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NEWTON’S LAWSMOTION IN A PLANE PHY1012F KINEMATICS IN 2D Gregor Leigh

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Presentation on theme: "NEWTON’S LAWSMOTION IN A PLANE PHY1012F KINEMATICS IN 2D Gregor Leigh"— Presentation transcript:

1 NEWTON’S LAWSMOTION IN A PLANE PHY1012F KINEMATICS IN 2D Gregor Leigh

2 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 2 MOTION IN A PLANE Learning outcomes: At the end of this chapter you should be able to… Extend the understanding, skills and problem-solving strategies developed for kinematics problems in one dimension to two dimensional situations. Define the terms specific to projectile motion; solve numerical problems involving projectile motion.

3 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 3 KINEMATICS IN TWO DIMENSIONS We shall apply vectors to the study of motion in 2-d. How the position of a body,, changes with time (  t ) is determined by its acceleration,, and depends on the body’s initial position,, and initial velocity,. We use vectors and their components to represent these directional quantities. In this chapter we shall concentrate on motion in which the x- and y- components are independent of each other. Later (in circular motion) we shall investigate motion in which the x- and y- components are not independent.

4 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 4 Non-zero and can only be either… KINEMATICS IN TWO DIMENSIONS In one dimensional (1-d) motion… parallel (body is speeding up), or antiparallel (body is slowing down). Or, in terms of components, v s and a s can only either… have the same sign (body is speeding up), or have opposite signs (body is slowing down).

5 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 5 can have components both parallel to and perpendicular to. KINEMATICS IN TWO DIMENSIONS In two dimensional (2-d) motion… The parallel components alter the body’s speed; The perpendicular components alter the body’s direction.

6 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 6 POSITION and DISPLACEMENT IN 2-D The curved path followed by an object moving in a (2-d) plane is called its trajectory. y x ( x 1, y 1 ) ( x 2, y 2 ) Position vectors can be written: and the displacement vector as: and i.e. WARNING:Distinguish carefully between y -vs- x graphs (actual trajectories) and position graphs ( x, or y -vs- t )!

7 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 7 yy, and thus also becomes tangent to the trajectory. Motion in 2-d may be understood as the vector sum of two simultaneous motions along the x- and y- axes. VELOCITY IN TWO DIMENSIONS y x Since, As seen in the diagram, as  t  0… Average velocity is given by: Instantaneous velocity by: xx it follows that and.

8 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 8 If ’s angle is measured relative to the positive x- axis, its components are: where is the body’s speed at that point. VELOCITY IN TWO DIMENSIONS y x Conversely, the direction of is given by  and

9 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 9 VELOCITY IN TWO DIMENSIONS Once again, distinguish carefully between position graphs and trajectories… y x s t Position graph: Tangent gives the magnitude of. Trajectory: Tangent gives the direction of.

10 NEWTON’S LAWSMOTION IN A PLANE PHY1012F y (m) x (m) t (s) x (m) y (m) A particle’s motion is described by the two equations: x = 2t 2 m and y = (5t + 5) m, where time t is in seconds. (a) Draw the particle’s trajectory

11 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 11 A particle’s motion is described by the two equations: x = 2t 2 m and y = (5t + 5) m, where time t is in seconds. (b) Draw a speed-vs-time graph for the particle. and t (s) v (m/s) v (m/s) t (s)

12 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 12 where is the change in instantaneous velocity during the interval  t. ACCELERATION IN TWO DIMENSIONS y x As we approach the limit  t  0… …the instantaneous acceleration is found at the same point on the trajectory as the instantaneous velocity.

13 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 13 Instantaneous acceleration can be resolved… ACCELERATION IN TWO DIMENSIONS y x We can resolve it into components parallel to and perpendicular to the instantaneous velocity… a  and a , however, are constantly changing direction, so it is more practical to resolve the acceleration into… Where… a  alters the speed, and a  alters the direction.

14 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 14 ACCELERATION IN TWO DIMENSIONS y x x- and y- components… Hence and. We now have the following parametric equations: v fx = v ix + a x  tv fy = v iy + a y  t x f = x i + v ix  t + ½a x (  t) 2 y f = y i + v iy  t + ½a y (  t) 2

15 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 15 x = v 0 cos  t y = v 0 sin  t + ½a y t 2 v x = v 0 cos  v y = v 0 sin  + a y t ACCELERATION IN ONE DIRECTION ONLY Let us consider a special case in the xy- plane, in which a particle experiences acceleration in only one direction… v fx = v ix + a x  tv fy = v iy + a y  t x f = x i + v ix  t + ½a x (  t) 2 y f = y i + v iy  t + ½a y (  t) 2 Letting a x = 0, and starting at the origin at t = 0 with v 0 making an angle of  with the x- axis, we get: y x v0yv0y v0xv0x 

16 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 16 Since the equations are parametric, we can eliminate t : x = v 0 cos  t Substituting in y = v 0 sin  t + ½a y t 2, we get i.e. Any object for which one component of the acceleration is zero while the other has a constant non-zero value follows a parabolic path. ACCELERATION IN ONE DIRECTION ONLY

17 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 17 y (m) x (m) A particle with an initial velocity of experiences a constant acceleration. (a) Draw a physical representation of the particle’s motion.

