Terminal Velocity of buoyant object

Question

I'm trying to figure out how much distance does a ball of balsa wood covers until reaching terminal velocity, being released from a bottom of a pool.

To my understanding, I need to first figure out the terminal velocity. For that, I take:

$$F_\text{drag} + mg = F_\text{buoyancy}$$

$$F_\text{drag} = C_d A \frac{\rho_m V_t^2}{2}$$ $$F_\text{buoyancy} = V \rho_m g$$ $$mg = V \rho_s g$$

$\rho_m$ is the liquid density, and $\rho_s$ is the ball. $C_d$ is the drag coefficient. A the surface of the ball, $V$ the volume of the ball.

From this equation I reached: $$v_\text{terminal}=\sqrt{\frac{2g}{\gamma}\left(1-\frac{\rho_s}{\rho_m}\right)}$$

$\gamma$ being defined as shape coefficient: $$ \gamma = C_d A / V$$

That looks good, and gives ok results. But when I try to go ahead and find the time / distance I get weird data. I start from this equation: $$m\frac{dv}{dt}=F_b-F_d-mg$$ Solving: $$\frac{dv}{dt}=g\left(\frac{\rho_m}{\rho_s}-1\right) - 0.5\frac{\rho_m}{\rho_s}\gamma v^2$$ with the integral border $v_t$ I found couple of lines before, I get an arctanh of 1, meaning infinity.

I checked the integral, can't find anything wrong. What am I missing here?

Answer 1

You're not missing anything. When an object moves through a uniform fluid under the influence of a constant force and drag, the closer it gets to terminal speed, the less the net force on it is and the less it will accelerate. So it only approaches terminal speed asymptotically, never actually reaching it.

As you know, the equation works out to

$$v = v_T\tanh\biggl(\frac{t}{t_0}\biggr)$$

and if you try to find the time at which $v = v_T$ (terminal speed), you do indeed get $\tanh^{-1}(1)$ which is $\infty$. But for any given precision $\delta v$, you can find a time after which $\lvert v - v_T\rvert \leq \delta v$, so given the precision of your measurements, you can tell when the falling object's speed will be indistinguishable from terminal speed.