The Drag Force and Terminal Speed — Question 6

Question 6

Wind changes ground velocity without changing the terminal slip balance

Take $g=10$ m/s$^2$ and neglect buoyancy unless explicitly specified. A 1 kg particle moves through a uniform horizontal wind $\vec u=6\hat{\mathbf i}$ m/s. Take right and downward as positive x and y. Its only forces are gravity and linear drag $\vec F_d=-b(\vec v-\vec u)$, with $b=2$ kg/s.

1. Find the terminal ground-frame velocity vector.

2. Find its ground speed and angle below horizontal.

3. At an instant when $\vec v=0$, find acceleration and explain why drag is then nonzero.

Question 6 - Solution

Wind changes ground velocity without changing the terminal slip balance

(a) Balance the two components.

Use right and downward positive. The terminal component equations are $$0=-b(v_{x,t}-u_x),\qquad0=mg-bv_{y,t}.$$ $$\boxed{\vec v_t=6\hat{\mathbf i}+5\hat{\mathbf j}\ \mathrm{m/s}},\qquad \vec v_t-\vec u=5\hat{\mathbf j}\ \mathrm{m/s}.$$ The terminal motion relative to the air is vertically downward.

(b) Combine the ground components.

The ground-frame magnitude and direction are $$\boxed{|\vec v_t|=\sqrt{6^2+5^2}=\sqrt{61}=7.810\ \mathrm{m/s}},$$ $$\boxed{\theta=\tan^{-1}\!\left(\frac56\right)=39.806^\circ\text{ below horizontal}}.$$

(c) Evaluate the actual relative velocity.

At ground-frame rest, $\vec v-\vec u=-6\hat{\mathbf i}$ m/s. Therefore $$\vec F_d=-2(-6\hat{\mathbf i})=12\hat{\mathbf i}\ \mathrm N,$$ $$\boxed{\vec a=\frac{\vec F_d+mg\hat{\mathbf j}}m=12\hat{\mathbf i}+10\hat{\mathbf j}\ \mathrm{m/s^2}}.$$ Zero ground speed does not eliminate drag when the surrounding air moves.