AP Physics C: Mechanics Retarding & Drag Forces

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about retarding and drag forces.0003

You have got it we are actually going to get to air resistance.0008

Our objectives are going to be to find the terminal velocity of an object.0011

To describe the motion of a particle under the effect of a retarding a drag force in terms of its displacement, velocity, and acceleration.0015

Use Newton’s second law to write a differential equations for the velocity of the object.0024

Derive the equation for velocity for Newton’s second law using separation of variables.0029

Determine the acceleration for an object falling under the influence of drag forces.0035

Let us start off by talking about these retarding a drag forces.0040

Sometimes the frictional force is a function of how fast an object is going.0044

Its velocity and air resistance is a great example of this.0048

These forces are called drag or retarding forces.0052

Now, as we go to our analysis of drag this is going to be our first lesson in a while that is going to get pretty heavy into the map and calculus.0056

Probably, seeing some things if you are taking calculus concurrently where you are seeing it in physics before you might in calculus.0064

That is ok, it is not uncommon.0071

Do your best to stick with it and then come back to this lesson later on once you are a little bit more solid0073

on the calculus and some of the integration we are going to be doing.0079

That is not uncommon at all.0082

If you leave this lesson a little fuzzy that is ok.0084

Alright, let us start by talking about a skydiver.0090

Assume we drop our dear friend Alex from an airplane.0094

Typically, the drag force on a freefalling object takes the form where the force of drag is some constant × velocity or constant × square of velocity.0097

Or sometimes even something in between that where b and c are constants.0104

For the purposes of this problem, let us assume that the drag force is to equal to a constant times the velocity.0111

Let us draw a free body diagram for our dear friend Alex who are about to push from an airplane.0118

We have the weight of Alex pulling down and the drag force backup.0123

We will call down the y direction since that is the way Alex was going to start his motion and finishes motion.0132

As I analyze this, the net force in the y direction using Newton’s second is going to be mg - the drag force0139

and that is all equal to mass times the acceleration in the y direction.0150

We just said that our drag force is equal to bv.0155

We can write that mg - bv = Ma.0162

Let us take a look at what happens when we first push Alex from the plane.0173

Initially at time T = 0 Alex’s velocity is 0, therefore, the drag force is 0 mg = Ma and a =g.0178

The acceleration = acceleration due to gravity.0194

Therefore, we can write at that point acceleration is equal to g.0197

As T increases as we head toward infinity eventually Alex reaches a maximum or terminal velocity.0205

Terminal velocity we will abbreviate that Vt and at that point acceleration is equal to 0.0228

At that point, all the forces must be balanced if there is no acceleration.0242

The drag force must equal mg at that point.0246

In Newton’s second law equation we have Mg – bv = ma with the condition of terminal velocity, acceleration is 0 and V = Vt.0251

Our equation becomes Mg – b Vt = 0.0268

Therefore, we can solve for terminal velocity and say terminal velocity is just going to be Mg divided by the constant b.0275

Our formula for terminal velocity based on that constant in our f drag = bv equation and we also know initially T =0.0285

Velocity 0 is equal to the acceleration due to gravity g.0294

I think that will set us up to get into our math using Newton’s second law of equation.0299

If we go back to our Newton’s second law equation which has velocity and velocities derivative acceleration in it,0306

we are really have 2 different forms of the same variable in the equation.0312

The variable itself and its derivative and that is called a differential equation.0318

We are going to solve that using a strategy known as separation of variables.0322

Let us start on the next page to give ourselves all kind of room here.0329

We will begin with our Newton’s second law of equation mg - bv = ma.0334

Like we said, acceleration is the derivative of velocity a = dv dt0343

We have mg - bv = m dv dt.0349

Our differential equation with V and its derivative in the same equation.0357

You can take entire courses on solving differential equations.0361

We are only to have to deal with a couple types, couple simple ones here in the course.0364

This one we can solve with that separation of variable strategy.0368

Let us walk through and see how we do this.0372

If mg – bv= m dv dt I can divide the whole thing by b so we would have mg / b - V = m/b dv dt.0374

Mg/b we just defined in the previous page that is what we called terminal velocity.0391

We can maybe pretty this up just a little bit by saying that VT = mg / b0397

Therefore our left hand side becomes Vt - V= m/ b dv dt.0403

Which implies then I will try to get all the variables V’s and derivative of these on the same side.0416

