Hi, this is module five of two dimensional dynamics. Today's learning outcomes are to describe the kinematic relationship of acceleration in a tangential and normal coordinate system. Last time we did pos, position and velocity. And then we're going to use those kinematic relationships to solve. A particle can emax problem in tangential and normal coordinates. So, going back to last class, we had this particle going along a curve linear path its velocity is going to be S dot for its magnitude the change in arc length with respect to time and the direction is always going to be tangential to the path e sub tangential, and so the acceleration is going to then be a derivative of the velocity, both S dot and E tangential, the direction will change as we go along the path, so I've got to use the product rule, so I've got S double dot times e tangential plus S dot times the derivative of e tangential, or S double dot, e tangential plus S dot, and I'm going to break. Similar to what I did before, I'm going to break this derivative of e tangential with respected time into two parts, a motion part, the theta dt, and a geometrical part, d theta t. A d of the e in the tangential direction, d theta. And so, here if I look at my particle at point p here, this is my x and my y coordinates, I have a tangential direction and I have a, a radial direction or a normal direction that I, I'll, dro, draw. That's, that's directed radially in, inward, inward for the curve path that it's on and so This is an overlay of those two coordinate system. And so, what I want do to know is I want to write ,e tangential in terms of i and j and e normal, in terms of i. And j. And so, I'd like you to do that put, fill in the blanks here and then come on back. So, what you should've found is okay for e tangential ,we're going to go out cosine theta, distance cosine theta in the i direction. And a distance sine theta ,in the j direction. And for e normal, here's e normal, I'm going to go minus sign theta in the i direction, and plus co sign theta in the j direction ,so what I can see from this relationship is real important if I if I take the derivative of e tangential with respect to theta, I get derivative of cosine with respect to theta is minus sine theta. The derivation of sine theta with respect to theta is cosine theta. So I know that ,e normal then is equal to the derivative of e tangential with respect to theta. And so that's important because that's the result I need up here. So, let's continue on. There's my velocity. I, I have now found that d theta, dt. D d theta d of e tangential. With respect to theta is e normal, so I know what this is. I still need to find, the theta dt. And so, for any infinitesimal segment of the curve along the arc length here, this goes along a small portion of the curve. Let's say that the radius of curvature here row, is constant, so s you should remember back from your math days, your basic math days, that this distance s that you go is equal to row times theta. Okay? And so if I take the derivative with respect to time for an infant, infinitesimal segment. ,row doesn't change, so I get the theta d t equals one over row d s d t and so again I can use that to put in here and so I get a sub p is equal to s double dot in the e tangential direction plus I've got s dot here. Here I've got the d theta which is s dot and one over row. So ,I've got s dot here. S dot there. So that's s dot squared. Over row and we know that d of e tangential d theta is equal to e normal. And so that's an expression for the acceleration in the tangential and normal coordinate system. Another way to express it would be, a sub p is equal to s double dot e tangential plus, s dot, remember? That's the speed, or it's the magnitude of the velocity. So, I can write this as the magnitude of the velocity of p squared over rho, in the normal direction. And so, this term is the rate of change of the magnitude of acceleration, rate of change of the magnitude of the velocity, excuse me. So this is, the rate, of change, of magnitude, of velocity and we call that the. Tangential acceleration because its in the tangential direction. This is the tangential acceleration, and this is the rate of change of the direction of the velocity. This is rate of, change of direction, of velocity and we call that the radial, or normal acceleration. This is the normal Or radial acceleration. So ,in normal and tangential coordinates, we have two components: a rate of change of the magnitude of the velocity, and that's" vection" but this should be Velocity and a rate of change of the direction. So let's look at a small demo here. Here's a, let's say that this is a particle here this, this ball. And it's moving along a curved path. And so as it moves along a curved path if I speed it up, there can be a change in the magnitude of the velocity and the tangential acceleration and the direction is constantly changing, and it's changing radially inwards, so that's the normal or radial acceleration that's inward. And so you see both components of the acceleration of