MATLAB Answers


How would you plot a graph which a ball then rolls down (say a y=x^2 graph)

Asked by Sena Clarke on 22 Aug 2019
Latest activity Edited by Jim Riggs
on 29 Aug 2019
How would you plot a graph which a ball then rolls down (say a y=x^2 graph)


This sounds like physics homework, so you should try your luck at first before worrying about how to implement the solution in MATLAB
But here's a generous hint. The velocity of the ball would depend on where you release it so start by assuming an initial condition (x,y position where you release the ball). Now if you draw the free-body diagram, you'll see that at any given moment, the ball should accelerate along the tangent to curve y=x^2 at g*cos(theta) where theta = arctan(1/2x) and g is 10m/s^2.
Ha, no its not homework, but I could see why you would think that. No, I was just trying to program brachistochrone curves and doing a write-up on the times with different curves, like straight lines vs y=x^2 and y=sqrt(x) etc... with a ball going down and I don't know how to program a model.

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2 Answers

Answer by David K.
on 27 Aug 2019

I attached a .m file I created that should work pretty well. It uses a numerical solution by calculating the slope at discrete time steps and determining the acceleration. I only considered a point mass so I ignored rotation. It can also take both static and kinetic friction into account (im slightly worried I messed up here but I think it is right). The function takes in a function handle for what you want the slope to be. I made it so that the part of the function you want the ball on needs to be between x = 0 and 10 but that should be easily changed by changing the function itself.
I did not get around to allowing arguments to be dropped from the function or personally checking arguments and throwing errors.
Let me know how it works for you!


OK. I see what you are doing. This might work for the frictionless case. (I regret to say that, for security reasons, I do not have Matlab on this machine, so I cannot download and run your code. I would like to try it out, but my Matlab is on an isolated network)
On a side note, I would sugest that you get in the habit of using vector forms whenever possible, avoiding trig forms, as vector calculations are much more robust, and do not have issues when changing quadrants. For example, I once worked on a missile program where the guidance law was coded using a lot of trig expressions. A navigation error in early flight tests caused the missile to believe that it went below the ground level while in terminal homing on the target, resulting in a negative altitude calculation. This caused an unexpected sign change in the guidance law, and the missule started flying away from the target, missing by a significant distance. This error would have been avoided using vector equations, and it would have hit the target even with the navigation error. You might argue that it's reasonable that missiles should't be required to work under the ground, but in the real world, these things happen, and it's always best to hit the target when possible.
For the problem at hand, I would define a unit tangent vector for the curve:
Ut = [Utx, Uty] = [dx, dy] / mag[dx, dy] (where dy is obtained from the numerical first derivative formula).
Now the unit normal vector Un = [Unx, Uny] = [ -Uty, Utx]
Use these two unit vectors and the dot product to obtain the tangent and normal quantities. No trig function required, and it works for all directions with no singularities.
This will also allow you to apply energy conservation principles because you are retaining the full velocity vector. In a frictionless environment the system energy remains constant, and the total velocity should be a function of the Y position, as potential energy is converted to kinetic energy, and vice-versa. You can use this relationship to test the accuracy of your solution, but only if you calculate the total velocity, and not just the x component.
So, I assure you that if you use vectors in place if trig functions, you are more likely to hit the target :)
Oh, and for the record, it does make a difference using central vs. fwd/bkwd difference for the slope calculation. Try it and see how it compares. Use conservation of energy as the accuracy metric and see how the step size effects it.
Welp, I went about trying to calculate frictionless case for checking the velocity. My velocity function based on conservation of energy was
mghmax-mgh = 1/2 mv^2
v = sqrt(2*g*(hmax-h));
Then I tried to calculate the velocity in the y direction and it went pretty terribly. I couldn't do the acceleration in the same way so I used the difference in y position. And the velocities were super off the expected values at some places. It seems that xvelocity was pretty close if I use the expected x part of the expected velocity but the y part is very off.
I guess to make this a decent answer I will need to change it at some point. Ug, I really thought modeling each time step as a slope would work.
btw, the only reason I would need to be in the habit of vector forms is for making better answers here since I do not do this stuff for work xD.
Yes, it's very important to have a way to check yourself. It sounds like you understand the problem and you are on the right track.
Another way to test your ramp assumption/approximation is to actually use a linear function for the curve, and see if it works there. If not, there is some implementation issue.

