Newtonian Physics to Predict the Speed of Gravity Racer - Report Example

Student Name(s) Student ID Course Hons Abstract This report contains MATLAB solutions for a racer moving down the slope. The solutions indicate that when the there is no air resistance as wells as no friction then speed of the race accelerates constantly. However the velocity was affected when parameters were changed. When weight and distance were varied velocity of racer was affected. This showed that the velocity of racer depends on its size, distance, and mass. When the drag force is introduced velocity is reduced. When air and friction are introduced velocity goes down. The racer slows down, approaching the new velocity. Note that the terminal speed is not the maximum possible speed; it is the speed that the falling object approaches, regardless of initial conditions, when the only forces acting are drag and gravity. Introduction Any movement down a slope is affected by gravitational force as well frictional force they act in opposite direction of the motion. This paper will look at the effect of air and rolling resistance of a racer. Then the pull of gravity makes the pulls the racer downward increasing the velocity by increasing the acceleration downward. During the downward trip velocity is still changing at the same constant rate with which it changed on the way up and at the top of the path. The acceleration has the same constant value magnitude and direction for the entire path. The displacement of the racer at any instant is the vector sum of the displacements in the two mutually perpendicular directions. The motion of a racer when air resistance is negligible is the superposition of horizontal motion with constant velocity and vertical motion with constant acceleration. The vertical and horizontal motions each proceed independently, as if the other motion were not present. The strobe photo shows snapshots of the same instant, another was projected horizontally. Method The Matlab code for the race car movement is as follows Distance=input('distance'); %given distance down the slope theta=input('degrees'); %changing degrees to radians Acceleration =10*sind(theta); %calculation of acceleration of the Time=((2*Distance)/Acceleration)*0.5;% computation of time v= Distance/Time; % computation of speed in m/s Velocity2= Velocity/2.24 %% computation of speed in mph T=0:Time/30:Time; y=Velocity x=Time plot(x,y,'-.dr'); xlabel('Time in seconds') ylabel('Velocity in m/s') title('Velocity vs Time of racer') the flowchart for making Matlab scripts is as follows. Answer: The required flowchart is given in Fig. 25.2. Results A) No air resistance and no rolling friction After running Matlab code as follows the Velocity is 0.6843m/s, Acceleration is 0.6843 and Time is 2.9226e+003 Distance= ('2000'); %given distance down the slope theta=input('4'); %changing degrees to radians Acceleration =10*sind(theta); %calculation of acceleration of the Time=((2*Distance)/Acceleration)*0.5;% computation of time v= Distance/Time; % computation of speed in m/s Velocity2= Velocity/2.24 %% computation of speed in mph T=0:Time/30:Time; y=Velocity x=Time plot(x,y,'-.dr'); xlabel('Time in seconds') ylabel('Velocity in m/s') title('Velocity vs Time of racer') Acceleration = 0.6843 Time = 2.9226e+003 Velocity = 0.6843 b) when air resistance Distance= ('2000'); %given distance down the slope theta=input('4'); %changing degrees to radians Acceleration =10*sind(theta)-mg*Cd; %calculation of acceleration of the Time=((2*Distance)/Acceleration)*0.5;% computation of time v= Distance/Time; % computation of speed in m/s Velocity2= Velocity/2.24 %% computation of speed in mph T=0:Time/30:Time; y=Velocity x=Time plot(x,y,'-.dr'); xlabel('Time in seconds') ylabel('Velocity in m/s') title('Velocity vs Time of racer') Acceleration = 7.4704 Time = 6.6931 6.4254 6.4254 6.4254 Velocity = 0.4704 Velocity2 = 3.3350 Both resistances of air and friction Distance= ('2000'); %given distance down the slope theta=input('4'); %changing degrees to radians Acceleration =10*sind(theta)-mg*Cd-mg*0.004; %calculation of acceleration of the Time=((2*Distance)/Acceleration)*0.5;% computation of time v= Distance/Time; % computation of speed in m/s Velocity2= Velocity/2.24 %% computation of speed in mph T=0:Time/30:Time; y=Velocity x=Time plot(x,y,'-.dr'); xlabel('Time in seconds') ylabel('Velocity in