# 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
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.

### Answers (2)

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.
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
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.

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
on 12 May 2022

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