Vt= sqrt (2mg/CpA}
d(m/A)/dr given =4/3
dVt/dr=1/2 * 1/Vt *g/Cp *d(m/A)/dr
= 1/2Vt*g*4/3Cp
= 2g/(3(Cp)Vt)
if I understand your problem statement correctly.
3. Terminal velocity of an object in freefall is given by Vt = root 2gMA/Cdñ
,
where g is gravitational acceleration,Cd is the coefficient of drag, p is air density, and MA is the ratio M/A of mass M to ‘downwards facing’ area A. For a falling sphere of radius r, we calculate that dMA dr = 4/ 3 . Find an expression for dVt /dr .
3 answers
LOL - she sure has tough ones. It took me a while to figure out the landing on an optimal radius plant one.
m = mass = rho * (4/3) pi r^3
A =cross section area = pi r^2
m/A = rho * (4/3) r
d/dr (m/A) = (4/3) rho
NOTE DISAGREE
I guess the density of your sphere is one?
Vt is when
m g = C * p (1/2)Vt^2 A
Vt^2 = 2 m g /(pCA)
Vt = [2 m g /(pCA)]^.5
let MA = m/A
then
Vt = [2 g/pC]^.5 [ m/A]^.5
dVt/dr =[2 g/pC]^.5 *.5 [m/A]^-.5 d/dr [m/A]
but we know
d/dr [m/A] = rho [4/3]
so
dVt/dr =[2 g/pC]^.5 *.5 [m/A]^-.5 (4/3) rho
so
(2/3)rho [2 g/pC]^.5/(rho*(4/3)r)^.5
That can be simplified
A =cross section area = pi r^2
m/A = rho * (4/3) r
d/dr (m/A) = (4/3) rho
NOTE DISAGREE
I guess the density of your sphere is one?
Vt is when
m g = C * p (1/2)Vt^2 A
Vt^2 = 2 m g /(pCA)
Vt = [2 m g /(pCA)]^.5
let MA = m/A
then
Vt = [2 g/pC]^.5 [ m/A]^.5
dVt/dr =[2 g/pC]^.5 *.5 [m/A]^-.5 d/dr [m/A]
but we know
d/dr [m/A] = rho [4/3]
so
dVt/dr =[2 g/pC]^.5 *.5 [m/A]^-.5 (4/3) rho
so
(2/3)rho [2 g/pC]^.5/(rho*(4/3)r)^.5
That can be simplified
2 answers