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PHY1012F KINEMATICS IN 2D Gregor Leigh

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**Learning outcomes: At the end of this chapter you should be able to…**

PHY1012F 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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**KINEMATICS IN TWO DIMENSIONS**

PHY1012F 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 .

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**KINEMATICS IN TWO DIMENSIONS**

PHY1012F KINEMATICS IN TWO DIMENSIONS In one dimensional (1-d) motion… Non-zero and can only be either… parallel (body is speeding up), or antiparallel (body is slowing down). Or, in terms of components, vs and as can only either… have the same sign (body is speeding up), or have opposite signs (body is slowing down).

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**KINEMATICS IN TWO DIMENSIONS**

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

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**POSITION and DISPLACEMENT IN 2-D**

PHY1012F POSITION and DISPLACEMENT IN 2-D y The curved path followed by an object moving in a (2-d) plane is called its trajectory. (x1, y1) Position vectors can be written: (x2, y2) and x and the displacement vector as: i.e. WARNING: Distinguish carefully between y-vs-x graphs (actual trajectories) and position graphs (x, or y-vs-t)!

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**VELOCITY IN TWO DIMENSIONS**

PHY1012F VELOCITY IN TWO DIMENSIONS y Average velocity is given by: Instantaneous velocity by: y x x As seen in the diagram, as t0… , and thus also becomes tangent to the trajectory. Since , it follows that and Motion in 2-d may be understood as the vector sum of two simultaneous motions along the x- and y-axes.

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**VELOCITY IN TWO DIMENSIONS**

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

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**VELOCITY IN TWO DIMENSIONS**

PHY1012F VELOCITY IN TWO DIMENSIONS Once again, distinguish carefully between position graphs and trajectories… s t y x Position graph: Tangent gives the magnitude of . Trajectory: Tangent gives the direction of .

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**(a) Draw the particle’s trajectory. t (s) x (m) y (m)**

PHY1012F A particle’s motion is described by the two equations: x = 2t2 m and y = (5t + 5) m, where time t is in seconds. (a) Draw the particle’s trajectory. 10 20 y (m) x (m) 30 t (s) x (m) y (m) 5 1 2 10 2 8 15 3 18 20 4 32 25

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**(b) Draw a speed-vs-time graph for the particle.**

PHY1012F A particle’s motion is described by the two equations: x = 2t2 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) 4 8 t (s) 16 1 2 3 12 5.0 1 6.4 2 9.4 3 13.0 4 16.8

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**ACCELERATION IN TWO DIMENSIONS**

PHY1012F ACCELERATION IN TWO DIMENSIONS y where is the change in instantaneous velocity during the interval t. As we approach the limit t0… x …the instantaneous acceleration is found at the same point on the trajectory as the instantaneous velocity.

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**ACCELERATION IN TWO DIMENSIONS**

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

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**ACCELERATION IN TWO DIMENSIONS**

PHY1012F ACCELERATION IN TWO DIMENSIONS x- and y-components… y Hence and x We now have the following parametric equations: vfx = vix + axt vfy = viy + ayt xf = xi + vixt + ½ax(t)2 yf = yi + viyt + ½ay(t)2

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**ACCELERATION IN ONE DIRECTION ONLY**

PHY1012F 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… xf = xi + vixt + ½ax(t)2 yf = yi + viyt + ½ay(t)2 vfx = vix + axt vfy = viy + ayt y x Letting ax = 0, and starting at the origin at t = 0 with v0 making an angle of with the x-axis, we get: v0y vx = v0 cos vy = v0 sin + ayt v0x x = v0 cos t y = v0 sin t + ½ayt2

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**ACCELERATION IN ONE DIRECTION ONLY**

PHY1012F ACCELERATION IN ONE DIRECTION ONLY Since the equations are parametric, we can eliminate t: x = v0 cos t Substituting in y = v0 sin t + ½ayt2, 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.

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**(a) Draw a physical representation of the particle’s motion.**

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

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**(b) Draw the particle’s trajectory.**

PHY1012F A particle with an initial velocity of experiences a constant acceleration (b) Draw the particle’s trajectory. –0.5 y (m) x (m) 1 2 3 4 –0.4 –0.3 –0.2 –0.1 –0.6 x (m) y (m) 0.000 1 –0.033 2 –0.130 3 –0.293 4 –0.522

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PHY1012F 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. vix = vi cos , viy = vi sin, and ay = –g. y x vi 0, although vix and viy may be < 0. The horizontal distance travelled by a projectile before it returns to its original height is called its range, R. R

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PHY1012F PROJECTILE MOTION Since ax = 0, and ay = –g , the kinematic equations for projectile motion are: xf = xi + vix t yf = yi + viy t – ½g (t)2 vfx = vix = constant vfy = viy – g t 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.

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PHY1012F RANGE y The horizontal distance travelled by a projectile before it returns to its original height is called its range. v0y v0x R x i.e. R = x – x0 if and only if y – y0 = 0 Starting at the origin at t0 = 0, y = v0 sin t – ½gt2 = 0 t (v0 sin – ½gt) = 0 t = 0 (trivial) or R = x = v0 cos t

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PHY1012F RANGE y 80° 70° 50° 20° 40° 10° R x 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)… I.e. a projectile fired at an elevation of (90° – ) will have the same range as a projectile launched at angle .

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PHY1012F 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?

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PHY1012F 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 x0P, y0P, t0 vxP, v0yP x1P, y1P, t1 vxP, v1yP ayP x x0B, t0, vxB x1B, t1, vxB x0P = t0 = v0yP = 0 y0P = +60 m y1P = 0 m x0B = ? x1B = x1P = ? vxP = +50 m/s ayP = –g = –9.8 m/s2 vxB = +30 m/s t1 = t = ?

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**y1P = y0P + v0yPt – ½gt2 0 = 60 + 0 – ½ 9.8 t2 t = 3.5 s**

PHY1012F 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? y1P = y0P + v0yPt – ½gt2 0 = – ½ 9.8 t2 t = 3.5 s x1P = x0P + v0xPt + ½axPt2 = x1P = x1B = 175 m x1B = x0B + v0xBt + ½axBt2 175 = x0B + 30 x0B = 70 m y0P x0B 70 m 60 m = 40.6°

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PHY1012F IN REALITY… y x 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. 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°.

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**trajectory without gravity**

PHY1012F HOW FAR WILL IT GO? Any projectile, irrespective of mass, launch angle or launch speed, loses 5t2 m of height every second in free fall (from rest). trajectory without gravity 5 m 5 m 1 s 1 s 20 m 20 m 5 m 20 m 1 s 2 s 2 s 2 s So, no matter how great a projectile’s initial horizontal speed, it must eventually hit the ground…

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**HOW FAR WILL IT GO? …except that the Earth is not flat!**

PHY1012F 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. 8 000 m 5 m 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… The object will be a satellite, in orbit around the Earth!

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**Learning outcomes: At the end of this chapter you should be able to…**

PHY1012F 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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Chapter 4 Lecture 6: Motion in 2D and 3D: I HW2 (problems): 2.70, 2.72, 2.78, 3.5, 3.13, 3.28, 3.34, 3.40 Due next Friday, Feb. 19.

Chapter 4 Lecture 6: Motion in 2D and 3D: I HW2 (problems): 2.70, 2.72, 2.78, 3.5, 3.13, 3.28, 3.34, 3.40 Due next Friday, Feb. 19.

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