WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton and in this lesson we are going to talk about retarding and drag forces.
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You have got it we are actually going to get to air resistance.
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Our objectives are going to be to find the terminal velocity of an object.
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To describe the motion of a particle under the effect of a retarding a drag force in terms of its displacement, velocity, and acceleration.
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Use Newton’s second law to write a differential equations for the velocity of the object.
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Derive the equation for velocity for Newton’s second law using separation of variables.
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Determine the acceleration for an object falling under the influence of drag forces.
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Let us start off by talking about these retarding a drag forces.
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Sometimes the frictional force is a function of how fast an object is going.
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Its velocity and air resistance is a great example of this.
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These forces are called drag or retarding forces.
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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.
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Probably, seeing some things if you are taking calculus concurrently where you are seeing it in physics before you might in calculus.
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That is ok, it is not uncommon.
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Do your best to stick with it and then come back to this lesson later on once you are a little bit more solid
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on the calculus and some of the integration we are going to be doing.
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That is not uncommon at all.
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If you leave this lesson a little fuzzy that is ok.
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Alright, let us start by talking about a skydiver.
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Assume we drop our dear friend Alex from an airplane.
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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.
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Or sometimes even something in between that where b and c are constants.
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For the purposes of this problem, let us assume that the drag force is to equal to a constant times the velocity.
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Let us draw a free body diagram for our dear friend Alex who are about to push from an airplane.
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We have the weight of Alex pulling down and the drag force backup.
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We will call down the y direction since that is the way Alex was going to start his motion and finishes motion.
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As I analyze this, the net force in the y direction using Newton’s second is going to be mg - the drag force
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and that is all equal to mass times the acceleration in the y direction.
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We just said that our drag force is equal to bv.
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We can write that mg - bv = Ma.
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Let us take a look at what happens when we first push Alex from the plane.
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Initially at time T = 0 Alex’s velocity is 0, therefore, the drag force is 0 mg = Ma and a =g.
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The acceleration = acceleration due to gravity.
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Therefore, we can write at that point acceleration is equal to g.
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As T increases as we head toward infinity eventually Alex reaches a maximum or terminal velocity.
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Terminal velocity we will abbreviate that Vt and at that point acceleration is equal to 0.
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At that point, all the forces must be balanced if there is no acceleration.
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The drag force must equal mg at that point.
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In Newton’s second law equation we have Mg – bv = ma with the condition of terminal velocity, acceleration is 0 and V = Vt.
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Our equation becomes Mg – b Vt = 0.
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Therefore, we can solve for terminal velocity and say terminal velocity is just going to be Mg divided by the constant b.
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Our formula for terminal velocity based on that constant in our f drag = bv equation and we also know initially T =0.
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Velocity 0 is equal to the acceleration due to gravity g.
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I think that will set us up to get into our math using Newton’s second law of equation.
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If we go back to our Newton’s second law equation which has velocity and velocities derivative acceleration in it,
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we are really have 2 different forms of the same variable in the equation.
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The variable itself and its derivative and that is called a differential equation.
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We are going to solve that using a strategy known as separation of variables.
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Let us start on the next page to give ourselves all kind of room here.
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We will begin with our Newton’s second law of equation mg - bv = ma.
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Like we said, acceleration is the derivative of velocity a = dv dt
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We have mg - bv = m dv dt.
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Our differential equation with V and its derivative in the same equation.
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You can take entire courses on solving differential equations.
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We are only to have to deal with a couple types, couple simple ones here in the course.
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This one we can solve with that separation of variable strategy.
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Let us walk through and see how we do this.
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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.
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Mg/b we just defined in the previous page that is what we called terminal velocity.
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We can maybe pretty this up just a little bit by saying that VT = mg / b
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Therefore our left hand side becomes Vt - V= m/ b dv dt.
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Which implies then I will try to get all the variables V’s and derivative of these on the same side.
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I have dv / Vt – V = b / m dt.
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It is just an algebraic rearrangement.
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I have dv / Vt – V I rather have V there than have V – Vt.
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I am going to multiply both sides by -1 to get the left hand side as dv / V – Vt = - b / m T.
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We have got all of V on one side and their variable in terms of T on the right hand side.
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What we are going to do is we are going to integrate both sides.
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As I integrate the left hand side I am going to integrate dv / V – VT.
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My variable of integration is that velocity.
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I am going to integrate from some initial velocity v= 0 to some final velocity V.
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I have to integrate do the same thing to the right hand side so that will be the integral of - b / dt.
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My variable of integration is T so we are going to integrate from T = 0 to some final value t.
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How do I integrate those?
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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.
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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.
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Since, we have the limits here we do not have to worry about our constant integration.
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That means that our left hand side is going to become the natural log of u V - Vt evaluated from 0 to V.
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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 T
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That is just going to be T and this becomes - b / MT.
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The left hand side here becomes log of V - VT substituting in V - VT - log of 0 – Vt plugging in 0 for V
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That is - Vt = - b / Mt.
