Mass of a prism using a 'While loop'

calculate the mass of a rectangular prism with a center hole as the hole diameter increases. the script must use a “while loop”. Allow the user to enter the following rectangular prism parameters (W=1m and L=2 m) starting hole diameter (r=0.05:0.1:<W/2) hole diameter increment size (step is 0.1) density of the prism material (2700 rho kg/m3) For each valid hole diameter, output the following values W, L, d (current value), rho, m (mass) how can do I print the d current value using vectors ?

[Geoff Hayes]

Samantha - you can use fprintf to write out values to the console. Though I suspect that you are asking for more than that (?).

[Samantha Cepeda]

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;

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)

I did this which is kinda similar but I just do not know how and where to use my 'i' for the if statement because I'm thinking that 'i' is the one that gives me my current values for wr?

[Samantha Cepeda]

I'm really confused with how to use fprintf and all the %g %f %s what if it's a vector w(i) do I still use f? or if it's a matrix ?

[Tom Wenk]

First of all, your calculation is wrong: mass = density * volume

Second, what do you mean by "hole" and "cylinder"?

As far as I understood the question, you should calculate the mass of a rectangular prism with a hole (cylinder) in it. To get the total mass you have to subtract the mass of the hole from the mass of the massive rectangular prism.

Although I have to admit, that the given task is very bad! It mixes diameter and radius of a hole/cylinder and the parameter for the height of the rectangular prism is missing.

If I assume, that it's the diameter, that is meant by "(r=0.05:0.1:<W/2)" and not the radius, and height = width (quadratic) for the prism, than the following code should do the trick:

% mass of rectangular prism with a hole (cylinder) in it
clear, clc
rho = 2700;

W = 1;              % Width of rectangular prism
L = 2;              % Length of rectangular prism
d0 = 0.05;          % Initial diameter of the hole (cylinder)
dStepwidth = 0.1;   % Stepwidth of the diameter of the hole
dMax = W/2;         % Maximum diameter of the hole

for d = d0 : dStepwidth : dMax
 mass_prism = W^2 * L * rho;
 mass_hole = (pi * (d/2).^2 * L) * rho;
 mass_total = mass_prism - mass_hole;
 fprintf ( 'd = %0.3f m \nW = %0.1f m \nL = %0.1f m \nRho = %0.3f kg/m^3 \nm = %0.5f kg\n\n',...
     d, W, L, rho, mass_total);
end

[Samantha Cepeda]

Thanks!! yes, you're right, the task itself is not clear and my mass formula is wrong. The task asks to use a 'while loop' and I don't know where would the while loop go? replacing the 'while loop' for 'for loop' the program just doesn't stop running. (when using 'while' diameter isn't bigger than the prism.) Thank you again!

[Tom Wenk]

Converting a for-loop into a while-loop is simple, you just do, what the for-loop does by hand:

1. create a variable d outside of the loop and initialize it with the starting value d0

2. think about the condition, when the while-loop should stop or continue to run. In your case this is d <= dMax. The loop continues as long as d is lower or equal dMax.

3. Increase variable d by your stepwidth dStepwidth inside the loop every iteration (I assume you forgot that step, when the loop doesn't end in your case).

d = d + dStepwidth;

Here is the while-loop version:

% mass of rectangular prism with a hole (cylinder) in it
clear, clc
rho = 2700;

W = 1;              % Width of rectangular prism
L = 2;              % Length of rectangular prism
d0 = 0.05;          % Initial diameter of the hole (cylinder)
dStepwidth = 0.1;   % Stepwidth of the diameter of the hole
dMax = W/2;         % Maximum diameter of the hole

d = d0;             % changing diameter              
while d <= dMax
 mass_prism = W^2 * L * rho;
 mass_hole = (pi * (d/2).^2 * L) * rho;
 mass_total = mass_prism - mass_hole;
 fprintf ( 'd = %0.3f m \nW = %0.1f m \nL = %0.1f m \nRho = %0.3f kg/m^3 \nm = %0.5f kg\n\n',...
     d, W, L, rho, mass_total);
 d = d + dStepwidth;
end

Answer 1

[Tom Wenk]

I copied the solution I wrote as a comment here as an answer:

First of all, your calculation is wrong: mass = density * volume

As far as I understood the question, you should calculate the mass of a rectangular prism with a hole (cylinder) in it. To get the total mass you have to subtract the mass of the hole from the mass of the massive rectangular prism.

Although I have to admit, that the given task is very bad! It mixes diameter and radius of a hole/cylinder and the parameter for the height of the rectangular prism is missing.

If I assume, that it's the diameter, that is meant by "(r=0.05:0.1:<W/2)" and not the radius, and height = width (quadratic) for the prism, than the following code should do the trick:

% mass of rectangular prism with a hole (cylinder) in it
clear, clc
rho = 2700;

W = 1;              % Width of rectangular prism
L = 2;              % Length of rectangular prism
d0 = 0.05;          % Initial diameter of the hole (cylinder)
dStepwidth = 0.1;   % Stepwidth of the diameter of the hole
dMax = W/2;         % Maximum diameter of the hole

for d = d0 : dStepwidth : dMax
    mass_prism = W^2 * L * rho;
    mass_hole = (pi * (d/2).^2 * L) * rho;
    mass_total = mass_prism - mass_hole;
    printf ( 'd = %0.3f m \nW = %0.1f m \nL = %0.1f m \nRho = %0.3f kg/m^3 \nm = %0.5f kg\n\n',...
        d, W, L, rho, mass_total);
end 

Converting a for-loop into a while-loop is simple, you just do, what the for-loop does by hand:

1. create a variable d outside of the loop and initialize it with the starting value d0

2. think about the condition, when the while-loop should stop or continue to run. In your case this is d <= dMax. The loop continues as long as d is lower or equal dMax.

3. Increase variable d by your stepwidth dStepwidth inside the loop every iteration (I assume you forgot that step, when the loop doesn't end in your case).

d = d + dStepwidth;

Here is the while-loop version:

% mass of rectangular prism with a hole (cylinder) in it
clear, clc
rho = 2700;

W = 1;              % Width of rectangular prism
L = 2;              % Length of rectangular prism
d0 = 0.05;          % Initial diameter of the hole (cylinder)
dStepwidth = 0.1;   % Stepwidth of the diameter of the hole
dMax = W/2;         % Maximum diameter of the hole

d = d0;             % changing diameter              
while d <= dMax
    mass_prism = W^2 * L * rho;
    mass_hole = (pi * (d/2).^2 * L) * rho;
    mass_total = mass_prism - mass_hole;
    fprintf ( 'd = %0.3f m \nW = %0.1f m \nL = %0.1f m \nRho = %0.3f kg/m^3 \nm = %0.5f kg\n\n',...
        d, W, L, rho, mass_total);
    d = d + dStepwidth;
end