I have dv / Vt – V = b / m dt.0423

It is just an algebraic rearrangement.0432

I have dv / Vt – V I rather have V there than have V – Vt.0436

I am going to multiply both sides by -1 to get the left hand side as dv / V – Vt = - b / m T.0441

We have got all of V on one side and their variable in terms of T on the right hand side.0455

What we are going to do is we are going to integrate both sides.0461

As I integrate the left hand side I am going to integrate dv / V – VT.0465

My variable of integration is that velocity.0472

I am going to integrate from some initial velocity v= 0 to some final velocity V.0475

I have to integrate do the same thing to the right hand side so that will be the integral of - b / dt.0481

My variable of integration is T so we are going to integrate from T = 0 to some final value t.0495

How do I integrate those?0504

The left hand side fits the form du / u where if I said that u = v - Vt the differential of u would just be dv.0506

In the formula maybe you have learned or you have not got in there yet in the rule of d/u is the natural log of u + that constant of integration.0523

Since, we have the limits here we do not have to worry about our constant integration.0533

That means that our left hand side is going to become the natural log of u V - Vt evaluated from 0 to V.0537

In the right, inside -b / m that is a constant that can come out of the integral sign we are just integrating dt from 0 to T0549

That is just going to be T and this becomes - b / MT.0557

The left hand side here becomes log of V - VT substituting in V - VT - log of 0 – Vt plugging in 0 for V0568

That is - Vt = - b / Mt.0583

We are going to use the identity of the difference of two logs is the one of the quotient.0590

We can use a log a – log b = log a/b to say that the left hand side is log of V - Vt /- VT.0598

In the right hand side still - b / MT.0613

A little bit more manipulation to do here.0619

Implies that the log of V – Vt /- Vt.0623

Let us see if we can rearrange them a little bit and take that negative to the top.0631

We can write that as the log of c VT - V / dt = -b/MT which implies that the log of b1- V/VT = -b/MT.0636

If I want to get rid of that nasty log I can use this as a power we raised e2 on both sides to state then that 1 - V / VT0660

Even the natural log is whether you happen to have there = e^b/MT.0672

Alright, next up.0682

We are trying to get V all by itself so I could take and rearrange this add V / VT to that side and subtract that from the other and come up with V/ VT = 1 – e^b/MT,0684

which implies then that V = VT × 1 – e^-b/MT.0703

Or we said VT was mg / b so if we wanted to we can put that back in those terms as well that V =MG / b × (1-e^-b/MT).0716

Those are equivalent statements.0734

We found velocity as a function of time.0738

Alright, now knowing VT we can solve for the acceleration and I'm going to give ourselves more room for that.0743

If we wanted to do that let us take a look here.0748

A = dv dt which would be the derivative of what we just found from our velocity which is VT - VT e^-b/MT.0753

I will expand that out so that is going to be equal to the derivative with respect to T of the derivative of the constant is going to be 0.0766

We will just have – VT e^-b/MT.0779

We can pull our constant out -VT derivative with respect to T of e^-b/MT0785

which implies then that a = - VT × derivative e^u is e^u du.0796

We have our e^u e^-b/MT × du is going to be another – b/m0806

which implies then that a is going to be equal to - Vt was mg /b replace that there we have - b/m and we still have e^-b/MT0815

which implies then that a = we can do some simplifications.0835

We got some m there and there.0838

B there and there our negative signs.0840

I'm just going to get that a = g × e^-b/MT.0843

You are going to see forms of the solutions like this for differential equations quite regularly where0854

you have a constant times e^-b/MT raised to some power times T.0860

Or a constant ×1 – e^{negative power}.0866

It will come up again and again.0870

These are all of the form where you have 1 – e^-T / some time constant.0872

Here time would be m/b or something times e^-T / time itself.0879

It is going to come up here.0886

We are going to see it in the electricity and magnetism course.0887

When we talk about capacitors and get into inductors that form keeps coming up again and again.0890

It is really nice that once you start to get a feel for that form you can almost guess the answer to these problems0896

before you go all the way through the math to actually prove it.0901

Or look at the initial and final conditions to help draw graphs of these before you actually go solve them.0906

