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

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How would you plot a graph which a ball then rolls down (say a y=x^2 graph)
  11 Comments
Kaushik Lakshminarasimhan
Kaushik Lakshminarasimhan on 26 Aug 2019
Edited: Kaushik Lakshminarasimhan on 26 Aug 2019
This sounds like physics homework, so you should try your luck at https://physics.stackexchange.com/ 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.
Sena Clarke
Sena Clarke on 27 Aug 2019
Edited: Sena Clarke on 27 Aug 2019
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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Answers (2)

David K.
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!
  5 Comments
David K.
David K. on 29 Aug 2019
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.
Jim Riggs
Jim Riggs on 29 Aug 2019
Edited: Jim Riggs on 29 Aug 2019
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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Jim Riggs
Jim Riggs on 29 Aug 2019
Edited: 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;
cnt=cnt+1
% 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
figure;
plot(X,fX,'r'); % truth Y vs. X
hold on;
plot(X,Y,'ob'); % calculated Y vs X
grid on;
legend('fX','Y');
%.... (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.
  1 Comment
Jack McCarthy
Jack McCarthy on 12 May 2022
If you wouldn't mind, could you provide a brief explanation as to how this would work in 3 dimensions? I am unsure as to how I could implement the vector method above. Thanks!

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