I know that "perfect" pendulums would be able to swing forever, unperturbed by air resistance. However, since there is air resistance around us, pendulums (swinging bobs) slow down and move closer and closer to a halt. Say we have a metal sphere of mass m and radius r as the bob, suspended at length l from a point. The bob is allowed to rise to a height $h_{0}$ above its equilibrium position (with the string remaining fully stretched), and is then released. After one swing, the bob reaches a new height $h_{1}$ above equilibrium, and so on, until after swing n, it reaches height $h_{n}$ above equilibrium. At what rate will this damping take place (i.e. how can one theoretically calculate $h_{n}$)? What are the factors that affect it?
[Physics] Rate at which a pendulum bob slows due to air resistance
dragforceshomework-and-exercisesnewtonian-gravitynewtonian-mechanics
Best Answer
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.