dragnewtonian-gravityvelocity
Inspired by this question on Gaming.SE
Using actual in-real-life physics, what would the terminal velocity of a sheep actually be? I would assume it would be around 50m/s, but I might be wrong.
Bonus question:
What would the terminal velocity of a chicken be?
Animal-friendly answers preferred.
If the drag force is being modeled as a linear function of velocity $(\vec{F}_D=-b\vec{v})$, then the problem is straightforward. The vertical force balance for a falling droplet is $$\Sigma F_y=mg-bv=m\dot{v},$$ which gives the following differential equation for the velocity: $$\boxed{\dot{v}+\frac{b}{m}v=g}.$$ In the limiting case of the maximum velocity/zero acceleration $(\dot{v}=0)$, the force balance simplifies to $$mg=bv_{max},$$ or $$\boxed{v_{max}=\frac{mg}{b}}.$$ Going back to our differential equation, if the initial velocity $v(0)=0$, then the solution to this ODE is $$v(t)=\frac{mg}{b}\left[1-e^{-bt/m}\right].$$ By defining the time constant as $\tau=\frac{m}{b}$ and using the definition of the terminal velocity, the time evolution of the velocity simplifies to $$\boxed{v(t)=v_{max}\left[1-e^{-t/\tau}\right]}.$$ The position, if desired, is found easily enough by performing another integration: $$y(t)=\int{v}dt=v_{max}\int{\left(1-e^{-t/\tau}\right)}dt.$$ Assuming that the initial position $y(0)=0$ and simplifying, the solution for vertical position is then $$\boxed{y(t)=v_{max}t+v_{max}\tau\left[e^{-t/\tau}-1\right]}.$$ So we now have analytical solutions for the acceleration, velocity, and position of the falling object as a function of time and the system parameters, all of which are known (except for $b$). Note, however, that the requested time to reach a speed of $0.63v_{max}$ is not arbitrary. After one time-constant has passed, we will have $$\frac{v(\tau)}{v_{max}}=1-e^{-1}=0.63212=\boxed{63.212\%}.$$ Thus, we simply need to calculate the value of the time constant and the resulting value will be your answer. As regards your classmates, they are not wrong. Our goal is to calculate $\tau$, and if you look carefully at our earlier math you will see that $\tau$ does indeed equal the terminal velocity divided by $g$. Octave plots of the position, velocity, and acceleration functions are included below for reference (replace $k$ with $b$ in the second plot).
Yes, the particle would continue to accelerate and would never reach a terminal velocity. But that is not what this equation tells you. This equation tells you what the terminal velocity is, given the parameters of the function. When in a vacuum, there is no terminal velocity. It is not zero, it is not infinity. A terminal velocity literally does not exist and that is exactly what the equation tells you.
Best Answer
That's a very hard question to answer theoretically because the aerodynamic drag, and therefore the terminal velocity, is highly dependant on the shape of the falling object.
Assuming we're in the turbulent regime, the aerodynamic drag is given by the equation:
$$ F_d = \tfrac{1}{2}\rho v^2 C_d A $$
where $\rho$ is the density of the air, $v$ is the velocity, $A$ is the cross sectional area in the direction of motion and $C_d$ is the drag coefficient. The terminal velocity is just the velocity at which the drag force is equal to the weight $mg$, and a quick rearrangement gives:
$$ v_t = \sqrt{\frac{2mg}{\rho C_d A}} $$
In the UK a popular breed is the Texel, which weighs around 80kg. I can't find any info on their size, but from my memories of sheep I'd guess they're about a metre long and the diameter of the body is about half a metre. The legs are pretty spindly, so the cross sectional area is going to be around $A = \pi/16$ m$^2$ if the sheep is falling head first or around 0.5 m$^2$ if it's falling sideways. Putting in the relevant figures we get:
$$ v_t \approx \frac{80}{\sqrt{C_d}} $$
The problem is what value to use for $C_d$. A sphere has $C_d = 0.47$, which would give $v \approx 117$ m/sec. But a cylinder (with sharp rims) has $C_d = 0.82$, which would give $v \approx 88$ m/sec. I'd guess the truth is somewhere in between.
Both these figures are higher than the terminal velocity of a human falling, which varies from around 60 to 90 m/sec depending on your orientation. But then sheep are pretty dense. If you've ever tried to pick one up you were probably surprised by how heavy they seem for their size. Bear in mind the Texel sheep I used as an example weighs 10kg more than I do and I'm 5'10". It's quite reasonable that the terminal velocity of a sheep would be higher than that of a human.