# How can I find the fall velocity of a sphere when it is falling in a fluid?

#### Chemist116

The problem is as follows:

A steel sphere falls inside of a beaker containing a fluid whose force exerted on the sphere is given by $kv$. Assuming that the sphere starts from rest, calculate the falling speed.

The alternatives given by my book are as follows:

$\begin{array}{ll} 1.&\frac{mg}{4K}\left(1-e^{-\frac{k}{m}t}\right)\\ 2.&\frac{mg}{2K}\left(1-e^{-\frac{k}{m}t}\right)\\ 3.&\frac{mg}{K}e^{-\frac{k}{m}t}\\ 4.&\frac{mg}{2K}e^{-\frac{k}{m}t}\\ 5.&\frac{mg}{K}\left(1-e^{-\frac{k}{m}t}\right)\\ \end{array}$

How exactly should I assess this problem? Since all alternatives appear an exponential, I believe the approach does involve the use of drag.

The forces acting in the object would be as follows?

$F_{net}=mg-kv$

But I'm stuck there. I'm assuming that there's an integral which will cause the exponential to appear. But I don't know how to get there. Can someone help me?

I don't know exactly how to relate it with the fact that the object begins from rest. How would it be? Help please!

#### skeeter

Math Team
$m \dfrac{dv}{dt} = mg - kv$

$\dfrac{dv}{dt} = g - \dfrac{k}{m} v$

let $b = \dfrac{k}{m}$ ...

$\dfrac{dv}{dt} = g - bv$

$\dfrac{-b}{g-bv} dv = -b \, dt$

$\log|g-bv| = -bt + C_1$

$g-bv = e^{-bt+C_1}$

$v = \dfrac{g}{b} - C_2 e^{-bt}$

$v(0)=0 \implies C_2 = \dfrac{g}{b}$

$v = \dfrac{g}{b} \left( 1-e^{-bt} \right) = \dfrac{mg}{k} \left(1 - e^{-\frac{k}{m} t} \right)$

#### skeeter

Math Team
You could also determine a solution based on the terminal velocity of the mass in the fluid medium ...

terminal velocity occurs when $a = 0$
$F_{net} = mg - kv = 0 \implies v_T = \dfrac{mg}{k}$

$\displaystyle v_T = \lim_{t \to \infty} v(t)$

take the limit of each answer choice ...

topsquark

#### Chemist116

You could also determine a solution based on the terminal velocity of the mass in the fluid medium ...

terminal velocity occurs when $a = 0$
$F_{net} = mg - kv = 0 \implies v_T = \dfrac{mg}{k}$

$\displaystyle v_T = \lim_{t \to \infty} v(t)$

take the limit of each answer choice ...
I think your first order differential equation is more straightforward. But I'm intrigued by the solution which goes by the terminal velocity using the limit.

$v_{T}=\frac{mg}{k}$ doesn't have term depending of the time. How am I suppose to take the limit here?

Can you help me with this approach?

#### skeeter

Math Team
Evaluate each limit ...

$$\displaystyle \lim_{t \to \infty} \dfrac{mg}{4k} \left(1 - e^{-\frac{k}{m} t} \right) = \, ?$$

$$\displaystyle \lim_{t \to \infty} \dfrac{mg}{2k} \left(1 - e^{-\frac{k}{m} t} \right) = \, ?$$

$$\displaystyle \lim_{t \to \infty} \dfrac{mg}{k} e^{-\frac{k}{m} t} = \, ?$$

$$\displaystyle \lim_{t \to \infty} \dfrac{mg}{2k} e^{-\frac{k}{m} t} = \, ?$$

$$\displaystyle \lim_{t \to \infty} \dfrac{mg}{k} \left(1 - e^{-\frac{k}{m} t} \right) = \, ?$$

... all these limits are very simple if you understand the behavior of the term $e^{-\frac{k}{m} t}$ as $t$ gets large

topsquark

#### skipjack

Forum Staff
$$\displaystyle \frac{dv}{dt} + bv = g$$, where $b = \large\frac{k}{m}$ and $v = 0$ when $t = 0$.

As multiplying by $e^{\large bt}$ and integrating from 0 to $t$ gives $e^{\large bt}v = \large\frac{g}{b}e^{bt} - \frac{g}{b}$,
$v = {\large\frac{g}{b}}\!\left(1-e^{\large-bt}\right) = {\large\frac{mg}{k}}\!\left(1 - e^{-{\large\frac{k}{m}t}}\right)$.