the particle. Alright, let's go ahead and use that that theory for the kinematic relationships in normal and tangential components, er, coordinates, to go ahead and solve a problem. We've got a racing car. Which can be treated as a, as a particle. It starts from rest and at a starting point here it goes for 30 seconds, through point A, a distance of one mile. And it accelerates at a constant rate. Immediately after it reaches point A, it starts into a curve. The greatest of curvature of that cure is .2 miles and we want to find the normal and the tangential acceleration just before the car reaches Point A and just after the car reaches Point A. And so, this is Point A right at the, at the beginning of the curve. And so, for Part A, The normal component of acceleration, we just found was the, magnitude of the velocity squared over rho. Or the magnitude of the velocity squared ,row before the car reaches point A that the radius of curvature, since this is a straight line, is infinity. So we find out, that the normal component of acceleration before point A is zero. All we have is a tangential component of acceleration. Let's go ahead and find that. Lets let ,the tangential component of acceleration equal c one and we're given that its constant. So we'll say that that's constant ,and so to then go to velocity I have to integrate so I've got the velocity is equal to the integral of a t d t, which is equal to the integral C 1 d t, we call that the constant t 1, or C 1 times time, since it's an indefinite integral plus a constant of integration C 2. How do we find the constant of integration? I have to apply the I c. So we're going to apply, the IC and we know that the velocity at time zero equals zero cause it starts from rest and so that's C1 times time at time zero plus C2 so we get c two Equals 0. And so ,now we have the velocity is equal to C1 times, t. To find the distance. We have to integrate again ,so we got S equals the integral of velocity, dt so that's equal to the integral if i integrate C1t that's C12 squared over and then we don't need the we don't need the integration sign here. Its going to be CI t squared over two and then I'm going to get since this is a this is going to be an indefinite integral again so I'm going to get another constant integration C3. And again, I can ply the I the IC to find the constant of integration. So, apply the IC for X at time 0 which is 0. X of 0 =0, therefore CI times zero squared over two, plus C3, or again, C3 equals zero. And so now, I can imply the over condition for S which is S at time t is equal to t is 30 seconds and that's we've gone a distance of one mile so I've got s at time 30 seconds equals uh,C1 times t again 30 seconds squared over two or C1 equals 2.22 times to the minus three miles Per second squared. You can see one is that constant acceleration rate that we wanted to find and so a tangential then is equal to C1 or 2.22 times ten to the minus three miles per second squared. That' s the magnitude. [COUGH] And the direction is tangent to the path, so it's going to be to the right prior to getting to point A. And that's the answer for part A. Now let's go ahead and do part B. So far part B, the first question I have to ask you is, what is the tangential acceleration just instantaneously after point A? And so what you should've answered is just instantaneously after point A, that tangential acceleration is going to continue to be 2.2 times ten to the minus three, miles per second squared. And it's going to be tangent to the path. Last thing we need to do is find the normal acceleration after point A. Again, we know that A, normal is v squared over row. We need to find v at this point A, well, we found that v was equal to the transcendental acceleration times time, and so v equals two point two, two times ten to the minus three times thirty seconds.A And so, v We get the point a is 6.67 times ten to the minus two miles per second squared. And so finally, a normal equals, we've got v now, 6.67 times ten to the minus two squared over the radius of curvature. The radius of curvature at point a, we said it goes through a turn with a radius of curvature of 0.2 miles. And so a normal, equals 2.22, times ten to the minus two miles per second squared. And I can write a normal as a vector, it's equal to 2.22 times ten to the minus two miles per second squared. And it's going to be directed radially inward, to the center, of the curvature of the curve. So that's it's magnitude. That's it's direction. And finally, for practice for you, I'd like you to do a,a, Can a max problem in normal tangential coordinates. We've got a racing car again. This time it starts at rest and increase its speed around our circular track at a constant rate of sixty per second squared. Its travelling counter clockwise and we want to determine the position and time at which the car's accelerations Magnitude reaches 20 feet per seconds squared, and, that's an exercise on, for you on your own, and you'll find the solution in the module handouts, and we'll start back up again, next time.