Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

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?

2 Answers

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  • hfshaw
    Lv 7
    1 decade ago
    Best Answer

    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!

  • alia
    Lv 4
    3 years ago

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