1. Atmospheric drag is proportional to Su2 where S is surface area and v is velo

Question

1. Atmospheric drag is proportional to Su2 where S is surface area and v is velocity. Kinetic energy is proportioanl to mu2 where m is mass. Terminal velocity is that velocity where the drag force balances the force due to gravity (mg). (a) Show that the terminal velocity, v scales like m1/6 for similarly shaped objects. (b) Show that the kinetic energy per unit area scales as m2/3. (c) Use this to comment on what happens to small vs large animals if they fall down a great height. (d) Show that the height at which terminal velocity is reached scales like m1/3 (Hint recall from HS physics that potential energy is mgh.)

Explanation / Answer

1. given drag, D = -Skv^2

S is surface area

v is velocity

KE = 0.5mv^2

for an object moving under influence of gravity

total energy = TE

d(TE)/dt = Dv = -Skv^3 = TE'

now

TE = KE + PE

PE = mgy

y' = -v

hence

PE' + KE' = TE'

mgv + KE' = -Skv^3

KE' = -Skv^3 + mgv

a. at terminal velocity KE' = 0

Skv^2 = mg

v = sqrt(mg/Sk)

now for similiarly shaped objects

considering constant density

m/S^3/2 = m'/S'^3/2

hence

we can say

S = k'm^2/3

hence

v = sqrt(m^1/3 * g/k'*k)

hencev scales as m^1/6 for similiarly shaped objects

b. KE per unit area = k*mv^2/S

v^2 scales as m^1/3,

hence

KE/S scales as m^(1/3 + 1 - 2/3) = m^2/3

c. if small/large animals fall from great height

energy per unit area is much more for large animals than the small animals and hence they have more danger of getting squishjed upon impact

d. terminal velocity is reached at height h

now, at terminal velocity

PE' = -Skv^2 = mgv

hence

v scales at m^1/6 ( terminal velocity)

hence

PE would scale by v^2 = m^2/6 = m^1/3

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