18 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 18 A particle with an initial velocity of experiences a constant acceleration. (b) Draw the particle’s trajectory. x (m) y (m) –0.033 –0.130 –0.293 – –0.5 0 y (m) x (m) 1234 –0.4 –0.3 –0.2 –0.1 –0.6

19 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 19  PROJECTILE MOTION A projectile is an object upon which the only force acting is gravity. I.e. a projectile is an object in free fall. The start of a projectile’s motion is called its launch. The angle above the x-axis at which a projectile is launched is called the launch angle, or elevation. The horizontal distance travelled by a projectile before it returns to its original height is called its range, R. y x R v ix = v i cos , v iy = v i sin , and a y = –g. v i  0, although v ix and v iy may be < 0.

20 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 20 PROJECTILE MOTION Since a x = 0, and a y = –g, the kinematic equations for projectile motion are: v fx = v ix = constant v fy = v iy – g  t x f = x i + v ix  t y f = y i + v iy  t – ½g (  t) 2 Notes:  t is the same for the x- and y- components. We determine  t using one component and then use that value for the other component. Although the x- and y- components are linked in this way, the motion in one direction is completely independent of the motion in the other direction.

21 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 21 RANGE The horizontal distance travelled by a projectile before it returns to its original height is called its range. i.e. R = x – x 0 if and only if y – y 0 = 0 v0xv0x y x v0yv0y R  R = x y = v 0 sin  t – ½gt 2 = 0 Starting at the origin at t 0 = 0,  t (v 0 sin  – ½gt) = 0  t = 0 (trivial) or = v 0 cos  t

22 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 22  RANGE y x R Notes: R is a maximum when sin(2  ) = 1 i.e. 2  = 90° i.e.  = 45° for maximum range. Since sin(180° –  ) = sin , it follows that sin(2(90° –  )) = sin(2  ) … 20° 70° 10° 80° 40° 50° I.e. a projectile fired at an elevation of (90° –  ) will have the same range as a projectile launched at angle .

23 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 23 A small plane flying at 50 m/s, 60 m above the ground, comes up behind a bakkie travelling in the same direction at 30 m/s and drops a package into it. At what angle to the horizontal should the “bomb sights” (a straight sighting tube) be set?

24 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 24 A small plane flying at 50 m/s, 60 m above the ground, comes up behind a bakkie travelling in the same direction at 30 m/s and drops a package into it. At what angle to the horizontal should the “bomb sights” (a straight sighting tube) be set? y x  x 0P, y 0P, t 0 v xP, v 0yP x 0B, t 0, v xB a yP x 1B, t 1, v xB x 1P, y 1P, t 1 v xP, v 1yP x 0P = t 0 = v 0yP = 0y 0P = +60 m v xP = +50 m/sa yP = –g = –9.8 m/s 2 x 0B = ?x 1B = x 1P = ?y 1P = 0 m v xB = +30 m/st 1 = t = ? 

25 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 25 A small plane flying at 50 m/s, 60 m above the ground, comes up behind a bakkie travelling in the same direction at 30 m/s and drops a package into it. At what angle to the horizontal should the “bomb sights” (a straight sighting tube) be set? y 1P = y 0P + v 0yP t – ½gt 2 0 = – ½  9.8 t 2  t = 3.5 s x 1P = x 0P + v 0xP t + ½a xP t 2 =   x 1P = x 1B = 175 m x 1B = x 0B + v 0xB t + ½a xB t = x 0B + 30   x 0B = 70 m   = 40.6°  y 0P  x 0B 70 m 60 m

26 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 26 IN REALITY… Air resistance does affect the motion of projectiles, destroying the symmetry of their trajectories. The less dense the projectile, the more noticeable the drag effect. Although vertical speed is independent of horizontal speed, the action of running helps to increase the initial vertical component of an athlete’s jump speed. y x If the landing is lower than the launch height, the “range equation” is not valid. Here, angles less than 45° produce longer ranges. When projecting heavy objects (e.g. shot puts), the greater the elevation, the greater the force expended against gravity, and the slower the launch speed. Thus maximum range for heavy projectiles thrown by humans is attained for angles less than 45°.

27 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 27 2 s 20 m 1 s 5 m 1 s 5 m HOW FAR WILL IT GO? Any projectile, irrespective of mass, launch angle or launch speed, loses  5 t 2 m of height every second in free fall (from rest). 20 m 2 s So, no matter how great a projectile’s initial horizontal speed, it must eventually hit the ground… trajectory without gravity

28 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 28 HOW FAR WILL IT GO? …except that the Earth is not flat! Because of its curvature, the Earth’s surface drops a vertical distance of 5 m every m tangent to the surface. So theoretically, in the absence of air resistance, tall buildings, etc, an object projected horizontally at m/s at a height of, say, 1 m will have the Earth’s surface dropping away beneath it at the same rate it falls and will consequently get no closer to the ground… 5 m m The object will be a satellite, in orbit around the Earth!

29 NEWTON’S LAWSMOTION IN A PLANE PHY1012F 29 MOTION IN A PLANE Learning outcomes: At the end of this chapter you should be able to… Extend the understanding, skills and problem-solving strategies developed for kinematics problems in one dimension to two dimensional situations. Define the terms specific to projectile motion; solve numerical problems involving projectile motion.


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