Projectile Motion: Equations and Examples, Lecture notes of Fluid Dynamics

Download Projectile Motion: Equations and Examples and more Lecture notes Fluid Dynamics in PDF only on Docsity! Projectile Motion Projectile motion is a special case of two-dimensional motion. A particle moving in a vertical plane with an initial velocity and experiencing a free-fall (downward) acceleration, displays projectile motion. Some examples of projectile motion are the motion of a ball after being hit/thrown, the motion of a bullet after being fired and the motion of a person jumping off a diving board. For now, we will assume that the air, or any other fluid through which the object is moving, does not have any effect on the motion. In reality, depending on the object, air can play a very significant role. For example, by taking advantage of air resistance, a parachute can allow a person to land safely after jumping off an airplane. These effects are very briefly discussed at the end of this module. Projectile Motion Analysis Before proceeding, the following subsection provides a reminder of the three main equations of motion for constant acceleration. These equations are used to develop the equations for projectile motion. Equations of Motion for Constant Acceleration The following equations are three commonly used equations of motion for an object moving with a constant acceleration. ... Eq. (1) ... Eq. (2) ... Eq. (3) Here, is the acceleration, v is the speed, is the initial speed, is the position, is the initial position and is the time. To begin, consider an object with an initial velocity launched at an angle of measured from the positive x-direction. The x- and y-components of the object's initial velocity, and , can be written as and ... Eq. (4) and Eq. (5) Here, the x-axis corresponds to the horizontal direction and the y-axis corresponds to the vertical direction. Since there is a downward acceleration due to gravity, the y-component of the object's velocity is continuously changing. However, since there is no horizontal acceleration, the x-component of the object's velocity stays constant throughout the motion. Breaking up the motion into x- and y-components simplifies the analysis. The Horizontal Component Horizontal Displacement Using Eq. (2), at a time , the horizontal displacement for a projectile can be be written as Maximum Vertical Displacement When an object is launched with a positive vertical velocity component, the vertical velocity becomes zero at the point where the maximum height is reached. Hence, substituting in Eq. (15) gives the following equation for the maximum vertical displacement of a projectile. ... Eq. (16) Also, from Eq. (16) observe that, to achieve the maximum possible vertical displacement for a fixed initial velocity, should be equal to 1. This happens when is 90 (-90 is also valid mathematically but not physically. Mathematically, with a launch angle of -90 , the velocity is zero at a negative time). Therefore, to achieve the maximum height, an object must be launched straight up, which is also what intuition would tell us. The Path Equation By combining Eq. (7) and Eq. (11), an equation for the projectile's trajectory can be obtained. Rearranging Eq. (7) to isolate for and substituting it into Eq. (11) gives ... Eq. (17) This is the equation for the path of a projectile. If the initial conditions are known, then the path of the projectile can be determined. Due to the quadratic form of the equation, each y-position has two corresponding x-positions. These different x-positions correspond to different time values. For example, when a ball is thrown, it can cross a certain height twice, once on the way up and once on the way down. It should also be noted that, due to the tan and cos terms, this path equation is not defined at 90 and -90 . This can also be understood by the fact that there will be many y-positions for the same initial x- position. In the following plot, the dashed line shows the trajectory of a projectile launched at an initial height of 1m, with an initial velocity of 4m/s and at an angle of 45 from the horizontal. The sliders can be used to adjust the initial conditions and observe the new trajectory (solid line). Initial height (m) Launch angle (deg) Initial velocity (m/s) 1.50.5 2.01.00.0 45.090.0-90.0-45.00.0 4.03.02.01.00.0 Horizontal Range The horizontal range can be defined as the horizontal distance a projectile travels before returning to its launch height. When the final height is equal to the launch height (i.e. ), Eq. (11) reduces to the following, ... Eq. (18) When is non-zero, this equation can be written as ... Eq. (19) By substituting Eq. (18) into Eq. (7), the following equation is obtained ... Eq. (20) Using the identity gives .... Eq. (21) This shows that the horizontal displacement, when the projectile returns to the launch height, is greatest when This means that the range is maximum when the launch angle is 45 . The following plot shows the trajectory of a projectile launched with an initial velocity of 10 m/s, at an angle of 45 and with no initial height (dashed line). The launch angle can be varied to observe the effect on the range. Launch angle (deg) 90.060.030.00.0 Part b) Determining if the ball clears the net it if leaves the racquet at an angle 10 below the horizontal. Using the same approach as Part a), the time that the ball takes to reach the net can be solved for using Eq. (7). = at 5 digits 0.41462 The height of the center of the ball when it reaches the plane of the net can be calculated using Eq. (11). = Since y2 is negative, it means that the ball bounced on the ground before reaching (2.1.1.3.1) the net. A negative value is obtained since the equations do not account for the presence of the ground. In the following plot, the solid line shows the trajectory of the ball for this case and the dashed line represents the net. Trajectory - Part b). Part c) Finding the minimum angle required to clear the net. For the ball to hit the top of the net, the height of the ball should be equal to the height of the net when the ball reaches the plane of the net. By combining Eq. (7) and Eq. (11), and substituting the parameters specific to this problem, the following equation is obtained. Due to the form of the equation, there are many angles that satisfy it. For this case, only the smallest angle in the first and fourth quadrant are of interest. The two angles in the first and fourth quadrant are 1.504 rad and -0.063 rad. The smaller of the two angles is -0.063 rad, which is approximately equal to -3.6 . Therefore, the ball must leave the racquet at an angle slightly greater than -3.6 for it to just clear the net. In the following plot, the solid line shows the trajectory of the ball of this case. The dotted lines show the x- and y- positions of the ball when it reaches the plane of the net and the dashed line represents the net. Trajectory - Part c). MapleSim Simulation Constructing the model Step1: Insert Components Drag the following components into the workspace: Table 1: Components and locations Compon ent Location Multibody > Body and Frames Fig. 3: Parameters Note: Step 3 and Step 4 are not essential and can be skipped. The parameter values can be directly entered for each component instead of using variables. However, creating a parameter block as described above makes it easy to repeatedly change the parameters and play around with the model to see the effects on the simulation results. Step 5: Change the initial conditions Return to the main diagram and click the Rigid Body component. As shown below, enter the initial conditions in the Inspector tab in terms of the parameters. 