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Relations for 2-D motion Motions that occur in 2-dimensions are more complicated since now the direction of the acceleration, velocity, force, etc. need to be taken into account. F ma These vectors, F , and a , are still proportional to each other by the proportionality constant, mass, and the two vectors are in the same direction. F ma mr Another important difference is that instead of taking time derivatives of a single valued function x, (i.e. a scalar), we now have to take successive time derivatives of a vector r . 5-1 How do we take time derivatives of a vector? dr r dt The answer is that it depends….. It depends on which coordinate system we choose as convenient for a specific problem. For 2-D motion, we will work in Cartesian and Polar coordinates. First, let’s consider how to write functions and take derivatives in Cartesian coordinates FORCE ACCELERATION xˆ yˆ F Fx xˆ Fy yˆ a ax xˆ ay yˆ Are unit vectors in the x- and y- directions (i.e. having magnitude of 1 and whose purpose is only to indicate direction) 5-2 Time derivatives in Cartesian Coordinates Why is it so convenient to work in Cartesian coordinates? The answer is that the unit vectors do not change in time. They always point in the same direction. Consider the vector: r xxˆ yyˆ dr v r xxˆ xxˆ yyˆ yyˆ xxˆ yyˆ dt d xxˆ yyˆ ar xxˆ xxˆ yyˆ yyˆ xxˆ yyˆ dt Thus, if the force is only in the x- or y- direction, then the trajectory of the object can be computed the same as for 1-dimensional motion. 5-3 Polar Coordinates In certain types of motion, such as revolution about an axis, it is sometimes more convenient to use polar coordinates. The main difficulty in polar coordinates is that the unit vectors can change direction as a function of position or time. y ˆ y r sin rˆ r x, y r, x r cos x The transformation between Cartesian and Polar unit vectors can be written as follows: rˆ xˆ cos yˆ sin ˆ xˆ sin yˆ cos rˆ cos sin xˆ ˆ sin cos yˆ 5-4 Converting back to Cartesian from Polar Coordinates rˆ cos sin xˆ Given ˆ sin cos yˆ The transformation between Polar and Cartesian unit vectors can be written as follows: cos sin rˆ cos sin cos sin xˆ xˆ sin cos ˆ sin cos sin cos yˆ yˆ 1 1 xˆ cos sin rˆ yˆ sin cos ˆ Given these forward and reverse transformations between Cartesian and polar coordinates, the next question is how to take time derivatives of vectors written in polar coordinates? Suppose there is a vector, written as follows: What is r ? r rrˆ 5-5 Time derivatives in Polar Coordinates: I r rrˆ rrˆ d r rrˆ r xˆ cos yˆ sin dt r rrˆ r xˆ sin yˆ cos ˆ r rrˆ rˆ Hence the velocity in polar coordinates can be given in terms of an object’s radial position, radial velocity, and rotation frequency. 5-6 Time derivatives in Polar Coordinates: II r d rrˆ rˆ dt r rrˆ rrˆ rˆ rˆ rˆ rˆ r 2 rˆ r r r2 rˆ 2r r ˆ d xˆ sin yˆ cos dt ˆ xˆ cos yˆ sin rˆ ˆ This is a very interesting result, but is it useful? Why can’t we solve everything in the much simpler Cartesian coordinates? It turns out that a certain class of forces, known as “central forces” are most easily described using polar coordinates. 