Terminal velocity

Consider a small metallic sphere released from rest and falling freely through a large column of a highly viscous fluid. Three forces act on the sphere throughout its fall: (i) its own weight, acting vertically downward; (ii) the upthrust U due to buoyancy, acting vertically upward; and (iii) the viscous drag, also acting upward (viscous force always opposes the direction of the sphere's actual motion). Initially, since the sphere is at rest, the downward force exceeds the upward forces, so the sphere accelerates downward. But as its speed increases, the viscous drag -- which grows with speed -- increases too, so the net downward force keeps shrinking. Eventually a stage is reached where the net downward force exactly balances the total upward force, the resultant force on the sphere becomes zero, and from that point on the sphere moves with a CONSTANT velocity. This maximum constant velocity attained by a body falling freely through a viscous medium is called its TERMINAL VELOCITY, $v_t$. A graph of the sphere's velocity (y-axis) against time (x-axis) rises steeply at first as the sphere accelerates, then flattens into a horizontal line at the terminal-velocity value once the force balance is reached. DERIVATION. For a sphere of radius r, density $\rho$, falling through a fluid of density $\sigma$ and coefficient of viscosity $\eta$: the gravitational (downward) force is $F_G=mg=\dfrac{4}{3}\pi r^3\rho g$; the upthrust (upward) is $U=\dfrac{4}{3}\pi r^3\sigma g$; and the viscous force (upward, by Stoke's law -- see the next section) is $F=6\pi\eta r v_t$. At terminal velocity, the net downward force equals the upward force: $F_G=U+F$, i.e. $\dfrac{4}{3}\pi r^3\rho g-\dfrac{4}{3}\pi r^3\sigma g=6\pi\eta r v_t$. Solving for $v_t$ gives $v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}$, so $v_t\propto r^2$: the terminal speed of a falling sphere is directly proportional …