PHY183-Lecture08 - Physics for Scientists& Engineers 1 Spring Semester 2011 Lecture 8 Realistic Trajectories and Relative Motion Physics for

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Unformatted text preview: Physics for Scientists & Engineers 1 Spring Semester 2011 Lecture 8 Realistic Trajectories and Relative Motion January 27, 2011 Physics for Scientists&Engineers 1 1 1 Clicker Quiz Which of the following is included in the formulas for the ideal projectile motion of an object: A. B. C. D. Wind resistance Its spin Its weight The acceleration due to gravity January 27, 2011 Physics for Scientists&Engineers 1 2 2 Realistic Projectile Motion We have been studying ideal projectile motion To calculate realistic trajectories we must take into account the effect of the air on the motion of the object • Air resistance • Spin Here is a comparison of ideal versus realistic for the flight of of a baseball January 27, 2011 Physics for Scientists&Engineers 1 3 3 Hitting a Home Run Suppose a batter hits a baseball so that it has an initial launch angle of 35º and an initial speed of 110 mph (realistic) How far will the baseball go? We call the distance the ball travels the range R We have derived the range in terms of the launch angle and initial initial speed v0 = 110 mph = 49.2 m/s R= v sin2θ0 = sin (2 ⋅ 35° ) = 231.5 m 2 g 9.81 m/s 2 0 ( 49.2 m/s ) 2 R = 231.5 m = 759.5 ft Not realistic! Air resistance has a big effect January 27, 2011 Physics for Scientists&Engineers 1 4 4 Relative Motion So far: we moved the origin of the coordinate system to a location that is convenient for calculations • Example: shifting x0 so that x0 = 0 at the start of a projectile’s trajectory Until now we have always kept the coordinate system in the same place during the motion that that we want to describe But there are some situations for which a moving coordinate system is necessary • Example: Airplane landing on a moving aircraft carrier January 27, 2011 Physics for Scientists&Engineers 1 5 5 Example: Airport Walkway Person walking with a velocity vw, as measured by an observer moving along with him on the walkway. Walkway surface moves with vwt relative to terminal. Two velocities add as vectors Velocity of person as measured by someone standing in the terminal: vt = vwt + vw January 27, 2011 Physics for Scientists&Engineers 1 6 6 Moving Reference Frames Assume that reference frame (= coordinate system) moves with constant velocity relative to a coordinate system that is at rest Then accelerations measured in both reference frames are the same Airport walkway example, again: • If vwt = const . dvwt / dt = 0 • From vt = vwt + vw we then obtain: dv dvt d (vwt + vw ) dvwt dvw = = + =0+ w dt dt dt dt dt at = aw January 27, 2011 Physics for Scientists&Engineers 1 7 7 3 Dimensions Two coordinate systems (xl , yl , zl ) (laboratory) and (xm , ym , zm ) (moving) that have their axes parallel to each other and coincide at t = 0 Origin of (xm , ym , zm ) moves with constant velocity vml relative to (xl , yl , zl ) After some time t , the origin of (xm , ym , zm ) is located at rml = vmlt Vector addition gives us the transformation between frames rl = rm + rml = rm + vmlt January 27, 2011 Physics for Scientists&Engineers 1 8 8 3 Dimensions (2) Velocities rl = rm + vmlt d d d rl = (rm + vmlt ) = dt rm + vml dt dt vl = vm + vml because vm is constant in time Accelerations d d d vl = (vm + vml ) = dt vm + 0 dt dt al = am January 27, 2011 Physics for Scientists&Engineers 1 9 9 Clicker Quiz One should never jump off a moving vehicle (train, car, bus, etc.). Assuming, however, that one does perform such a jump, from a physics standpoint, what would be the best direction to jump from a train in order to minimize the impact of the landing? A. In the same direction as the moving train B. In the opposite direction as the moving train C. Perpendicular to the direction of the moving train Speed with respect to the ground should be minimized Jump backward January 27, 2011 Physics for Scientists&Engineers 1 10 10 Driving Through Rain When driving through the rain, one notices that the rain almost always comes straight at us. Why? Again, this is a consequence of moving reference frames. View of observer standing on street View from the car -vcar is the velocity of reference frame January 27, 2011 Physics for