aerodynamicsdragfluid dynamics
The magnitude of the air resistance for objects with Reynolds numbers greater than 1000 is given by the formula:
Why it does not hold for objects with lower Reynolds numbers? Can I use this equation safely to determine the terminal velocity of the tennis ball falling from the top of a skyscraper and generally for objects travelling in air?
Finally in what case can a drag coefficient have value greater than 1?
Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. According to this paper, the viscous torque on a spinning sphere of radius $R$ in a fluid with viscosity $\eta$ spinning with constant angular velocity $\vec\Omega$ is $$ \vec\tau = -8\pi R^3\eta\vec\Omega $$ The paper goes on to describe how to generalize this relationship to the case of a sphere rotating with arbitrary time-varying angular velocities (under some other assumptions).
If we, however, assume that to good approximation, the viscous torque is linear in the instantaneous angular velocity, then the sphere under the influence of a constant external torque $\vec\tau_0$ will, as you point out, reach terminal angular velocity when the external torque and viscous torque are equal in magnitude but opposite in direction $$ \vec\tau=-\vec\tau_0 $$ This follows from the fact that the net torque $\vec\tau_{\mathrm{net}}$ on the sphere (undergoing fixed axis rotation) is related to its angular acceleration $\vec\alpha$ and moment of inertia $I$ about it's rotation axis by $$ \vec\tau_\mathrm{net} = I\vec\alpha $$ so if the net external torque is zero (which happens when the viscous torque equals the other external torque), then $$ \vec\alpha = 0 $$ the angular acceleration vanishes, so the sphere will no longer speed up in its spinning.
$C_d$ is a function of speed via the Reynolds number. See here and here.
For some numeric values for the $C_d$ vs. $\rm Re$ use the following from here of which you take the log values to interpolate.
The kinematic viscosity of Air is $\nu = 14.8\; \rm{cSt} = 14.8 \cdot 10^{−6}\; \rm{m^2/s}$. At a speed of $v=20\;\rm{m/s}$ a ball with diameter $d=5\,\rm{cm} = 0.05\;\rm{m}$ has Reynolds number of $\rm{Re} = \frac{v d}{\nu} = 67400 $.
When you look at the $C_d$ vs. $\rm Re$ graph you get $C_d = 0.47$. The area of the ball is $A=\pi \frac{d^2}{4} = 0.001964\;\rm{m^2}$ so the drag force is
$F_d = \frac{1}{2} \rho A C_d v^2 = \frac{1}{2} (1.2\;\rm{kg/m^3}) (0.001964\;\rm{m^2}) (0.47) (20\;\rm{m/s})^2 = 0.2215\;\rm{N}$
Best Answer
At low speeds (more precisely low Reynold's numbers) where the flow is laminar or only partly turbulent the drag varies as $v^\alpha$ where $1 \le \alpha \le 2$. Under most conditions the air flow is turbulent and you can assume a $v^2$ dependance. This will certainly be the case for the terminal velocity of a tennis ball. The Reynolds number for a sphere is:
$$ R_e = \frac{vd\rho}{\eta} $$
The density of air is about 1.2kg/m$^{-3}$ and the viscosity about 1.8 $\times$ 10$^{-5}$Pa.s so for a tennis ball 10cm diameter a Reynolds number of 1000 corresponds to 0.17m/s.
The drag equation is phenomenological rather than derived from any rigorous theoretical treatment, and the drag coefficient is basically a fudge factor. It's only approximately constant. For example the drag coefficient of a sphere can vary depending on the speed and can have values greater than 1 at low Reynolds numbers.