FLUIDS BYU 31

1. Determine Drag Force Dependencies:
The resistive force is given as $F_d = k r^\alpha v^\beta$. For an air bubble rising at constant velocity, the Buoyant Force equals the Drag Force (neglecting weight of air). $$ F_B \propto r^3 \implies k r^\alpha v^\beta \propto r^3 $$

2. Use Bubble Data to find $\alpha$ and $\beta$:
Bubble 1: radius $r$, velocity $v$. Bubble 2: radius $2r$, velocity $4v$.
From the proportionality $r^3 \propto r^\alpha v^\beta$: Case 1: $C r^3 = r^\alpha v^\beta$ Case 2: $C (2r)^3 = (2r)^\alpha (4v)^\beta$
Dividing Case 2 by Case 1: $$ 8 = 2^\alpha 4^\beta = 2^{\alpha + 2\beta} $$ $$ 2^3 = 2^{\alpha + 2\beta} \implies \alpha + 2\beta = 3 $$ Since this usually follows Stokes’ Law ($\alpha=1, \beta=1$) or quadratic drag ($\alpha=2, \beta=2$), only $\alpha=1, \beta=1$ fits the condition $\alpha+2\beta=3$. Thus, Drag Force $F_d \propto r v$.

3. General Equation for Terminal Velocity:
At terminal velocity, Net Force = 0. $$ |F_g – F_B| = F_d $$ $$ \frac{4}{3}\pi r^3 g |\rho_{body} – \rho_{water}| = C r v $$ $$ v \propto r^2 |\rho_{body} – \rho_{water}| $$

4. Compare Metal Ball and Plastic Ball:
Metal Ball: $r_1 = 1$ mm, $\rho_1 = 5$ g/cm$^3$, $v_1 = 8$ cm/s. $\rho_w = 1$ g/cm$^3$. $$ v_1 = K r_1^2 (\rho_1 – \rho_w) $$ $$ 8 = K (1)^2 (5 – 1) = 4K \implies K = 2 $$ Plastic Ball: $r_2 = 3$ mm, $\rho_2 = 2/3$ g/cm$^3$ (Wait, question says radius 3.0 mm and density 2/3? No, question: radius 3.0 mm, density 2/3?). Wait, let’s check density of water vs plastic. $\rho_p = 2/3$ g/cm$^3$. It is lighter than water, so it rises. The driving force is Buoyancy – Weight. Effective density difference $|\rho_p – \rho_w| = |2/3 – 1| = 1/3$. $$ v_2 = K r_2^2 |\rho_p – \rho_w| $$ $$ v_2 = 2 \times (3)^2 \times \frac{1}{3} $$ $$ v_2 = 2 \times 9 \times \frac{1}{3} = 6 \, \text{cm/s} $$