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We are going to use the identity of the difference of two logs is the one of the quotient.
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We can use a log a – log b = log a/b to say that the left hand side is log of V - Vt /- VT.
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In the right hand side still - b / MT.
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A little bit more manipulation to do here.
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Implies that the log of V – Vt /- Vt.
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Let us see if we can rearrange them a little bit and take that negative to the top.
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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.
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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 / VT
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Even the natural log is whether you happen to have there = e^b/MT.
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Alright, next up.
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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,
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which implies then that V = VT × 1 – e^-b/MT.
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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).
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Those are equivalent statements.
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We found velocity as a function of time.
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Alright, now knowing VT we can solve for the acceleration and I'm going to give ourselves more room for that.
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If we wanted to do that let us take a look here.
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A = dv dt which would be the derivative of what we just found from our velocity which is VT - VT e^-b/MT.
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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.
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We will just have – VT e^-b/MT.
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We can pull our constant out -VT derivative with respect to T of e^-b/MT
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which implies then that a = - VT × derivative e^u is e^u du.
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We have our e^u e^-b/MT × du is going to be another – b/m
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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/MT
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which implies then that a = we can do some simplifications.
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We got some m there and there.
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B there and there our negative signs.
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I'm just going to get that a = g × e^-b/MT.
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You are going to see forms of the solutions like this for differential equations quite regularly where
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you have a constant times e^-b/MT raised to some power times T.
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Or a constant ×1 – e^negative power.
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It will come up again and again.
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These are all of the form where you have 1 – e^-T / some time constant.
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Here time would be m/b or something times e^-T / time itself.
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It is going to come up here.
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We are going to see it in the electricity and magnetism course.
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When we talk about capacitors and get into inductors that form keeps coming up again and again.
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It is really nice that once you start to get a feel for that form you can almost guess the answer to these problems
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before you go all the way through the math to actually prove it.
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Or look at the initial and final conditions to help draw graphs of these before you actually go solve them.
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Let us take a look at how these would look graphically.
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I'm going to draw a couple graphs here and we are going to draw the acceleration time graph, velocity time graph,
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and there is our position or displacement time graph.
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Let us see what all these are going to look like.
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We will start up here in acceleration time graph.
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We will do a velocity time graph and also a y displacement time graph.
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We said initially as far as acceleration goes to moment Alex left the airplane when velocity was 0 the acceleration of Alex was g.
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We are going to start at a maximum value and over time as Alex goes faster and faster the acceleration
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is going to decrease and decrease until reaching terminal velocity when there is no more acceleration.
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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.
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That should be right on the line there.
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As far as velocity goes, we know when we first push Alex out of the plane the velocity is 0
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We know that point and after a long time we know we eventually get to some value of terminal velocity which is mg / b.
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We will write that in there as an asymptote and we will have an exponential increase there following that same basic shape.
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For displacement, displacement begins at 0 and increases as speed increases
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until reaching a constant rate of increase when the velocity reaches VT.
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As far as displacement goes we are going to start at 0.
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Increase and increase until we get to some point where it is just going to be linear as
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we have a constant velocity when we are running it just terminal velocity.
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There would be graphs of acceleration, velocity, and displacement as functions of time that correspond to those calculations we just did.
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Let us take a look and see how this would look in a free response problem.
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You can download the problem yourself if you want 2005 free response 1 from the mechanics exam.
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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.
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2005 mechanics 1 looking at part A, we have a ball that is thrown vertically upward at some initial speed.
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It has some air resistance given by - kv the positive direction we are going to call up.
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Is the magnitude of the acceleration of the ball increases or decrease or remain the same as it moves upward?
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We are calling up the positive y direction and if I were to draw a free body diagram of the ball
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as it moves up we have its weight down and we have that drag force kv down.
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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.
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If we look at that as V goes down, as it slows down, as it gets higher and higher, V gets smaller.
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It looks to me like our acceleration, the magnitude of our acceleration in the y direction must decrease based on our formula.
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There is A must decrease.
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Let us take a look at part B.
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Write but do not solve the differential equation for the instantaneous speed of the ball in terms of time as it moves upward.
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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.
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We can write it as dv dt = - g –kv /m.
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That is a differential equation that would probably give us credit but let us clean it up just a little bit.
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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 – kv
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but either one of those you have your different equation down.
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For C, find the terminal speed of the ball as it moves downward.
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For C, looking at terminal speed.
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At terminal velocity we know the net force =0
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which implies then that our free body diagram is going to have something like this kv,
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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.
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Let us move on to check out part D.
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For part D, we are asked does it take longer for the ball to rise to its maximum height
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or to fall from its maximum height back to the height from which it was thrown?
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A tricky question.
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Let us see.
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On the way up, friction brings the ball to a stop quickly.
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This helps bring it to a stop more quickly than if they were no fiction.
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On the way, down the friction slows the ball down so it has more time in the air on the way down.
