0
$\begingroup$

As you know if I want to find the force for an accelerated object I will use the law $F_o=ma$ so I can get the affecting force of it.

But there is another force affecting against the object. It's the air resistance force, So I will have to calculate the drag force (Air resistance force) and the subtract it from the accelerated object force to get the exact affecting force $F_{net}=F_o-F_d$ while $F_o$ is accelerated object force and $F_d$ is drag force (Air resistance force).

Now, how can I calculate the drag force (Air resistance force)?

OK, Actually I know the formula below:

$F_d=\dfrac{\rho\,\nu^2AC_d}{2}$

While: $\rho$ is the air density, $\nu$ is the speed of the object relative to the air, $A$ is the cross sectional area of the object and $C_d$ is the drag coefficient.

So my problem is: I don't know what is the $A$ cross sectional area of the sphere (in my state I used a ball) and I don't know what is the $C_d$ drag coefficient of a sphere (ball)?

$\endgroup$
3
$\begingroup$

$C_d$ is a function of speed via the Reynolds number. See here and here.

enter image description 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.

Example

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

$F_d = \frac{1}{2} \rho A C_d v^2 = \frac{1}{2} (1.2\;\rm{kg/m^3}) (0.001964\;\rm{m^2}) (0.47) (20\;\rm{m/s})^2 = 0.2215\;\rm{N}$

$\endgroup$
1
$\begingroup$

If the density $\rho _{ball} $ of the ball (or, equivalently, the name of the material from which it is made) is known, it is possible to find its volume $V $ via:

$$\rho_{ball}=\frac {m}{V}$$

Once $ V $ is obtained, it is possible to find the ball's radius $ r $ via:

$$ V=\frac {4}{3}\pi r^3$$

Once the expression for $ r $ is obtained, it is possible to substitute it into the expression for $ A $:

$$A=\pi r^2 $$

According to this Wikipedia article, the drag coefficient for a shpere is $0.47 $.

$\endgroup$

protected by AccidentalFourierTransform Aug 11 '18 at 14:01

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.