The Drag Force and Terminal Speed — Question 4

Section 6-3 · Physics pages 818–819

Question 4 Acceleration data reveal a quadratic drag coefficient A 2 kg body descends in still fluid under gravity and quadratic drag only. Two exact measurements are (v, a) = (2, 9.2) and (6, 2.8) in SI units, with downward acceleration positive. Treat local g and c in D = cv2 as unknown. v2 (m2/s2) a (m/s2) (4, 9.2) (36, 2.8) Two exact measurements constrain the hypothesized linear relation between acceleration and squared speed. (a) Use the measurements to find g and c. (b) Infer terminal speed. (c) Predict acceleration at 8 m/s and explain which plot would test the model using additional measurements.
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Question 4 - Solution Acceleration data reveal a quadratic drag coefficient v2 (m2/s2) a (m/s2) (4, 9.2) (36, 2.8) (a) Subtract the force equations. Use a = g −(c/m)v2 with m = 2 kg: 9.2 = g −2c, 2.8 = g −18c (SI numerical values). Subtracting gives −6.4 = −16c, so c = .40 kg/m , g = 9.2 + .20(4) = 10 m/s2 . (b) Find the zero-acceleration speed. Set the fitted acceleration to zero: 0 = 10 −.20v2 t =⇒vt = √ 50 = 5 √ 2 = 7.071 m/s . (c) Extrapolate and propose a diagnostic. At the proposed speed, a(8) = 10 −.20(8)2 = −2.8 m/s2 . The body is descending but slowing. Additional data should satisfy a = g −(c/m)v2: a straight a-versus-v2 plot with intercept 10 m/s2 and slope −.20 m−1.

Question 4

Acceleration data reveal a quadratic drag coefficient

A 2 kg body descends in still fluid under gravity and quadratic drag only. Two exact measurements are (v,a)=(2,9.2)(v,a)=(2,9.2) and (6,2.8)(6,2.8) in SI units, with downward acceleration positive. Treat local g and c in D=cv2D=cv^2 as unknown.

Diagram shown in the original SVG worksheet.

  1. Use the measurements to find g and c.

  2. Infer terminal speed.

  3. Predict acceleration at 8 m/s and explain which plot would test the model using additional measurements.

Question 4 - Solution

Acceleration data reveal a quadratic drag coefficient

Diagram shown in the original SVG worksheet.

(a) Subtract the force equations.

Use a=g−(c/m)v2a=g-(c/m)v^2 with m=2m=2 kg: 9.2=g−2c,2.8=g−18c(SI numerical values).9.2=g-2c,\qquad2.8=g-18c\quad\text{(SI numerical values)}. Subtracting gives −6.4=−16c-6.4=-16c, so c=.40kg/m,g=9.2+.20(4)=10m/s2.\boxed{c=.40\ \mathrm{kg/m}},\qquad\boxed{g=9.2+.20(4)=10\ \mathrm{m/s^2}}.

(b) Find the zero-acceleration speed.

Set the fitted acceleration to zero: 0=10−.20vt2⇒vt=50=52=7.071m/s.0=10-.20v_t^2\Longrightarrow\boxed{v_t=\sqrt{50}=5\sqrt2=7.071\ \mathrm{m/s}}.

(c) Extrapolate and propose a diagnostic.

At the proposed speed, a(8)=10−.20(8)2=−2.8m/s2.\boxed{a(8)=10-.20(8)^2=-2.8\ \mathrm{m/s^2}}. The body is descending but slowing. Additional data should satisfy a=g−(c/m)v2a=g-(c/m)v^2: a straight aa-versus-v2v^2 plot with intercept 1010 m/s2^2 and slope −.20-.20 m−1^{-1}.