Terminal velocity of a body of mass

1. Sep 24, 2008

musicfairy

A small body of mass m located near the Earth’s surface falls from rest in the Earth's gravitational field. Acting on the body is a resistive force of magnitude kmv, where k is a constant and v is the speed of the body.

a) Write the differential equation that represents Newton's second law for this situation.

a = dv/dt
F = mg - kmv
ma = mg - kmv
a = g - kv
dv/dt = g - kv

I'm pretty sure I got this one right.

b) Determine the terminal speed vT of the body.

F = 0
g - kv = 0
g = kv
vT = g / k

I think I got this one right too.

c) Integrate the differential equation once to obtain an expression for the speed v as a function of time t. Use the condition that v = 0 when t= 0.

What I did so far:

dv/dt = g - kv
dt = dv / (g - kv)

So I would integrate dt = dv / (g - kv). The problem is I don't know how. Can someone please explain how to do this kind of integration, why I would do it that way?

2. Sep 24, 2008

cepheid

Staff Emeritus
Often times if you have something like:

dv/dt = v,

you'll think to yourself, "oh, the equation is separable", and you'll go:

dv/v = dt

and then "integrate"

to get ln(v) = t + C

notice, however, that you integrated the left side wrt v and the right hand side wrt t, which makes NO sense! So what is actually going on here? Why does this nonsensical operation of splitting dv/dt actually work? It works because you are implicitly using the chain rule as follows

$$\frac{dv}{dt} = v$$

$$\frac{1}{v}\frac{dv}{dt} = 1$$

notice that the left hand side looks like something that *has been* differentiated using the chain rule, and we can write it as that something:

$$\frac{d}{dt}(\ln{v}) = 1$$

Now we can integrate BOTH sides wrt time (which makes waaay more sense)

$$\int { \frac{d}{dt}(\ln(v)) \, dt } = \int{\, dt}$$

$$\ln(v) = t + C$$

In your more complicated example, it is important to realize that this is what is going on, otherwise you'll get stuck.

We have

$$\frac{1}{g-kv}\frac{dv}{dt} = 1$$

We want the left hand side to look like something that has been differentiated by the chain rule. It SORT OF does. The only problem is that the derivative of ln(g-kv) wrt time actually differs from what's on the left hand side by a factor of -k. Once you get that sorted, it's the same procedure.

3. Sep 24, 2008

musicfairy

Thanks for the explanation. I'm starting to understand this more.

What do you mean that the left differs by a factor of -k?

I have the answer. It's -(1/k)ln(g - kv) = t + C

Where did that come from? Where did the other k come from?

I thought doing this problem would help me understand how these weirder intergrations work, but it's confusing me even more.

4. Sep 24, 2008

cepheid

Staff Emeritus
alright, here goes:

The goal is to recognize that the left hand side can be expressed as an exact derivative, so that integrating it becomes trivial.

the equation has a term 1/(g-kv) multiplied by v'(t), which suggests that it is the result of differentiating ln(g-kv) with respect to time using the chain rule. If that were true, then the left hand side would reduce to the time derivative of the natural log of g-kv, and integrating it would be easy (integral of a derivative is just the function itself...so you're just left with the log term). But that's NOT quite right. Check it out:

$$\frac{d}{dt} (\ln(g-kv)) = \frac{1}{g-kv} \frac{d}{dt}(g-kv) \frac{dv}{dt}$$

$$= \frac{1}{g-kv} (-k) \frac{dv}{dt}$$

Notice that this doesn't QUITE match the left hand side of our differential equation. There's that extra factor of (-k). Which means that the left hand side of our equation is this thing divided by -k. THAT's where the 1/(-k) comes from.

5. Sep 24, 2008

musicfairy

I stared at the problem and answer while waiting for an answer and had the same train of thought. That was very helpful. Thanks a lot.