Question

An object of mass falls from rest under gravity subject to an air resistance proportional to its speed. Taking the yaxis as positive down, show that the differential equation of motion is$\mathit{m}\left(dv/dt\right)\mathbf{=}\mathit{m}\mathit{g}\mathbf{-}\mathit{k}\mathit{V}$ , where kis a positive constant. Find vas a function of t , and find the limiting value of vas tends t to infinity; this limit is called the terminal speed. Can you find the terminal speed directly from the differential equation without solving it?

Hint:What is $\mathit{d}\mathit{v}\mathbf{/}\mathit{d}\mathit{t}$after vhas reached an essentially constant value?

Consider the following specific examples of this problem.

(a) A person drops from an airplane with a parachute. Find a reasonable valueOf k .

(b) In the Millikan oil-drop experiment to measure the charge of an electron, tiny electrically charged drops of oil fall through the air under gravity or rise under the combination of gravity and an electric field. Measurements can be made only after they have reached terminal speed. Find a formula for the time required for a drop starting at rest to reach $\mathbf{99}\mathbf{\%}$of its terminal speed.

Short answer

The particular solution is$v\left(t\right)=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}}\right)$

The terminal speed is $v_{\max }=\frac{mg}{k}$

(a) When a person drops from an airplane with a parachute than a reasonable value of k is$k=12.96Kg/s$

(b)the time required for a drop starting at rest to reach 99% of its terminal speed is$t=\frac{m}{k}\ln \left(100\right)$

Step-by-step solution

 Step 1: Definition of Newton's Second Law of Motion

Newton's Second Law of Motion states that the rate of change of instigation of a body is directly commensurable to the applied force and takes place in the direction in which the force acts.

Given Parameters

It is Given that an object of mass mfalls from rest under gravity subject to an air resistance proportional to its speed.

Solving Differential Equations

Now the force that acts on that object is the gravitational force (acting downwards), that is$F_g=mg$ and the air resistance acting in the opposite direction of the motion (that is upward). Thus, it has been noticed that this force is proportional to the speed, so it can be written that, by using Newton's second law of motion we can write

$$\begin{gathered} ma=\sum F \\ m\frac{dv}{dt}=mg-kv \end{gathered}$$

Now solve the differential equation by using the separation of variables method

$$\begin{gathered} \int \frac{dv}{mg-kv}=\frac{1}{m}\int 1dt \\ \frac{\ln \left(mg-kv\right)}{k}=\frac{t}{m}+C \\ mg-kv=C_2{e}^{-\frac{kt}{m}} \end{gathered}$$

So the general solution is

$v\left(t\right)=\frac{mg-C_2{e}^{-kt/m}}{k}$

Find Terminal speed by using a differential equation

To find the particular solution use the initial condition, that is v=0 when t=0 therefore$C_2=mg$

And the particular solution is

$v\left(t\right)=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}}\right)$

The terminal speed when $t\to \infty$is

$v\left(t\to \infty \right)=\frac{mg}{k}$

Now the terminal speed is the maximum speed the object is falling, therefore directly from the differential equation by making it equal to zero is

$$\begin{gathered} \frac{dv}{dt}=0=mg-kv \\ {v}_{\max }=\frac{mg}{k} \end{gathered}$$

Finding reasonable value for k

In general, the value of the constant t is equal to the weight of the human divided by the terminal speed. On average, the terminal speed of the human body is $v_{\max }=53m/s$, and the average mass is$m=70kb$ Also the gravitational force near the earth's surface is $g=9.81m/s^2$,

Therefore, the value of the constant k is

$$\begin{gathered} 53=\frac{70\times 9.81}{k} \\ k=12.96kg/s \end{gathered}$$

 Find time for a drop starting at rest to reach 99% of its terminal speed

When the drop reaches the terminal speed the air resistance force and the gravitational force are equal, therefore, the terminal speed is

$$\begin{gathered} k{v}_{\max }=mg \\ {v}_{\max }=\frac{mg}{k} \end{gathered}$$

And 99% of the terminal speed is

$99\%v_{\max }=0.99\frac{mg}{k}$

Assuming the drops starts their motion from rest, therefore use the particular solution

$$\begin{gathered} 0.99\frac{mg}{k}=\frac{mg}{k}\left(1-e^{-kt/m}\right) \\ 0.01=e^{-kt/m} \\ -\ln \left(100\right)=-\frac{k}{m}t \\ t=\frac{m}{k}\ln \left(100\right) \end{gathered}$$

Hence,the particular solution is$v\left(t\right)=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}}\right)$and the terminal speed is $v_{\max }=\frac{mg}{k}$

(a) When a person drops from an airplane with a parachute than a reasonable value of k is $k=12.96Kg/s$

(b) the time required for a drop starting at rest to reach 99% of its terminal speed is

$t=\frac{m}{k}\ln \left(100\right)$