Drag on sphere
Introduction: We considered the drag force and argued that to a good approximation it might depend just on the radius of the sphere R, the density of the air ½, and the speed of the sphere through the air v. Only one combination of these three quantities has the dimensions of force:
We can't infer the dimensionless coefficient. Let's define b as the product of this unknown coefficient time’s ½R2, so
Note that the dimensions of bare M=L. If the sphere has mass m and falls in a gravitational field with gravitational acceleration g, then Newton's law (taking the down direction as positive) says
If the sphere begins at rest it will initially accelerate with acceleration g. As it speeds up the drag force increases until it balances the gravity force. This will happen asymptotically, i.e. in an infinite amount of time. Let's work out the details
To simplify the computation let's choose units so that m = g = b = 1.Can we really do this?! Yes, because there is no dimensionless combination of these three quantities. In particular, we can choose the units of mass, length, and time, as
m0= m, l0= m=b and t0= (m/bg)1/2
It characterizes when the mass of the air is comparable to the mass of the sphere. It will also be the scale over which the sphere comes close to its terminal velocity.
With the above choice of units, the equation of motion for the sphere becomes
Using the method of partial fractions this can be written:
And integrating both sides then gives:
If we choose t = 0 to be the time when the sphere is at rest v = 0, it follows that c1 = 0, so exponentiation yields
Finally, solving for v we find
That is, we found the nice solution v = tanh t.
To find the position as a function of time we integrate the differential equation
Rather than work this out \by hand", let's just look up the integral of tanh t. These yields
If we choose the value of y at t = 0 to be 0, then c2 = 0. This gives us the complete answer, expressed in terms of the units of mass m0, length l0 and time t0.