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Spherical Projectile Motion with DragTo calculate the motion of a simple projectile without drag, the equations for the x position and y position of the object on an xy plane are as follows:x=vxt+12at2+xoy= vyt+12at2+yoWhere vox and voy are the original speeds in the x and y directions, xo and yo are the coordinates of the initial position of the projectile, a is equal to the acceleration of the projectile, and t is equal to the time since the projectile was launched.The only force acting on the projectile in these equations is gravity. In order to account for forces such as drag, these equations will need to be altered to give better approximations. Drag equation:D=pCDA2The force of drag:FD=Dv2CD is the drag constant, determined experimentally, p is the density of the fluid (in this case air), v is the velocity of the projectile, and A is the cross-sectional area of the projectile. For simplicity, these equations will be modified using the drag coefficient and surface area of a sphere (though they can be easily changed to accommodate any drag coefficient and surface area). To calculate the motion of a spherical projectile in a plane with drag, the steps are as follows (explanations will follow the steps):Split FD into its component forces along the x and y axis (v is in the opposite direction of FD).Fx=FDcosθ=-Dv2cos(θ)=-Dv(vx)Fy=FDsinθ-mg=-Dv2sinθ-mg= -Dvvy-mgDetermine ax and aySince F=ma:Fx=max , therefore: ax=FxmFy=may , therefore: ay=Fymax=-Dv2cos(θ)may=-Dv2sin(θ)-mgm(vx=vcosθ, and vy=vsin(θ)) Calculate the value of DD=pCDA2D=1.225*0.5*πr22D=0.306πr2Incorporate the new accelerations into the equations that do not take into account drag. Since finding the integral of ay is much more challenging than ax, the equation for y position will be slightly inaccurate, as it assumes a constant acceleration. y=vyt+12-D*(v2sin(θ))-mgmt2ax, however, does not need to incorporate gravity, and can be solved for with respect to time using very basic differential equations:-Dv2cos?(θ)=mdvdt-Dcos(θ)mdt=1v2dv-Dcos(θ)mdt=1v2dv-Dcos(θ)mt= -1v+CC= 1vo (because v is a function of t, setting t = 0 returns one over the initial velocity, which is equal to 1vo)1v=Dtcos(θ)m+1vo1v=1voDtvocos?(θ)m+1solving for v results in:v= vomm+(Dt*vo*cos?(θ))now, taking the integral of v gives the equation for x position:x= vxmm+(Dt*vo*cos?(θ))dtwhich evaluates to: (C = 0)x= mD*ln1+Dvocosθ*tm(vx=vcosθ, D=0.306πr2)Explanation:Step 1:The total force in the x-direction is equal to the drag force times the velocity squared of the projectile multiplied by the cosine of the angle of the force below the x-axis. This force is negative because it is in the opposite direction of the projectile’s motion. The total force in the y-direction is equal to the drag force times the velocity squared multiplied by the sine of the angle of the drag force below the x-axis, minus the mass of the projectile times gravity. This force is also negative because it is pulling the projectile down.Step 2:Using force is equal to mass times acceleration, we can solve the equations of Fx and Fy for ax and ay. Step 3:CD, the drag coefficient, for spheres is about 0.5, the density for air, p, is around 1.225 kg/m3, and the equation for surface area of a sphere is 4πr2, so D for a sphere can be calculated to equal 1.225πr2. Step 4In order to solve for the parametric equations used to find the projectile’s location, some work must be done since the acceleration is not constant and is proportional to the velocity. For the purposes of simplicity in this paper, the equation for the y position will be simplified and assume a constant acceleration from drag. To find an accurate equation for x, however, the velocity needs to be found with regard to t so the equation for velocity can be integrated. Because acceleration is equal to the derivative with respect to t of velocity, a simple differential equation shown in step 4 can be set up. This is fairly simple to solve, and once it is solved there is a simple equation relating v to t. Taking the integral of this equation results in the final equation for the x position of the projectile, assuming non-constant acceleration.Modeling these equations for comparative graphing purposes can be done in Matlab using the following code:This code will graph the path of a projectile without air resistance in green and a projectile with air resistance in blue on the same axis. Changing the variables for velocity (vi), radius (r), mass (m), and θ (theta), as well as the limits of the graph (xlim and ylim), will allow you to graph the path of any spherical projectile of any size and mass against the path of the same projectile with drag applied. For example, with an initial speed of 34 m/s, a radius of 0.1 m, and a mass of 1kg, and a launch angle of 45 degrees, the paths look like this:With an initial velocity of 35m/s, at an angle of 65 degrees, a mass of .5kg, and a radius of 0.05m (5cm), the paths look like this:Unfortunately, this is still only an approximation (though a much more accurate one) due to the simplified equation for y which used constant acceleration. However, this program using these equations clearly demonstrates the effect of drag on a spherical object. With some fairly straightforward calculus, these equations for simple projectile motion can be used to derive equations which much more accurately model the true path of a projectile. Works Cited:"Drag of Cylinders & Cones."?, 2012, question/aerodynamics/q0231.shtml. Accessed 8 Dec. 2017."Projectiles with Air Resistance."?Dynamics, 2015, dynref.engr.illinois.edu/afp.html#afp-fd . Accessed 8 Dec. 2017."Projectile Motion with Air Resistance." wps.wps/media/objects/877/898586/topics/topic01.pdf. Accessed 8 Dec. 2017.Allain, Rhett.?Air Resitance with Quadratic Drag.YouTube, 2015, . Accessed 8 Dec. 2017. ................
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