Let us take a look at how these would look graphically.0913

I'm going to draw a couple graphs here and we are going to draw the acceleration time graph, velocity time graph,0917

and there is our position or displacement time graph.0927

Let us see what all these are going to look like.0932

We will start up here in acceleration time graph.0951

We will do a velocity time graph and also a y displacement time graph.0955

We said initially as far as acceleration goes to moment Alex left the airplane when velocity was 0 the acceleration of Alex was g.0966

We are going to start at a maximum value and over time as Alex goes faster and faster the acceleration0974

is going to decrease and decrease until reaching terminal velocity when there is no more acceleration.0981

We start at g and we have to decay down here to 0 when it is an exponential decay so we can radiate like that as we approach that asymptote.0987

That should be right on the line there.0996

As far as velocity goes, we know when we first push Alex out of the plane the velocity is 01000

We know that point and after a long time we know we eventually get to some value of terminal velocity which is mg / b.1005

We will write that in there as an asymptote and we will have an exponential increase there following that same basic shape.1014

For displacement, displacement begins at 0 and increases as speed increases1024

until reaching a constant rate of increase when the velocity reaches VT.1029

As far as displacement goes we are going to start at 0.1035

Increase and increase until we get to some point where it is just going to be linear as1038

we have a constant velocity when we are running it just terminal velocity.1046

There would be graphs of acceleration, velocity, and displacement as functions of time that correspond to those calculations we just did.1051

Let us take a look and see how this would look in a free response problem.1063

You can download the problem yourself if you want 2005 free response 1 from the mechanics exam.1067

A link to it there where you can google it and will take a few minutes look it over and give it a try and come back here and let us see how you did.1073

2005 mechanics 1 looking at part A, we have a ball that is thrown vertically upward at some initial speed.1083

It has some air resistance given by - kv the positive direction we are going to call up.1092

Is the magnitude of the acceleration of the ball increases or decrease or remain the same as it moves upward?1098

We are calling up the positive y direction and if I were to draw a free body diagram of the ball1105

as it moves up we have its weight down and we have that drag force kv down.1110

Net force in the y direction is just going to be - mg - kv is equal to MA y which implies that Ay was just going to be -g - kv/M.1119

If we look at that as V goes down, as it slows down, as it gets higher and higher, V gets smaller.1134

It looks to me like our acceleration, the magnitude of our acceleration in the y direction must decrease based on our formula.1142

There is A must decrease.1152

Let us take a look at part B.1157

Write but do not solve the differential equation for the instantaneous speed of the ball in terms of time as it moves upward.1162

From our free body diagram again Ay = -g - kv/ m which implies then since our A is the derivative of velocity with respect to time.1170

We can write it as dv dt = - g –kv /m.1185

That is a differential equation that would probably give us credit but let us clean it up just a little bit.1194

I'm going to write that in a manner that will be a lot more useful that is going to be M dv dt = - mg – kv1200

but either one of those you have your different equation down.1215

For C, find the terminal speed of the ball as it moves downward.1219

For C, looking at terminal speed.1224

At terminal velocity we know the net force =01227

which implies then that our free body diagram is going to have something like this kv,1235

Let us be careful it is M according to this problem Mg which implies that terminal velocity kv terminal must equal Mg or a terminal velocity Mg / k.1244

Let us move on to check out part D.1269

For part D, we are asked does it take longer for the ball to rise to its maximum height1273

or to fall from its maximum height back to the height from which it was thrown?1278

A tricky question.1283

Let us see.1284

On the way up, friction brings the ball to a stop quickly.1286

This helps bring it to a stop more quickly than if they were no fiction.1300

On the way, down the friction slows the ball down so it has more time in the air on the way down.1308

That means that average velocity on the way up has to be greater than1323

the average velocity on the way down because it happen more quickly.1328

And because the distance is constant distance traveled on the way up and the way down1334

the time to go down is greater than the time to go up because T = d/V.1341

I would say then that it takes longer to fall.1349

Some sort of explanation like that to go along with your answer.1360

Part E, on the graph, sketch a graph of velocity vs. Time for the upward and downward parts of the ball's flight.1365

We are going to need a graph here.1377

Here is our velocity, there is our time, and we are looking at what happens for the upward and downward parts.1387

It starts at some initial velocity V0.1402

It is going to cross the axis here to have a velocity of 0 at its highest point.1407

It is going to take less time than it does on the rest of the trip because it is going to take longer to fall.1412

On the way down, we are going to have some value of terminal velocity.1418

We will draw an asymptote in here for our V terminal.1427

I would think that our graph would probably look something like this where it is approaching terminal velocity.1434

