The Drag Force and Terminal Speed — Question 5

Question 5

Two drag mechanisms need one combined balance

Take $g=10$ m/s$^2$ and neglect buoyancy unless explicitly specified. A 1 kg body descends in still fluid. Drag magnitude is $D=bv+cv^2$, where $b=1$ kg/s and $c=0.5$ kg/m. Both terms act upward during descent.

1. Find terminal speed and reject any inadmissible root.

2. Find the contribution of each drag term at terminal speed.

3. Would dropping either term overestimate or underestimate terminal speed? Calculate both single-term predictions.

Question 5 - Solution

Two drag mechanisms need one combined balance

(a) Solve the total force balance.

Both upward drag terms balance the weight at terminal descent: $$mg=bv_t+cv_t^2\Longrightarrow10=v_t+.5v_t^2.$$ $$v_t=-1\pm\sqrt{21}\Longrightarrow\boxed{v_t=\sqrt{21}-1=3.5826\ \mathrm{m/s}}.$$ The negative root is inadmissible because this equation uses a downward speed and assumes both drag terms point upward.

(b) Resolve the total resistance.

The two contributions are $$\boxed{D_b=bv_t=\sqrt{21}-1=3.5826\ \mathrm N},$$ $$\boxed{D_c=cv_t^2=\tfrac12(\sqrt{21}-1)^2=11-\sqrt{21}=6.4174\ \mathrm N}.$$ Their sum is $D_b+D_c=10$ N, exactly the weight.

(c) Compare consistent simplified balances.

Removing either positive resistance term gives $$v_{t,\mathrm{linear}}=\frac{mg}{b}=10\ \mathrm{m/s},\qquad v_{t,\mathrm{quadratic}}=\sqrt{\frac{mg}{c}}=\sqrt{20}=4.4721\ \mathrm{m/s}.$$ Both exceed $3.5826$ m/s. With less resistance at each positive speed, a larger speed is needed to balance the same weight.