Scientists&Engineers 1 11 11 Example: Driving Through the Rain It is raining and there is no wind to speak of. Rain drops of 0.08 inches diameter (typical) are falling with a terminal velocity of 14 mph (6.26 m/s) A car is driving through the rain (direction does not matter!) with a speed of 25 mph (11.2 m/s). With which angle relative to the horizontal does the rain hit the car? vrain θ January 27, 2011 r − vcar vrain −1 14 θ = tan = tan = 29.2° vcar 25 −1 Physics for Scientists&Engineers 1 12 12 Projectiles in Moving Reference Frames Bow hunting: target is 25 m away, shoot an arrow with initial velocity 90 m/s, horizontal direction, exactly aimed at bull’s eye. Target moves with 3 m/s from left to right. Question: Where does the arrow hit? January 27, 2011 Physics for Scientists&Engineers 1 13 13 Projectiles in Moving Reference Frames Answer: • It takes t = (25 m)/(90 m/s) = 0.28 s for arrow to arrive at target • Arrow falls by (1/2)gt2 = 0.5 · 9.81 · 0.282 m = 0.38 m • Target movement is moving reference frame => sideward deflection of arrow by vt = 3 m/s · 0.28 s = 0.83 m Aim here! January 27, 2011 Physics for Scientists&Engineers 1 14 14 Ferry Crossing The captain of a ferry wants to travel directly across a river that flows due east with a speed of 1.07 m/s. He starts from the south bank of the river and wants to reach the north bank by traveling straight across the river. The boat has a speed of 6.34 m/s with respect to the water. What direction (in degrees) should the captain steer the boat? Note that 90º is east, 180º is south, 270º is west, and 360º is north. vr vb sin θ = vr vb vr 1.07 m/s = sin −1 = 9.71° 6.34 m/s vb θ θ = sin −1 Captain must steer 360° - 9.71° = 350.3° 15 January 27, 2011 Physics for Scientists&Engineers 1 15 Clicker Quiz An airplane flies with a speed of 200.0 mph with respect to the air. The pilot wants to fly from point A to point B (due east, 90º) with respect to the ground. The wind is blowing from north to south (180º) with a speed of 20.0 mph. What course course does the pilot have to steer? A. B. C. D. E. 67.2º 79.4º 84.3º 90.0º 97.1º 16 January 27, 2011 Physics for Scientists&Engineers 1 16 Clicker Quiz Solution The angle θ is given by sin θ = vwind vairplane −1 θ = sin vwind 20 mph −1 = sin = 5.7° vairplane 200 mph So the direction the pilot has to fly is Course = 90° - 5.7° = 84.3° January 27, 2011 Physics for Scientists&Engineers 1 17 17 Third Base to First Revisited Remember we discussed that a third baseman could throw a ball from third to first with an initial speed of 40.2 m/s at an angle of 6.81º or an angle of 83.2º. What is the difference in the time it takes for the ball to get to from third to first? There are several ways to approach this problem • Considering motion in the y-direction only, we could calculate the amount of time it would take for the ball to reach its maximum height • Considering motion in the x-direction only, we could calculate the amount of time it would take for the ball to travel from third to first • Let’s do both! January 27, 2011 Physics for Scientists&Engineers 1 18 18 Third Base to First Revisited (2) Let’s calculate the time it would take for the ball to reach its maximum height and return v y 0 = v0 sin θ0 v y = 0 = v y 0 − gt v y 0 = gt t= vy0 g 2v y 0 g g = 2v0 sin θ 0 g Physics for Scientists&Engineers 1 19 (at the top) ttotal = ttotal 2 ( 40.2 m/s ) sin θ 0 2 9.81 m/s = 8.20sin θ 0 s ttotal = 2t = ttotal = for θ 0 = 6.81°, ttotal = 0.972 s for θ 0 = 83.2°, ttotal = 8.14 s 2 (v0 sin θ0 ) January 27, 2011 19 Third Base to First Revisited (3) Let’s calculate the time it would take for the ball to travel from third to first analyzing the x direction vx = vx 0 = v0 cos θ 0 x = vx t total t total = t total = x x = vx v0 cos θ 0 x1 v0 cos θ 0 t total = t total = 1 38.8 m 0.965 s = 40.2 m/s cosθ 0 cosθ 0 0.965 s cosθ 0 for θ 0 = 6.81°, t total = 0.972 s for θ 0 = 83.2°, t total = 8.15 s We get the same answer whether we analyze the horizontal direction or the vertical direction That answer is: If you want to get the runner, throw the ball at 6.81º! January 27, 2011 Physics for Scientists&Engineers 1 20 20 ...
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This note was uploaded on 04/23/2011 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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