* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Lesson 2 – Vectors, more motion problems, using computers
Survey
Document related concepts
Photon polarization wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Tensor operator wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Equations of motion wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Bra–ket notation wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Classical central-force problem wikipedia , lookup
Four-vector wikipedia , lookup
Transcript
Lesson 2 – Vectors, more motion problems, using computers Vectors An engineer has to deal with physical quantities. Unlike plain numbers in math, physical quantities usually have dimensions (mass, length and time). In addition, physical quantities can have both direction and magnitude. Scalars Some physical quantities only have magnitude. Examples of such quantities would be things like mass and time. Such quantities are called Scalars. Adding scalar quantities is similar to adding numbers in math. The main difference is that scalar quantities have dimensions and only scalars with the same dimensions can be added. For example, it does not make sense to add 1 kilogram to 25 seconds. Furthermore, adding scalars also implies that you convert the scalar quantities so that they are represented in the same units. For example, suppose you want to add a half hour to 17 minutes: 1 1 60 min hour 17 min= hour × 17 min=30 min17 min=47 min 2 2 hour So, as long as you change the scalars to have the same units, you can add (or subract) them just like you add (or subtract) numbers. Some physical quantities have direction in addition to magnitude Vectors are physical quantities that have direction as well as magnitude (size). Common examples would be things like forces and velocities. The gravitational force on an object points towards the center of the earth. The gravitational force has a magnitude equal to the object's mass multiplied by the acceleration due to gravity. That force has a definite direction. This direction must be taken into account if you are trying to balance that force (and keep that object from falling). Speed is the magnitude of the velocity. So, speed is a scalar. If you look at the speedometer in a car, it registers the speed. It is possible to drive around the hairpin curve on the Pali highway with a constant speed. It is not possible to do so with a constant velocity (or you will go off the road). Adding vectors Because vectors have direction as well as magnitude, they must be added in a different way from how we add scalars and numbers. Go through the “Animated intro to adding vectors” link to see a step by step explanation of adding vectors. As you will see in that PDF file, the standard way of adding vectors is to 1 break the vectors into rectangular components. Adding the rectangular components of a vector is like adding scalar quantities. So, this simplifies the process of adding vectors. To add vector A to vector B to obtain a resultant vector R: 1. Break vector A and vector B into their x and y components. 2. Add the x components like scalars. That is R x = A x B x and R y =A y B y 3. Combine R x and R y to obtain the resultant vector R Although there are some important details left out, the above three steps summarize how to add vectors. Here are some sample problems. To check your answers, you can use the “Tool for adding vectors” link. Vector Addition Problems When solving these problems, sketch the vectors first. Check to see if your results match your drawing. Do not use the “Tool for adding vectors” link until you have solved the problem by hand using a calculator. Vector A is 100 N at 30º above the positive x axis. Vector B is 50 N at 20º above the negative x axis. Find R , where R = A B Vector A is 100 N at 30º to the right of the negative y axis. Vector B is 200 N in the positive y direction. Find R , where R = A B Make up your own problems, solve them, and check with the “Tool for adding vectors” link. Vector Addition using spreadsheets In the computer lab, we can set up vector addition using spreadsheets. Spreadsheet programs are valuable programs for an engineer to use. They can be used for many engineering calculations. Even more important, spreadsheet programs provide two ways of helping to visualize a solution to a problem involving calculations. We can set up a spreadsheet to add vectors. The main thing to keep in mind is that the spreadsheet treats angles as radians not degrees. So, we have to convert our angles in degrees to radians. In addition, we need to think of our angles as starting at zero for the positive x direction, and increasing counterclockwise. This means that the positive y direction is 90 degrees, the negative x direction is 180 degrees and the negative y direction is 270 degrees. Then, we have to convert those degrees to radians. The conversion of degrees to radians is done using this formula: 2 degrees× =radians 180 Particle Motion Problems Let's return to the previous topic of solving particle motion problems. Consider the following problem: A particle undergoes rectilinear motion with a constant acceleration of −10m / s 2 . The particle has an initial velocity of 8.45 m/ s . At what time(s) will the particle have a position of 2.65 m ? What will the velocity be at that time or times? You can solve this problem using the quadratic equation, but let's solve it using the spreadsheet instead. Start by coming up with the appropriate equation. The equation we want to start with is: 1 2 x =x o v o t a t 2 1 2 2.65=08.45 t −10 t Let's bring everything to one side. 2 5 t 2−8.45 t2.65=0 The key manipulation is to move everything to one side of the equation, so that the other side is zero. This means we have: f t=5 t 2−8.45 t2.65=0 So solutions are when f t=0 Next, we set up the spreadsheet so that we can evaluate f (t ) . Note in the formula bar, how we have used the formula: = 5*(A2)^2 – 8.65*(A2) + 2.65 Now, all we have to do is add more rows with the time value incremented, to see how f (t ) changes. 3 The formulas to do this are shown in the figure above. Filling these formulas down several rows will result in the following values. Note that f(t) crosses zero between 0.2 and 0.4 seconds, and also betwee 1.2 and 1.4 seconds. The approximate solution lies somewhere in those regions. This can be seen if you make a chart of the data. The arrows show the location of the approximate solutions. 4 You can get better approximations by decreasing the time increment value. Here are the data with a time increment of 0.1 sec. Now, you can see the answer is between 1.3 and 1.4. We can go to a smaller increment , like 0.01 sec. This time, we skip over some times, so that this table does not go over too many rows. 5 Now the approximate answers can be seen to be t = 0.40 sec, 1.33 sec. You could go to even smaller time increments to get greater accuracy if you wanted. See if you can figure out how to get approximate answers for the velocity at these times. (Did you get v =4.45 m/s and v =−4.85 m/s ?) Suppose you change the problem a bit. The particle still starts off with the same initial velocity of 8.45 m/ s and still undergoes a constant acceleration of −10 m/ s 2 . This time, find the velocities when the position is 2.47 m . Make sure you start with the right formula. (Did you get v =±4.69 m/s ?) In this case, is solving with a spreadsheet a good idea? 6