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If I drop equal volume of cotton and rock from a mountain, which one of them will have higher air resistance? I think it should be cotton, but the fact that they are both equivalent in volume confuses me.

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    $\begingroup$ Please view my updated and detailed answer below. $\endgroup$
    – Programmer
    Nov 15 '21 at 15:10
  • $\begingroup$ I am not clear that anyone answers the actual question, which does not mention acceleration, gravity or terminal velocity, and is specifically about air resistance. So forget the mountain, and consider the objects tethered in a (horizontal) wind tunnel. Assuming both objects have the same shape (such as a sphere of equal radius), does the texture of the surface make a significant difference to the air resistance, and why ? $\endgroup$ Nov 16 '21 at 9:38
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From Stoke's law the air resistance acting on a sphere of radius $r$ is is $$F=6\pi r \eta v$$ where $\eta$ is the viscosity of air and $v$ is velocity.

If the cotton and rock had the same volume, hence radius, the air resistance would be the same.

However the 'terminal velocity' occurs when the velocity is such that the air resistance matches the weight and so is much lower for cotton.

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    $\begingroup$ It really should be noted that the Stokes' law is only applicable to laminar flow which is very unlikely when dropping a macroscopic object from a mountain. Maybe for a small piece of cotton, but a piece of rock would have to be very small. It would not be called a rock, but a sand grain. $\endgroup$
    – Vladimir F
    Nov 16 '21 at 10:12
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You can model the drag force by $$F_{drag} = 1/2*\rho v^2C_DA$$ where v is the velocity, p is the density of air, A is the cross sectional area, and $C_D$ is the drag coefficient.

Hence, the net force is $$F_{net}=ma=-mg+\frac{\rho v^2C_DA}{2}$$

Therefore our final equation of motion (acceleration), $$a = -g+\frac{\rho v^2C_DA}{2m}$$

Cotton has a density of $448kg/m^3$, and rocks generally have densities between $1600kg/m^3$ and $3500kg/m^3$. Let's take the worst case and go with $1600kg/m^3$.

You mentioned they have equal volume, so let's say $1m^3$. The rock will have a mass of $1600kg$ and the cotton will be $448kg$. Let's also assume that $C_D$ and $A$ are the same for both rock and cotton (in other words, they are the same shape and size).

Hence, if you plug these numbers into $m$ in the acceleration equation, you'll notice that the second term $\frac{\rho v^2C_DA}{2m}$ will be larger for cotton and smaller for rock. Hence, $a_{rock} <a_{cotton}$ because $m_{rock}>m_{cotton}$, which is why rock falls faster (note that $|a_{rock}| >|a_{cotton}|$ as we are taking downwards to be the negative direction).

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Two things are going on. The weight of the object is a downward force. Gravity accelerates it downward. And air resistance is an upward force that slows it down.

The force of gravity does not change for an object, no matter what the speed. It only depends on the mass of the object.

Air resistance comes from two sources. Air is viscous to a degree, so there is some friction. And air must be pushed out of the way. It takes a force to push it. For both of these, air resistance depends on the shape of the object and the speed of the object, but not the mass of the object. Some details are below if you are interested.

When an object starts to fall, air resistance is $0$. As it falls faster and faster air resistance increases. When the upward force from air resistance gets to be as big as the downward force of gravity, the total force is $0$. The object does not accelerate further. It has reached terminal velocity.

For a rock, the force of gravity is large. The rock must fall fast for air resistance to balance the weight.

For cotton of the same size and shape, the force of gravity is smaller. A lower speed generates enough air resistance to balance gravity.


Stokes law describes friction. Friction depends on shape, and is proportional to velocity. John Hunter's answer gives the force for a sphere. It is similar for other shapes, though it can be hard to calculate.

Air is pushed out of the way as the object falls. Air has a surprising amount of mass - $1$ m$^3$ is $1$ kg. Air must be accelerated for this to happen, which takes energy. The speed of the air is proportional to the speed of the object. And the amount of air pushed in time $\Delta t$ is proportional to the cross sectional area and the speed. The kinetic energy is

$$E_{air} = 1/2 \space m_{air}v_{air}^2 $$ $$\propto 1/2 \space \left(A_{object} v_{ojbect} \Delta t\right) v_{object}^2 $$ $$ \propto v_{object}^3 \Delta t$$

From this, you find that force

$$F = E/ d$$ $$\propto (v^3 \space\Delta t) / (v \space\Delta t)$$ $$= v^2$$


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  • The answer given by @JohnHunter mentions the Stoke's law, according to which the resistance force in a liquid/gas is proportional to the object velocity, whereas @Programmer uses the force proportional to velocity squared (see Drag equation). In fact, both these forces are present, but it is the latter one that would come to dominate, see Explanation that air drag is proportional to speed or square speed?.
  • Terminal velocity. @Programmer further focuses on the acceleration experienced by the object. The acceleration is indeed different for the objects of the same shape but different mass, and thus affects the time that it takes for the object to reach the constant terminal speed: $$V_t=\sqrt{\frac{2mg}{\rho A C_d}}.$$ This speed is greater for heavier objects.
  • Finally

I think it should be cotton, but the fact that they are both equivalent in volume confuses me.

Much of our intuition regarding cotton vs. other materials comes from the fact that cotton is very pourous. This however, means that its shape and hence its drag coefficient is different from that of a rock of a seemingly equal shape. The answers discussed have therefore implicitly assumed that the cotton is pressed to high density and packed into, e.g., a plastic or a metallic foil. Otherwise it experiences much stronger drag, and hence falls much slower.

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