1. Fig. 4: Rigid Body Parameters Step 6: Set up the net This step creates a rod that helps visualize the top of the net in the 3-D view. The length of a tennis net is approximately 11m and the height of the top of the net is approximately 0.91m at the center. Click the Fixed Frame component and, as shown in Fig. 5, enter 2. 1. for the location of the Fixed Frame. Fig. 5: Fixed Frame Parameters Then, as shown in Fig. 6, click the Rigid Body Frame component and change the length in the z direction to 11m. This length is not important and does not affect the results of the simulation, it just aids with the visualization. Fig. 6: Rigid Body Frame Parameters Step 7: Add a plot window for the trajectory 1. 1. Due to the form of the equation, there are many angles that satisfy it. For this case, only the smallest angle in the first quadrant, which is 0.049 rad or 2.81 , is of interest. Therefore, the rifle needs to be pointed at an angle of 2.81 above the horizontal to hit the target dead center. Part b) Determining the height of the point above the target that has to be aimed at. If the rifle needs to be pointed at an angle of , it needs to be aimed at a point above the center of the target. Since, the height above the target to which the rifle should be pointed is, = Therefore, the rifle has to aimed at a point 4.9 m above the target for the pellet to hit the target dead center. The following plot contains the pellet trajectory and the line of sight of the rifle. Trajectory - Part c). Part c) Determining the maximum vertical displacement of the pellet for the required trajectory. 1. 1. Using Eq. (16), the maximum vertical displacement of the bullet is = Therefore, the maximum vertical displacement of the pellet for the required trajectory is 1.22 m. Part d) Determining the vertical velocity of the pellet 0.25 sec before impact. The total time it takes the pellet to strike the target can be found using Eq. (7). = The vertical velocity 0.25 sec before impact can be calculated using Eq. (13). = Therefore, the velocity of the pellet 0.25 seconds before impact is -2.45 m/s. MapleSim Simulation Constructing the Model Step1: Insert Components Drag the following components into the workspace: Table 2: Components and locations Comp onent Location (2 require d) Multibody > Body and Frames Multibody > Visualization Multibody > 1. 1. Sensors Signal Blocks > Routing > DeMultiplexer s Signal Blocks > Relational Signal Blocks > Boolean Multibody > Body and Frames Step 2: Connect the components Connect the components as shown in the following diagram. Fig. 8: Component diagram Step 3: Create parameters Click the Add a Parameter Block icon ( ), click on the workspace and double click the Parameters icon that will appear on the workspace. Create parameters for the initial speed of the ball , the distance to the target , and the launch angle (as shown below). 1. 1. 1. 2. 1. 2. Click Probe 1 and select 1 and 2 in the Inspector tab. In this case, 1 corresponds to the x-position and 2 corresponds to the y-position. Using the Plot tab, add a plot window for the trajectory of the pellet. To add a window, click the Default Plot Window drop down menu and click --Add Window--. After giving the window a name, click OK and then click Empty to show the plot options. Give the plot a title and set up the plot to show Probe1: value[1] on the x-axis and Probe2: value[2] on the y-axis. Step 8: Run the simulation Click Probe 2 and select 1 and 2 in the Inspector tab. Also, click on the Greater Equal Threshold component and enter d in the threshold textbox in the Inspector tab. Click Run Simulation ( ). The following figure shows the 3-D visualization for the simulation. Fig. 12: 3-D visualization One step further: Incorporating Drag Drag force refers to the force that opposes the relative motion of an object in a fluid. In the case of the tennis ball, it is resistance offered by the air. The force due to drag is expressed using the following equation: 1. 1. ... Eq. (22) where, is the coefficient of drag, is the density of the fluid, is the characteristic area of the object and is the speed of the object relative to the fluid. In the case of a ball, the characteristic area is the area of cross-section. The coefficient of drag depends on a lot of different factors and is usually very difficult to calculate analytically. Hence the coefficient of drag is usually found through experiments or using numerical methods. For the case of a sphere, coefficient of drag curves are commonly available in textbooks and the internet. Fig. 13 is obtained from the www.nasa.gov website (available at: http://www.grc.nasa. gov/WWW/k-12/airplane/dragsphere.html). Fig. 13: Coefficient of drag for spheres. As can be seen, the coefficient of drag is plotted as a function of the Reynolds number Re. This a dimensionless number which is very commonly used in the field of fluid dynamics. It gives a measure of the ratio of inertial forces to viscous forces. For a sphere the Reynolds 1. 1. number is expressed as, ... Eq. (23) where, is the density of the fluid, is the velocity of the flow, D is the diameter of the sphere and is the viscosity of the fluid. If the ball is moving through the air with a velocity of then the direction of the drag force will be in a direction opposite to this vector. The magnitude of this drag force will be given by Eq. (22). In vector form this force will be ... Eq. (24) This can be simplified and written as, ... Eq. (25) where, is the speed of the ball. This can be incorporated into the MapleSim model as shown in the following diagram. Since the coefficient of drag is not constant, points from the curve are input into a Microsoft Excel spreadsheet in the form of a table and attached to the model. For the case of the tennis ball, the curve that corresponds to a rough sphere has been used. This data can then be used by the 1D Lookup Table component which interpolates between the points.