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Answer by Floris for Rate at which a pendulum bob slows due to air resistance?

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Drag for a sphere is roughly proportional with velocity squared over a wide range of velocities (as long as the Reynolds number is reasonably large) and given by

$$F= \frac12 \rho v^2 A C_D$$

Where $\rho$ is th density of the medium (about 1.2 kg/m$^3$ for air), $v$ is the velocity, A the cross sectional area ($\pi r^2$) and $C_D$ the drag coefficient which varies with Reynolds number but which can be approximated to 0.5 for a wide range of velocities.

Since velocity squared is proportional to the height from the top of the swing, this suggests that the work done by the force of drag is roughly proportional to the height of the swing multiplied by the arc of the swing.


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