The Drag Force and Terminal Speed — Question 2

Question 2

Geometric similarity changes terminal-speed scaling

Take $g=10$ m/s$^2$ and neglect buoyancy unless explicitly specified. Similar solid spheres have the same material density and dimensionless drag coefficient C. Assume quadratic drag $D=\tfrac12 C\rho_f A v^2$ remains valid with constant C, where A is frontal area. Sphere B has twice the radius of sphere A.

1. Find the ratios of masses, frontal areas and terminal speeds.

2. At the same downward speed, find the ratio of drag accelerations $D/m$.

3. If instead the masses were equal but B had four times the area, find the terminal-speed ratio.

Question 2 - Solution

Geometric similarity changes terminal-speed scaling

(a) Scale geometry before balancing forces.

For equal-density similar spheres, $m\propto R^3$ and $A\propto R^2$. Therefore $$\boxed{\frac{m_B}{m_A}=8},\qquad\boxed{\frac{A_B}{A_A}=4}.$$ The drag balance gives $v_t=\sqrt{2mg/(C\rho_fA)}$, hence $$\boxed{\frac{v_{tB}}{v_{tA}}=\sqrt{\frac{8}{4}}=\sqrt2}.$$

(b) Divide drag by the body’s own mass.

At equal speeds, the common fluid and drag coefficient give $D_B/D_A=A_B/A_A=4$. Dividing by each sphere’s mass, $$\boxed{\frac{(D/m)_B}{(D/m)_A}=\frac{D_B/D_A}{m_B/m_A}=\frac48=\frac12}.$$ The larger sphere experiences half the drag acceleration.

(c) Change the stated constraint, not the earlier ratio.

Under the new equal-mass constraint, $$\boxed{\frac{v_{tB}}{v_{tA}}=\sqrt{\frac{m_B/m_A}{A_B/A_A}}=\sqrt{\frac14}=\frac12}.$$ This comparison replaces the equal-density similarity condition. Both results assume the stated constant-$C$ quadratic-drag model applies.