Something like that final velocity.1446

Something like that is your approach of final velocity at time Tf.1451

I think that covers that one.1454

Let us take a look at one more free response problem.1462

Let us go to the 2013 APC mechanics exam.1466

You can find it at this address or google it.1470

It will take a few minutes to look it over and print it out, give it a try, and come back here and hit play.1474

We will see how it worked for you.1478

In this problem, we have a box of mass M at rest in the constant applied force being applied1483

there is a frictional drag force proportional to kv where V is the speed of the box,1489

k is some positive constant, and we are given a dot to draw on label the forces.1495

Draw our free body diagram actually.1499

Let us draw our free body diagram first.1501

There is our box.1504

We know we have the normal force, we have the weight of the box, the force of gravity.1507

We have some applied force which we are calling Fa.1513

We must have our frictional force our drag force kv.1518

For part B, it asks us to write but do not solve the differential equation that can be used to determine the speed of the box and that sounds familiar.1528

We have done that sort of thing.1536

Let us take a look net force in the x direction is going to be the applied force - kv assuming we are calling to the right positive.1539

All that must be equal to Ma but as a differential equation A = dv dt therefore Fa – kv = M dv dt.1550

That will work.1571

There is our different equation.1572

We have velocity and its derivative in the same equation.1574

Moving on to part C, determine the magnitude of the terminal velocity of the box.1583

At terminal velocity acceleration is 0, f net is 0 therefore we know that the applied force in the x direction must equal kv in magnitude.1591

Therefore, the applied force = kv terminal or solving for V terminal that is just going to be our applied force divided by k.1606

Part D, use the differential equation to derive the equation for the speed.1624

Alright, we have to do some math and let us give ourselves some room.1630

Starting with our equation f - kv = M dv dt, we are going to do the separation of variables again.1636

This implies then that dv / f - kv = dt / M.1646

Let us see how it would integrate that.1657

The integral of the left hand side dv / F - kv must equal the integral of dt / M.1659

We are going to integrate here from our velocity V = 0 to some final value V in the right hand side from T = 0 to some final value T.1668

If we are going to fit this into the form du / u we would need -dv in the top and we would also need –k.1678

You need –k on the top.1686

If we are going to put –k on the top to fit that form we have to multiply by -1/k so that we maintain the same value.1689

We cannot just arbitrary throw things in there.1696

In the right hand side it looks ok to integrate.1699

This implies then, that we have -1/k integral of du/U is going to be the natural log of our U which was f-kv evaluated from 0 to V.1702

The right hand side is just going to be t/M.1709

Expanding out our left hand side we have if we do this log I am going to take a moment and I am going to put our –k over the right hand side.1726

If I multiply both sides by –k let us put –k there.1738

It can go away and can make this a little bit simpler to see.1743

Our left hand side becomes the log of f –k and I plug in V for my variable v - the log of f-0 for our V.1747

That is going to be – the log of f = - kt/M.1758

Which implies that the difference of the logs is the log of the quotient so we have on the left hand side our log of f-kv/f = -kt/M.1768

If I raise both sides to the e, the left hand side becomes f-kv/f = e^-k/Mt.1784

A little bit more rearrangement here.1798

F – kv / f let us multiply both sides by f to get f –Kv= f e^-k/Mt.1801

We will get V all by itself.1813

Let us get kv= f-fe^-k/Mt.1816

We can factor out that f ÷ k so the velocity is going to be f/k × (1- e^-k/Mt).1824

There is part D.1847

Finally for part E, on the axis sketch a graph as the speed as a function of time and label the asymptotes things like that.1851

We are getting pretty good at graph and these sorts of things by now.1863

Let us give it a shot.1864

We have V on the y axis, time on our x, and we know it is going to start at some velocity 0.1877

We can also plug that in for T in our formula.1886

If T is 0, e⁰ is 1.1890

1-1 Is 0 so the velocity would be 0.1892

We will start here at 0 and as T gets big that whole term goes to 0 so we have f/k as our asymptote.1895

Let us mark that here f/k.1906

The shape of our graph is something like that.1913

Retarding forces and drag forces, air resistance.1920

Hopefully that gets you a good start.1923

Thank you so much for watching here at www.educator.com.1924

I look forward to seeing you soon and make it a great day everyone.1927