MATLAB Answers

Why the energy value is constant?

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MG NN
MG NN on 23 Jun 2021
Commented: Walter Roberson on 23 Jun 2021
clear
clc
I=0.0308;
p=7850;
X=1:1:10 ;
y = zeros(size(X));
d = zeros(size(X));
g = zeros(size(X));
for i = 1:length(X)
x = X(i) ;
fprintf("\nWhen the ratio , x = %.1f", x)
radius=(((2*I*x)/(p*3.142))^0.2);
fprintf("\nThe radius is %.3fm", radius)
thickness=radius/x;
mass=(p*(3.142*(radius*radius)*thickness));
fprintf("\nThe mass is %.3fkg", mass)
energy=((mass)*((radius*radius)*(W*W)))/2;
fprintf("\nThe energy is %.3fW\n", energy)
y(i) = radius;
d(i) = mass;
g(i) = energy;
end
figure
plot(y,g,'-*'), xlabel('Radius'), ylabel('Energy'), title('Energy vs Radius')
figure
plot(d,g, '-*'), xlabel('Mass'), ylabel('Energy'), title('Energy vs Mass')

Answers (2)

Walter Roberson
Walter Roberson on 23 Jun 2021
x cancels out.
radius involves x^(1/5)
Your energy calculation has radius*radius*thickness*radius*radius.
thickness = radius/x
So your energy calculation has radius*radius*radius/x*radius*radius . Which is radius^5/x . And since radius involves x^(1/5) that means radius^5 involves x^(1/5)^5 which is x . Which is then divided by x, giving something independent of x.
%I = sym(600);
%W = sym(3000);
%p = sym(7850);
X = (1:1:10).' ;
syms I W p x positive
Pi = sym(pi)
Pi = 
π
radius = expand(((2*I*x)/(p*Pi))^0.2)
radius = 
thickness = expand(radius/x)
thickness = 
mass = expand(p*(Pi*(radius*radius)*thickness))
mass = 
energy = expand((p*(Pi*(radius*radius)*thickness))*(radius*radius)*(W*W))/2
energy = 
Y = radius;
D = mass;
G = energy;
y = subs(Y, x, X)
y = 
d = subs(D, x, X)
d = 
g = subs(G, x, X)
g = 
  5 Comments
Walter Roberson
Walter Roberson on 23 Jun 2021
I showed the symbolic calculations.
Your mass has radius*radius*thickness. Your energy has mass * radius*radius . Your thickness is radius/x . Multiply all those together and you have radius^5/x . Your radius has x^(1/5) and when you take that ^5 because of all the *radius then you get x^(1/5)^5 which is x^1 . Which is then being divided by x because of the thickness term. The end result is independent of x.
There are two possibilities here:
  1. That you made a mistake in the equations; or
  2. That the energy really is independent of x.
Verify that radius is proportional to x^(1/5) . If it is, then count how many times radius is multiplied together in creating energy. Then take into account any ratio between radius and x in creating thickness, that is also being multiplied to create energy. For the purpose of this calculation, you can ignore the actual value of all of the variables: you just need to look at powers of x being multiplied or divided together.

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VBBV
VBBV on 23 Jun 2021
@Walter Roberson explanation seems correct, the values for x in the numerator and denominator cancel out each other

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