Physics 1110 Fall 2004L7 - 1Lecture 7 8 September 2004 Announcements• Reading Assignment for Fri, Sept 10: Knight 3.4, 4.1-4.2Continuation of Example ProblemLast time we discussed started a problem of a cart rolling up and thendown a ramp. We started with a motion diagram, then a picture todefine our variables and coordinate system. Before turning to theformulas we should think just a bit more about what we expect for ananswer by using a graphical representation. Since the acceleration ispositive and constant, a graph of the acceleration vs time is just ahorizontal line. Because the acceleration give the slope, we thereforeexpect a velocity vs time plot which just shows a straight line with aconstant slope. Since we know that the initial velocity is negative, theline should start at –2.0 m/s at time zero. Where this line crosses zerovelocity is the turning point. Finally the position plot for constantacceleration is always a parabola. We want a parabola that starts outwith a negative slope (because our initial velocity is negative) andthen comes to a minimum at the turning point (slope equal to zero)and then begins to increase with a positive slope (positive velocity).Plots showing this graphical representation are shown below.At this point we are ready to look at the formulas. Our first question(part a) asks for the time to get to the turning point. To choose the“best” formula we first look for a formula with t in it and this rules outthe equation with v2 in it. Then we look to see in which equation weknow everything BUT the t. To do this we need to keep in mind thatour initial variables are the one with the 0 subscript and our finalvariables are the one with a T subscript. Since we know the velocitiesinitially and at the top and we know the acceleration, it looks like the2nd equation is the best bet: vsf= vsi+ ast . Now we need to plug ourvariables into this generic formula. This is a very common place tomake a mistake.vT= v0+ gsintTPhysics 1110 Fall 2004L7 - 2At this point, many people put all of their numbers in, but this isactually not a good idea. Any arithmetic mistake becomes completelylost and not easy to find. Also, a general formula usually is useful. Soat this point, I usually only put constants in which are equal to zero. Inthis case that’s vT.0 = v0+ gsintTNow I solve the equation algebraically for tT to gettT=v0gsin Physics 1110 Fall 2004L7 - 3At this point I can insert the data with unitstT=2.0 m/s(9.81 m/s2)sin10°= 1.2 sNow we want to answer part b, which is to find the position of theturning point, which is xT in my variable scheme. Again, we look forequations with xT in them, which gives the first or third. Now that wehave solved part a), either one will work. Just because it lookssimpler, I will use the 3rd equation, vsf2= vsi2+ 2ass. Inserting myvariables givesvT2= v02+ 2gsin(xT x0)Once again, I will only put in zeros, which gives0 = v02+ 2gsin(xT x0)Solving for xT givesxT= x0v022gsin,and now putting in numbers givesxT= 2.0 m (2.0 m/s)22(9.81 m/s2)sin10°= 0.83 m.(By the way, make sure you agree that my units come out correctly!)Finally, we look to part c), which ask us to find the velocity when thecart has returned to its starting position, that is, when xB= x0. I don’tneed to know the time, so once again the 3rd formula looks like agood bet, since I know the initial velocity and the acceleration and thedisplacement (which is zero). Now I put my variables (and the zero) inthe generic equation, and I getvB2= v02+ 2gsin(xB x0) = v02Physics 1110 Fall 2004L7 - 4vB=±v0.I have two solutions to the equation; which is correct? At this point, Ireturn to my motion diagram; this is motion going down the ramp, sogiven my definition of the positive direction as pointing down theramp, the positive solution is the correct answer here. So finally weget vB=+2.0 m/s . As a general rule, when you get two solutions tosome equation always use reasoning about the motion to pick thecorrect answer. Don’t fall into the trap of using some general rulesuch as “take the positive solution.”Once final question remains; do you really need to do all of thesesteps to solve problems? After some practice, probably not. But somesteps, such as the pictorial representation where the variables aredefined, are really necessary every time. A better statement might be,that if you fully understand a problem, you should be able to giveevery representation fully as well as final numerical answers.Certainly someone who can just get numerical answers alone wouldnot be considered to have learned the physics behind the problem.In any case, these methods are most important in problems whereyou have almost no idea how to start. The first representations youdraw are in fact the correct way to start. The formulas should alwayscome last!Vectors and Vector ComponentsBefore we can move on to two- and three-dimensional physics, weneed to develop further our vector skills. While the intuitivedescription of vectors as having a magnitude and direction is themost easy to comprehend, it does not lend itself to precise calculationas soon as we need to add or subtract vectors. In order to do that, weneed to learn a method of breaking vectors down into smaller 1Dpieces, adding them, and then retransforming them back tomagnitude and direction. This process is called decomposing a vectorand it has a simple but powerful basis.In general, if we add two vectors, we will have some arbitrary lookingtriangle that we must try to sort out using things like the laws ofcosines or sines. If on the other hand, we can break it into rightPhysics 1110 Fall 2004L7 - 5triangles, then we can use trigonometry to find the vectors. Here’show we do it.In general I can always make up two vectors which add up to a giventhird vector, just like I can make up two numbers which add up to100, for example. If I chose the two vectors so that they are parallel tothe axes of my coordinate system (that is perpendicular for regularxyz systems, known as Cartesian systems), then the two vectors areunique and perpendicular. These are called vector components, andthe figure below shows an example. Make sure you see how the twovector components add up to the main vector using the tip-to-tailmethod.The component vectors now give me a right triangle, so I can usetrigonometry to find their magnitudes. We’ll do that next
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