Aerodynamic Drag

Calculus Level 5

An object falling vertically downwards is experiencing two forces: air resistance and gravity.

Suppose that the position p p of the falling object obeys the following differential equation: p ′ ′ ( t ) + k m ( p ′ ( t ) ) 2 − g = 0 , p''(t)+\frac{k}{m}{\left(p'(t)\right)}^2-g=0, where k , m , g k,m,g are constants representing the coefficients of air resistance, the mass of the falling object, and the acceleration due to gravity, respectively, and t t represents time.

Given that, at time t = 0 , t=0, velocity p ′ ( 0 ) = 0 p'(0)=0 and position p ( 0 ) = 0 p(0)=0 , one can find the position of the falling object at any time t t with the following equation: p ( t ) = m k ln ⁡ ( 1 A ( e B t k g m + C ) ) − D t m g k + E , p(t)=\frac{m}{k}\ln{\left(\frac{1}{A}\Big(e^{Bt\sqrt{\frac{kg}{m}}}+C\Big)\right)}-Dt\sqrt{\frac{mg}{k}}+E, where A , B , C , D , E A,B,C,D,E are non-negative integers.

Find A + B + C + D + E A+B+C+D+E .

Bonus:

  1. Find the object's terminal velocity, or lim ⁡ t → ∞ p ′ ( t ) \displaystyle\lim_{t\to\infty}p'(t) .
  2. The above differential equation only works when there is turbulence involved (typical for objects at relatively high speeds). Otherwise, a better approximation would be p ′ ′ ( t ) + k m p ′ ( t ) − g = 0 p''(t)+\frac{k}{m}p'(t)-g=0 . Find p ( t ) p(t) , p ′ ( t ) p'(t) , and lim ⁡ t → ∞ p ′ ( t ) \displaystyle\lim_{t\to\infty}p'(t) for this case, assuming p ′ ( 0 ) = 0 p'(0)=0 and p ( 0 ) = 0 p(0)=0 .

If you enjoyed this problem, you might also want to try this one .


The answer is 6.

Nick Turtle by Nick Turtle , 21, USA

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1 solution

Nick Turtle
May 11, 2018

The solution is A = 2 B = 2 C = 1 D = 1 E = 0 \begin{aligned} A&=2\\ B&=2\\ C&=1\\ D&=1\\ E&=0 \end{aligned}

Starting from the original differential equation, we can reduce the order by substituting v = p ′ ( t ) v=p'(t) and v ′ = p ′ ′ ( t ) v'=p''(t) . Then, we can solve it simply like a separable equation:

v ′ ( t ) + k m v 2 ( t ) − g = 0 v ′ ( t ) = g − k m v 2 ( t ) d t = d v g − k m v 2 ( t ) = 1 2 g ( 1 1 + k m g v ( t ) + 1 1 − k m g v ( t ) ) d v t = 1 2 m k g ln ⁡ ( 1 + k m g v ( t ) 1 − k m g v ( t ) ) + C C e 2 t k g m = 1 + k m g v ( t ) 1 − k m g v ( t ) \large \begin{aligned} v'(t)+\frac{k}{m}v^2(t)-g&=0\\ v'(t)&=g-\frac{k}{m}v^2(t)\\ dt&=\frac{dv}{g-\frac{k}{m}v^2(t)}\\ &=\frac{1}{2g}\left(\frac{1}{1+\sqrt{\frac{k}{mg}}v(t)}+\frac{1}{1-\sqrt{\frac{k}{mg}}v(t)}\right)dv\\ t&=\frac{1}{2}\sqrt{\frac{m}{kg}}\ln{\left(\frac{1+\sqrt{\frac{k}{mg}}v(t)}{1-\sqrt{\frac{k}{mg}}v(t)}\right)}+C\\ Ce^{2t\sqrt{\frac{kg}{m}}}&=\frac{1+\sqrt{\frac{k}{mg}}v(t)}{1-\sqrt{\frac{k}{mg}}v(t)} \end{aligned}

Now, solving for v ( t ) v(t) gives

v ( t ) = g m k C e 2 t k g m − 1 C e 2 t k g m + 1 \large v(t)=\sqrt{\frac{gm}{k}}\frac{Ce^{2t\sqrt{\frac{kg}{m}}}-1}{Ce^{2t\sqrt{\frac{kg}{m}}}+1}

Apply the condition p ′ ( 0 ) = v ( 0 ) = 0 p'(0)=v(0)=0 to get that C = 1 C=1 :

v ( t ) = g m k e 2 t k g m − 1 e 2 t k g m + 1 \large v(t)=\sqrt{\frac{gm}{k}}\frac{e^{2t\sqrt{\frac{kg}{m}}}-1}{e^{2t\sqrt{\frac{kg}{m}}}+1}

Note here that the terminal velocity lim ⁡ t → ∞ v ( t ) = lim ⁡ t → ∞ p ′ ( t ) = g m k \displaystyle\lim_{t\to\infty}v(t)=\lim_{t\to\infty}p'(t)=\sqrt{\frac{gm}{k}} .

Now, simply integrate v ( t ) = p ′ ( t ) v(t)=p'(t) (substitute u = e 2 t k g m u=e^{2t\sqrt{\frac{kg}{m}}} ) to get that

p ( t ) = m k ln ⁡ ( e 2 t k g m + 1 ) − t m g k + D \large p(t)=\frac{m}{k}\ln\left(e^{2t\sqrt{\frac{kg}{m}}}+1\right)-t\sqrt{\frac{mg}{k}}+D

Since p ( 0 ) = 0 p(0)=0 , we have that D = − m k ln ⁡ ( 2 ) D=-\frac{m}{k}\ln\left(2\right) .

Thus, the final answer is

p ( t ) = m k ln ⁡ ( 1 2 ( e 2 t k g m + 1 ) ) − t m g k \large p(t)=\frac{m}{k}\ln\left(\frac{1}{2}\left(e^{2t\sqrt{\frac{kg}{m}}}+1\right)\right)-t\sqrt{\frac{mg}{k}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Air resistance being modelled as proportional to v 2 v^2 tends to represent more turbulent air flow, while air resistance proportional to v v would represent less turbulent flow. This may be due to the shape of the object or the speed at which the particle travels, but it is not directly due simply to the mass of the object, as you state in the question.

Mark Hennings - 3 years ago

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Ok, thanks for pointing that out. I changed it in the question.

I guess I misunderstood Wikipedia when it stated that "Terminal velocity for very small objects (pollen, etc.) at low Reynolds numbers is determined by Stokes law" by thinking that "small" stood for low mass instead of volume…

Nick Turtle - 3 years ago

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