m/s') title('Velocity vs Time of racer') Question 2 m = 216 % mass of 216kg A = 0.36 % area of chasis b = 0.31 % drag of coefficient of G = tf([b k],[m b k]) % creating Transfer function figure(1); step(G) % creating graph Increasing the drag coefficient will lead to decreased comfort due to increased resistance changes thus changing the frequencies for an compensated time. This uncomfortable oscillatory response will make the passenger. When decreasing the damping ratio there will be reduced uncomfortable oscillatory response leading to increased comfort. It shows that the relationship between damping coefficient and the stiffness of the spring is inverse that is an increase in damping coefficient will lead to a decrease in the number of oscillation that are experienced by the passengers. Once there are changes in the damping ratio there is need to consider the cause of an error if it is a shock absorbers then they should be replaced. The damping ratio affects the time‐domain of system are again important thus affecting smooth ride of a car. Matlab was used to assess frequency response and to simulate the system transient response. System performance specification was based upon closed-loop time domain criteria of rise time and damping ratio Question 3 Distance=('2000'); %given distance down the slope Distance1=('0:0:500'); %given distance down the slope Distance2=('0:500:1250'); %given distance down the slope Distance3=('0:1250:2000'); %given distance down the slope Distance4=('0:2000:3000'); %given distance down the slope theta=('4'); %changing degrees to radians theta1=('5'); %changing degrees to radians theta2=('2'); %changing degrees to radians theta3=('7'); %changing degrees to radians theta4=('3'); %changing degrees to radians Acceleration =10*sind(theta); %calculation of acceleration of the Acceleration1 =10*sind(theta1); %calculation of acceleration of the Acceleration2 =10*sind(theta2); %calculation of acceleration of the Acceleration3 =10*sind(theta3); %calculation of acceleration of the Acceleration4 =10*sind(theta4); %calculation of acceleration of the Time=((2*Distance)/Acceleration)*0.5;% computation of time v= Distance/Time; % computation of speed in m/s Velocity2= Velocity/2.24 %% computation of speed in mph T=0:Time/30:Time; y=Velocity x=Time plot(x,y,'-.dr'); xlabel('Time in seconds') ylabel('Velocity in m/s') title('Velocity vs Time of racer') Discussion When air resistance and friction are taken into consideration is taken the velocity of the race is reduced. The race takes 6.7s reach its maximum destination before as compared to 4.32s when retarding force of air resistance are not in existence. The mass does not affect the maximum speed, the time of travelled, or the range, if we are given the initial velocity. Where mass does matter is in the ability of the racer to accelerate from rest to its launch velocity. The racer has an easier time imparting a particular velocity to a less massive projectile. Conclusion From the results it can be noted that retarding force of air resistance and friction with direction opposite to the direction of motion. This detrimental to movement of the racer, when it is moving, air drag contributes to the net force on the object. Once it reaches terminal velocity, the drag force is equal in magnitude to the weight of the object, but acts upward. The net force on the object is zero, so the acceleration is zero. The direction of the terminal velocity is always downward if there are no forces other than air resistance and gravity. The magnitude of the terminal velocity, the acceleration gets smaller and smaller. The acceleration is zero when the object falls at its terminal velocity. The net force on an object falling at its terminal velocity is zero, so the drag force is equal in magnitude to the weight the terminal speed, and the magnitude of the drag force at any speed. The terminal speed of an object depends on its size, shape, and mass. The drag force is larger in magnitude than the weight, making the acceleration upward. The slows down, approaching the new terminal speed. Note that the terminal speed is not the maximum possible speed; it is the speed that the falling object approaches, regardless of initial conditions, when the only forces acting are drag and gravity. qu Works Cited Hsu, K, Soong, R, Chen, K & Lan, T, (2013). 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