Linear Motion. Miroslav Mihaylov. February 13, 2014

Transcription

1 Linear Motion Miroslav Mihaylov February 13, Vector components Vector A has manitude A and direction θ with respect to the horizontal. On Fiure 1 we chose the eastbound as a positive x direction and the northbound as a positive y direction so that the x and y components A x and A y of the vector A are both positive. For the manitude of vector A and its components we have the relations. A = A 2 x + A 2 y tan(θ) = A y A x sin(θ) = A y A cos(θ) = A x A Fiure 1: Vector components 2 One dimensional motion Equations of motion in one dimension are. x = x i + v i t at2 v = v i + at v 2 f v2 i = 2a x The sin in front of the velocity could be neative dependin on the choice of the positive direction. Acceleration bein the rate of chane of the velocity its sin is dependent on both the direction and the rate of chane. With other words the sin of the acceleration is a product of the sin of the chane with the sin of the velocity that is chanin. Example: if we have deceleration ( ) of an object movin in the positive (+) direction then the sin in front of the acceleration would be ( ) (+) = ( ) neative. 1

2 3 Motion in two dimensions When we deal with linear motion in two dimensions we typically reduce the problem into two one dimensional problems. Consider the example of projectile launched with initial velocity v i and at an anle θ i with respect to the horizontal. See Fiure 2 Fiure 2: Projectile motion For the initial velocity in each direction we have v ix = v i cos(θ) v iy = v i sin(θ) In the x direction we have simple uniform linear motion with velocity v ix x = v ix t In the y direction we have the earths acceleration slowin down the object. The equations of motion for the y component are: y = v iy t 1 2 t2 v y = v iy t v 2 y = v 2 iy 2 y 2

3 Note that the neative sin in the y direction equations of motion already takes into account the direction of the earths acceleration. You should pluin just the absolute value of i.e. 9.8m/s 2 when solvin problems. We can use the fact that at its peak of the trajectory v y is zero so we can find h max. v 2 y = v 2 iy 2 y 0 = v 2 iy 2h max v 2 iy = 2h max h max = v2 iy 2 h max = v2 i sin2 (θ) 2 The point where the projectile hits the round has a y = 0 coordinate. Puttin that in the equation of motion for the y direction we et the total travel time: y = y i + v iy t 1 2 t2 0 = 0 + v iy t 1 2 t2 v iy t = 1 2 t2 v iy = 1 2 t t = 2v iy t = 2v 0 sin θ The expression for t above is the time when the projectile hits the round. Puttin that into x = x i + v ix t we can find the rane of the projectile R. x = x i + v ix t 2v iy R = 0 + v ix R = v i cos(θ) 2v i sin(θ) R = v2 i 2 cos(θ) sin(θ) Note that the total travel time equals twice the time needed for projectile to reach the peak. Also the final velocity and its components equals the initial velocity. v f = v i v yf = v iy θ f = θ i 3

4 4 Appendix 4.1 Problem with forces An object acted on by three forces moves with constant velocity. One force actin on the object is in the positive x direction and has a manitude of 6.5 N. A second force has a manitude of 4.4 N and points in the neative y direction. What is the direction and manitude of the third force actin on the object? Solution The combined force due to the 4.4N and 6.5N forces is And it acts in a direction θ below the x axis. F = = 7.85N tanθ = θ = 34 Fiure 3: Problem Forces Object movin with constant velocity has acceleration a = 0, so the third force has to have the same manitude as force F i.e. 7.85N opposite direction to that of the force F i.e. at anle 34 w.r.t y axis or 124 o and w.r.t. x axis. 4

5 4.2 Problem on one dimensional motion You are drivin at 12 m/s and all of a sudden a ball rolls out in front of you. Then you immediately apply the brakes and bein to decelerate at rate of 3.4 m/s 2. What is the time from the moment you hit the brakes till the complete stop? How far did you travel before stoppin? What is your speed when you have traveled half the distance needed to stop? Solution: Let us choose as positive the direction that we are drivin. Complete stop translates to zero final velocity i.e.v f = 0 and decelerate means that the acceleration has neative sin. 0 = v i at v i = at t = v i a = 3.5s To et the travel distance we can use the travel time that we ve already found. We already know the acceleration and initial velocity so we can write. x = v i t 1 2 at2 x = (3.5) 2 = 21.2m Alternatively we can use this relation to find the distance traveled. v 2 f v2 i = 2a x 0 v 2 i = 2a x x = v2 i 2a = 21.2m For the third question if we replace x with 1 2 x from the answer above we et. v 2 = v 2 i 2a 1 2 v 2 i v 2 = v 2 i 2a x 2a = v2 i 1 2 v2 i = 1 2 v2 i v = 1 2 v i = 8.5m/s 5

6 4.3 Problem on projectile motion Find the launch anle for which the rane R and maximum heiht h max of a projectile are the same. Solution Takin the ratio h max /R h max /R = v2 i sin2 (θ) / v2 i 2 cos(θ) sin(θ) 2 h max R = vi 2 sin2 (θ) 22vi 2 sin(θ) cos(θ) vi 2, and one of the sin θ cancels out so we are left with When h max = R we et h max R = sin(θ) 4 cos(θ) = 1 4 tan(θ) 1 = 1 4 tan(θ) 4 = tan(θ) θ =