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Answer by Jim Riggs
on 29 Aug 2019
Edited by Jim Riggs
on 29 Aug 2019

Here is a model for the kinematics.
clear data % I use data to save values in the time loop
func = @(x) x.^2; % this is the user function
% set model parameters and initial conditions
dx = 0.001; % step used to compute numerical derivatives
dt = .01; % integration time step
tstart = 0; % starting value of time
tstop = 12; % time to stop
x = 0.1; % starting value of X
y = func(x) % starting value of Y
speed = 40; % initial speed
grav = 9.806; % acceleration due to gravity
G = [0, -grav]; % gravity vector
nstep = (tstop-tstart)/dt; % numer of calculation steps
cnt = 1; % itteration counter
% initial energy state (per unit mass)
Ep = gravity*y; % potential energy
Ek = 0.5*speed^2; % kinetic energy
Etot = Ep + Ek; % total system energy
% initialize saved data table
% (you can modify this to save the parameters that you want)
data = zeros(nstep,8);
data(1,1) = tstart;
data(1,2) = x;
data(1,3) = func(x); % truth value of y
data(1,4) = y; % numerical approximation of y
data(1,5) = speed;
data(1,6) = Ep;
data(1,7) = Ek;
data(1,8) = Etot;
% calculation time loop
cnt = 1;
while cnt <= nstep
time = cnt*dt + tstart;
dy = (func(x+dx/2) - func(x-dx/2)) / dx; % first derivative
deltax = dx; % step change in X value
deltay = dy*dx; % corresponding change in Y value
mag = sqrt(deltax^2+deltay^2); % magnitude of step change
% compute the unit tangent vector
Tx = deltax/mag;
Ty = deltay/mag
T = [Tx, Ty]; % unit tangent vector
% compute accelerations
At = dot(G, T); % acceration in the tangential direction
% update states (numerical integration)
speed = speed + At*dt;
delta = speed*dt; % dstance traveled along curve
x = x + delta*Tx; % updated X position
y = y + delta*Ty; % updated Y position
% update energy states
Ep = gravity*y;
Ek = 0.5*speed^2;
Etot = Ep + Ek;
% save data
data(cnt,1) = time;
data(cnt,2) = x; % X position
data(cnt,3) = func(x); % truth Y position
data(cnt,4) = y; % calculated Y position
data(cnt,5) = speed;
data(cnt,6) = Ep; % potential energy
data(cnt,7) = Ek; % kinetic energy
data(cnt,8) = Etot; % total energy
end % end of time loop
% extract data vectors from data table
time = data(:,1);
X = data(:,2);
fX = data(:,3);
Y = data(:,4);
speed = data(:,5);
Ep = data(:,6);
Ek = data(:,7);
Etot = data(:,8);
% Plot data
plot(X,fX,'r'); % truth Y vs. X
hold on;
plot(X,Y,'ob'); % calculated Y vs X
grid on;
%.... (add additional plots as desired)
Note that for the function Y = X^2, with X starting near zero there must be a nonzero initial velocity or it won't do anything.
Also note the use of unit vectors and dot product (no trig functions).
Speed is a signed quantity. Positive speed causes x to increase. As the point rises along the curve, the speed drops to zero, and then goes negative as it falls backward.
Numerical integration errors will result in errors in the conservation of energy. Etot will not be constant, but will drift due to integration errors. Try different values for dt and you will see different ammounts of error in the energy conservation. (smaller dt should produce smaller errros). The plot of Y vs X will also become closer to the plot of fX vs X as the time step becomes smaller.


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