5-7 F Fx xˆ Fy yˆ Fx m ax 0 Fy m ay m g ay g dvy dt dy v y gt v y 0 gt dt a x t 0 v x t v x 0 dvx t dt dx dt 1 2 dy g tdt y y0 y 2 gt 1/ 2 2y t g 2y dx v x0 dt x x0 x v x0 t v x0 g 1/ 2 5-8 Solution 1/ 2 2y x vx 0 g g 9 .8 m s2 1/ 2 g vx 0 x 2y x 40 m y 16m 26m 10m vx 0 28m / s 5-9 F Fx xˆ Fy yˆ v vx xˆ vy yˆ where With Initial Conditions Fx m ax 0 Fy m ay m g vx v cos v y v sin vx 0 vx v cos v y g v y gt v sin x v cos xt v cos t x0 1 2 y gt v sin y t gt v sin t y0 2 5 - 10 Given that we know the y position of the projectile for t=0 and t=final are at the same position, we can solve for t. 1 2 1 yt y0 0 gt v sin t t gt v sin 2 2 2v sin t 0; t g 2v 2 sin cos v 2 sin 2 xt v cos t x0 x0 x0 g g When is xt x0 a maximum? Answer: It is maximal when sin2 1; 45 o In other words, at this angle the projectile can travel a maximal distance. So, what initial velocity is required for the projectile to travel 12km? v2 xt x0 12km v g 12kmg 343ms 5 - 11 Initial Conditions: r 70 sft2 0.02 rad s 2 r 30000ft 60o r r r2 rˆ 2r r ˆ r rrˆ rˆ Terms we don’t know x const. d r cos dt d2 x 0 2 r cos dt x 0 5 - 12 x 0 d r cos r cos r sin 0 dt sin r r r tan cos d2 d x 0 2 r cos r cos r sin dt dt x 0 rcos r sin r sin r cos 2 r sin r cos r cos 2 2 r sin r sin r cos cos 2 2 2 tan r sin sin 5 - 13 Solving for velocity magnitude r r 2 r 2 2 r tan r 2 r tan2 1 r cos rad r 30000ft 0.02 s cos 1200 fts 3 Solving for acceleration magnitude r r r 2r r r r 2r tan r r r rcot r cot r 2 2 2 2 2 2 2 2 r 70 ft2 30000ft 0.02 rad 2 s s r 67 ft2 s r cot 2r tan 2 r cot 2 2 2 2 r r 2 1 cot2 r r 2 sin 3 sin 3 5 - 14 Problem 1 As observed from a ship moving due east at 8 km/h, the wind appears to blow from the south. After the ship has changed course, and as it is moving due north at 8 km/h, the wind appears to blow from the southwest. Assuming that the wind velocity is constant during the period of observation, determine the magnitude and direction of the true wind velocity. 5 - 15 Problem 1 As observed from a ship moving due east at 8 km/h, the wind appears to blow from the south. After the ship has changed course, and as it is moving due north at 8 km/h, the wind appears to blow from the southwest. Assuming that the wind velocity is constant during the period of observation, determine the magnitude and direction of the true wind velocity. 1. To solve relative motion problems: a. Write the vector equation that relates the velocities of two particles. vB = vA + vB/A vB and vA are the velocities of particles B and A relative to a fixed frame, respectively. vB/A is the velocity of B relative to A. 5 - 16 Problem 1 As observed from a ship moving due east at 8 km/h, the wind appears to blow from the south. After the ship has changed course, and as it is moving due north at 8 km/h, the wind appears to blow from the southwest. Assuming that the wind velocity is constant during the period of observation, determine the magnitude and direction of the true wind velocity. b. Express all velocity vectors in terms of their rectangular components. c. Scalar equations. Substitute the velocity vectors in vB = vA + vB/A and solve the two independent sets of scalar equations obtained. 5 - 17 Problem 1 Solution First observation Write the vector equation that relates the velocities of two particles. y N vs1 vw = vs1 + vw/s1 vw/s1 E x Express all velocity vectors in terms of their rectangular components. vw = vwx i + vwy j Scalar equations. vs1 = 8 i km/h vwx = 8 + 0 vw/s1 = vw/s1 j vwx = 8 km/h vwy = 0 + vw/s1 5 - 18 Problem 1 Solution Second observation Write the vector equation that relates the velocities of two particles. y N vs2 vw/s2 vw = vs2 + vw/s2 o 45 E x Express all velocity vectors in terms of their rectangular components. vw = vwx i + vwy j vs2 = 8 j km/h vw/s2 = vw/s2 cos 45o i + vw/s2 sin 45o j Scalar equations. vwx = 0 + vw/s2 cos 45o vw = 8 i +16 j km/h vwy = 8 + vw/s2 sin 45o N vwy = 16 km/h vw = 17.89 km/h 63.4o E 5 - 19