For loop and if statement in MATLAB

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Samantha Cepeda
Samantha Cepeda on 10 Apr 2018
Edited: Jim Riggs on 14 Apr 2018
Please, I need a quick help with this problem. If you can guide me on how I can use a "for loop" and "if statement " to solve the problem I would really really appreciate it. Even if you just try to let me know how to use 'if and for' in anything similar. Thank you
  8 Comments
Samantha Cepeda
Samantha Cepeda on 13 Apr 2018
got it! thank you. I ended up doing this but I'm not sure how to print the Wr(i) like the current value for the side of my square.
clear, clc
rc = input ('Enter the radius of the cylinder in meters: '); lc = input ('Enter the length of the cylinder in meters: '); wo = input ('Enter the initial width of the hole in meters: '); ws = input ('Enter the step size of the hole in meters: ');
rho = 2700; %kg/m^3 wr = wo:ws:rc;
if wo < rc
for i = 1:length(wr)
mass_cyl = (pi * (rc)^2 * lc)/ rho;
mass_hole = ((wr).^2 * lc)/ rho;
tmass_cyl = mass_cyl - mass_hole;
end
end
fprintf ( '\nThe radius and length are %0.1f m and %0.1f m respectively.\nThe density of the cylinder material is %0.3f kg/m^3.\nThe rectangular hole width is %0.3f m.\nThe total mass remaining is %0.5f kg.\n\n', rc, lc, rho, wr(i), tmass_cyl)

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Answers (2)

Jim Riggs
Jim Riggs on 13 Apr 2018
Edited: Jim Riggs on 13 Apr 2018
First of all, when calculating the masses you have to MULTIPLY the volume by the density.
You should always verify this using dimensional analysis. For example: the calculation pi*rc^2*lc is the volume of the cylinder and has units of (m^3) if you divide m^3 by kg/m^3, that's the same as multiplying m^3 by m^3/kg and the units you end up with are m^6/kg. Now if you multiply m^3 by Kg/m^3 you end up with Kg, which is the correct unit for mass. so,
volume_cyl = (pi * rc^2 * lc); % cylinder volume in m^3
mass_cyl = (pi * rc^2 * lc)* rho; % cylinder mass in Kg
% (or)
mass_cyl = volume_cyl * rho; % Cylinder mass in Kg
Now, this cylinder volume and mass represents the cylinder without the hole, so it does not change, and therefore, this calculation can be placed outside the FOR loop.
The FOR loop is intended to vary hole size, so the loop index 'i' must be used to define the size of the hole, in this case it is the length of one side of the square hole. You are instructed to make the step size 0.1m, so this can be built into the loop. But, rather than use "i" as the loop index, you might just as well use the variable name for the hole width, "Wr". This gives you
for Wr = 0:.1:1 % start at 0, step by 0.1 up to 1
Inside the for loop, you calculate the mass of the hole and subtract it from the mass of the cylinder
mass_cyl = (pi * rc^2 * lc)* rho; % cylinder mass (Kg)
% put a print statement before the loop to show the problem variables that don't change
fprintf('lc=%f, rho=%f, mass_cyl=%f ...etc \n ',lc, rho, mass_cyl, ...etc)
for Wr = W0:Ws:rc % Wr starts at W0, steps by Ws up to rc
mass_hole = Wr^2*lc*rho; % mass of hole (Kg)
tmass_cyl = mass_cyl - mass_hole % mass of the cylinder with the hole in it (Kg)
% put a print statement inside the loop to see each loop calculation
fprintf('Wr=%f, mass_hole=%f, tmass_cyl=%f \n ',Wr, mass_hole, tmass_cyl)
end
  11 Comments
Samantha Cepeda
Samantha Cepeda on 14 Apr 2018
I put it betwen the two last 'end' and it worked well. I'm just unsure if the i that I'm using because I couldn't print all your the values out in your code, it's correctly used. Also I read that using array I need to use the .^ or .* or ./. the control + C did work out! thanks haha I was going crazy! I'm gonna do a while loop now I think I undertand how the for loop works using an if statement now. it's just the operators (symbols) and know how and where to place f printf.

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Samantha Cepeda
Samantha Cepeda on 14 Apr 2018
No, you're good. When I tried to use the fprintf inside the for loop it doesn't print all the values but the last one. I ended up using the i symbol that I'm still unsure if I'm using it right and I got the same last mass as your code gives me but I see that the wr varies sooo much and if I use numbers from rc=3,lc=5,wo=1 and ws=0.5 I should get wr= 1, 1.5,2,2.5,3 nothing else but I got 54000.00. DO you have an idea what's going on there? here's my code:
clear, clc
rc = input ('Enter the cylinder radius: ')
lc = input ('Enter the cylinder length: ')
wo = input ('Enter the initial width of the hole: ')
ws = input ('Enter the increment of the hole size: ')
rho = 2700
wr = wo:ws:rc
mass_c = pi * (rc)^2*lc*rho
if wo < rc
for i = 1:length(wr)
mass_h(i) = (wr(i)).^2 *lc*rho
tmass = mass_c - mass_h
fprintf(' wr=%f, masshole=%f, totalmass=%f \n', wr(i),mass_h,tmass )
end
end
  2 Comments
Jim Riggs
Jim Riggs on 14 Apr 2018
One other observation: The statement (wr(i)).^2 is a matrix statement that indicates each element in the matrix is to be squared (that's what the dot "." is for). However, using the subscript (i) means it is only referencing one element (the ith element) so this is a scalar. This means that you don't need the "." and can simply write wr(i)^2 as this is performing a scalar operation.

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