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That means that average velocity on the way up has to be greater than
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the average velocity on the way down because it happen more quickly.
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And because the distance is constant distance traveled on the way up and the way down
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the time to go down is greater than the time to go up because T = d/V.
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I would say then that it takes longer to fall.
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Some sort of explanation like that to go along with your answer.
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Part E, on the graph, sketch a graph of velocity vs. Time for the upward and downward parts of the ball's flight.
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We are going to need a graph here.
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Here is our velocity, there is our time, and we are looking at what happens for the upward and downward parts.
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It starts at some initial velocity V0.
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It is going to cross the axis here to have a velocity of 0 at its highest point.
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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.
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On the way down, we are going to have some value of terminal velocity.
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We will draw an asymptote in here for our V terminal.
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I would think that our graph would probably look something like this where it is approaching terminal velocity.
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Something like that final velocity.
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Something like that is your approach of final velocity at time Tf.
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I think that covers that one.
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Let us take a look at one more free response problem.
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Let us go to the 2013 APC mechanics exam.
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You can find it at this address or google it.
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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.
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We will see how it worked for you.
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In this problem, we have a box of mass M at rest in the constant applied force being applied
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there is a frictional drag force proportional to kv where V is the speed of the box,
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k is some positive constant, and we are given a dot to draw on label the forces.
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Draw our free body diagram actually.
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Let us draw our free body diagram first.
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There is our box.
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We know we have the normal force, we have the weight of the box, the force of gravity.
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We have some applied force which we are calling Fa.
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We must have our frictional force our drag force kv.
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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.
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We have done that sort of thing.
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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.
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All that must be equal to Ma but as a differential equation A = dv dt therefore Fa – kv = M dv dt.
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That will work.
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There is our different equation.
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We have velocity and its derivative in the same equation.
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Moving on to part C, determine the magnitude of the terminal velocity of the box.
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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.
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Therefore, the applied force = kv terminal or solving for V terminal that is just going to be our applied force divided by k.
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Part D, use the differential equation to derive the equation for the speed.
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Alright, we have to do some math and let us give ourselves some room.
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Starting with our equation f - kv = M dv dt, we are going to do the separation of variables again.
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This implies then that dv / f - kv = dt / M.
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Let us see how it would integrate that.
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The integral of the left hand side dv / F - kv must equal the integral of dt / M.
00:27:48.000 --> 00:27:57.900
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.
00:27:57.900 --> 00:28:06.200
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.
00:28:06.200 --> 00:28:08.800
You need –k on the top.
00:28:08.800 --> 00:28:16.400
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.
00:28:16.400 --> 00:28:19.600
We cannot just arbitrary throw things in there.
00:28:19.600 --> 00:28:22.300
In the right hand side it looks ok to integrate.
00:28:22.300 --> 00:28:29.000
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.
00:28:29.000 --> 00:28:45.900
The right hand side is just going to be t/M.
00:28:45.900 --> 00:28:58.600
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.
00:28:58.600 --> 00:29:03.300
If I multiply both sides by –k let us put –k there.
00:29:03.300 --> 00:29:07.400
It can go away and can make this a little bit simpler to see.
00:29:07.400 --> 00:29:18.400
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.
00:29:18.400 --> 00:29:28.300
That is going to be – the log of f = - kt/M.
00:29:28.300 --> 00:29:44.600
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.
00:29:44.600 --> 00:29:58.200
If I raise both sides to the e, the left hand side becomes f-kv/f = e^-k/Mt.
00:29:58.200 --> 00:30:01.600
A little bit more rearrangement here.
00:30:01.600 --> 00:30:13.400
F – kv / f let us multiply both sides by f to get f –Kv= f e^-k/Mt.
00:30:13.400 --> 00:30:15.700
We will get V all by itself.
00:30:15.700 --> 00:30:24.000
Let us get kv= f-fe^-k/Mt.
00:30:24.000 --> 00:30:47.000
We can factor out that f ÷ k so the velocity is going to be f/k × (1- e^-k/Mt).
00:30:47.000 --> 00:30:51.400
There is part D.
00:30:51.400 --> 00:31:02.700
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.
00:31:02.700 --> 00:31:04.500
We are getting pretty good at graph and these sorts of things by now.
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Let us give it a shot.
00:31:17.500 --> 00:31:26.400
We have V on the y axis, time on our x, and we know it is going to start at some velocity 0.
00:31:26.400 --> 00:31:29.700
We can also plug that in for T in our formula.
00:31:29.700 --> 00:31:31.900
If T is 0, e⁰ is 1.
00:31:31.900 --> 00:31:34.900
1-1 Is 0 so the velocity would be 0.
00:31:34.900 --> 00:31:45.800
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.
00:31:45.800 --> 00:31:52.800
Let us mark that here f/k.
00:31:52.800 --> 00:31:59.900
The shape of our graph is something like that.
00:31:59.900 --> 00:32:03.000
Retarding forces and drag forces, air resistance.
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Hopefully that gets you a good start.
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