# Mass of a prism using a 'While loop'

7 views (last 30 days)
Samantha Cepeda on 13 Apr 2018
Edited: Tom Wenk on 16 Apr 2018
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 ? Tom Wenk on 16 Apr 2018
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

Tom Wenk on 16 Apr 2018
Edited: Tom Wenk on 16 Apr 2018

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
```