u = r^3

then

du/dt = 3*k*u^(2/3)/D

u^(-2/3) du = 3*k/D dt

Integrate this to obtain

3(u^1/3 -u0^1/3) = 3*k*(t-t0)/D

where u0 and t0 are the constants of integration, and in this problem, we can let t0 = 0.

Substituting back for u and simplifying:

r - r0 = k*t/D

r(t) = k*t/D + r0

This gives the radius as a function of time, which is the desired result.

In the absence of friction and neglecting the change in gravitational acceleration as a function of height, the drop would have a constant acceleration of g (9.807 m/s^2). If we include air friction (and continue to assume a spherical raindrop) then the net acceleration is:

a = g - 0.5 * C * D * A * v^2/M

where C is the drag coefficient (= 1/2 for a sphere),

A is the cross sectional area

M is the mass

v is the velocity

D is the density

Substituting expressions for the mass and area in terms of the radius, we get:

a = g - (0.5 * C * D * pi * r^2 * v^2)/(4/3 * pi * r^3 * D)

a = g - 3/8 * C * v^2/r

if C = 0.5 (for a sphere)

a(t) = dv(t)/dt = g - (3*v(t)^2)/(16*r(t))

From here, one would need to substitute the expression derived previously for r as a function of t, solve the differential equation to find v(t), then take the derivative of v(t) with respect to t to get a(t) in terms of t. This is straightforward, but involves some tedious algebra that I'm not inclined to do at the moment!

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