Physics 1110 Fall 2004L14 - 1Lecture 14 24 September 2004Announcements• Reading Assignment for Mon, Sept 27: Knight 6.4, 7.1-7.2Two-Dimensional MotionIn the last two weeks we have focused on problems in which themotion ended up being one-dimensional, along a line. This meantthat in general, the acceleration and velocity had to lie along this line.If they were in opposite directions, the object would slow down; if theywere in the same direction, the object would speed up. Notice thatthis quality of the motion is not related to the coordinate system. Infact, many of the problems were solved with a two-coordinate system(usually x-y). However, the motion was nonetheless along a line. Nowit is time to explore the consequences of having acceleration that isnot along the velocity. In general, this will mean that the accelerationvector will have components perpendicular to the velocity (as well aspossibly parallel to the velocity). We looked at such motion inqualitative terms in Chapter 1, and you’ll recall that having suchcomponents led to motion with curvature. The motion was not along astraight line.The figure below compares these types of motion: case A is simpleconstant velocity motion, while case B is 1D constant acceleration.Case C shows an example with constant acceleration withcomponents perpendicular to the motion.Physics 1110 Fall 2004L14 - 2The most obvious feature of case C is that the motion is curved andno longer lies along a line. This means the motion really takes placein two dimensions (in a plane, in other words). Notice also that thecurve is in the direction of the acceleration vector.Two-Dimensional Motion with Constant AccelerationAlthough we have a more general case than in the previous lectures,nothing is actually changed in terms of analysis. We still find the netforce, and then use that to find the acceleration, using Newton’s 2ndlaw: r a =r F netm.However, now we will have to analyze all components of this vectorequation. If the forces are constant, then the net force is a constantand therefore the acceleration is a constant. It will have twocomponents in general. (As a side note, when acceleration andvelocity vectors were parallel or antiparallel, the two vectors onlydefined a line in space, so the motion was 1D; we could have alwaysfound a coordinate system so that all motion was in 1 coordinate.Now that the acceleration and velocity vectors aren’t along a line,they define a plane, which means the motion will be 2D. This meanswe can always find a coordinate system so that we only have twocomponents to solve for.) Writing Newton’s law explicitly for eachcomponent gives: ax=r F net()xmay=r F net()ym.In general, each acceleration components will be non-zero and wesolve the equations for each component separately. But this is not abig deal, since it is no different (equation-wise) than solving the 1Dconstant acceleration problems. To make this clear, below are thestandard kinematic equations from Ch. 2, but now written for the xand y coordinates separately:Physics 1110 Fall 2004L14 - 3xf= xi+ vxit +12axt2yf= yi+ vyit +12ayt2vxf= vxi+ axtvyf= vyi+ aytvxf2= vxi2+ 2axxvyf2= vyi2+ 2ayySo at this point, we really don’t have new analysis formulas to learn,as much as we need to just do a few examples.Example 1: A ball in free fall. Consider a ball that is launched fromthe ground with an initial speed of 10 m/s at an angle 30° above thehorizontal. Ignore air resistance. A) How high does the ball go? B)How long is the ball in the air?From a motion diagram standpoint, we expect something looking likecase C in the figure above. Since we are neglecting air drag, there isonly one force on the ball: gravity, which is straight down. If wechoose a standard coordinate system with the positive y-direction up,and the positive x-direction in the same direction as the horizontalcomponent of the initial velocity, then we have the following pictorialrepresentation:From what we are given in the problem we know the initial positionand velocity. I’ll take the initial position as the origin of my coordinatesystem and the initial time to be 0. This gives the following for theinitial variable:Physics 1110 Fall 2004L14 - 4 ti= 0r r i= xi) i + yi) j = 0) i + 0) j r v i= vxi) i + vyi) j = vicos) i + visin) j = (10 m/s)cos(30°)) i + (10 m/s)sin(30°)) j = (8.7 m/s)) i + (5.0 m/s)) j .Since there is only a single force, the weight w = mg in the negative y-direction, then the acceleration is constant and with this coordinatesystem has the components: ax= 0 and ay= g . This immediately tellsus that in the horizontal direction, the speed is constant, while in thevertical direction we have the same free fall motion we had for a ballthrown straight up. Thus, all the solutions will be the same as wefound in our previous work on Ch. 2 material. Before looking at theequations, let’s remind ourselves of the graphical representation forthe motion in each coordinate, shown below.Make sure you understand these graphs!Now finally we obtain the formulas for our motion. In the x direction,we have constant x component of velocity sox = vxitvx= vxiPhysics 1110 Fall 2004L14 - 5In the y direction we have constant acceleration, soy = vyit 12gt2vy= vyi gtvy2= vyi2 2gy .Keep in mind that these formulas were derived for this particularproblem, where I chose the initial position to be the origin and theinitial time to be zero!Trajectories Before finding the solutions to the questions, we shouldalso discuss the trajectory. The trajectory of an object is its paththrough space. The path is only a limited piece of information since itdoes not tell us the speed along the path, just the direction of motion.To find the trajectory in this case, we eliminate the time variable fromour formulas for x and y position, and then get a formula relating xposition to y position, which is what a trajectory commonly is. Firstthen we solve the x equation for t to findt =xvxi.Then we substitute this into our y equation to gety =vyivxix g2vxi2x2.This is the equation for a parabola, which is in agreement with theshape of trajectory we are familiar with in everyday life!Finally now we find the answers to the specific questions asked. Wefind the maximum height using the y motion, just as in the case of aball going straight up and down. Although the velocity of the ball doesnot go to zero at the highest point, the y-component of the velocitydoes go to zero
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