# Why the energy value is constant?

1 view (last 30 days)
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)
fprintf("\nThe mass is %.3fkg", mass)
fprintf("\nThe energy is %.3fW\n", energy)
d(i) = mass;
g(i) = energy;
end
figure
figure
plot(d,g, '-*'), xlabel('Mass'), ylabel('Energy'), title('Energy vs Mass')

Walter Roberson on 23 Jun 2021
x cancels out.
%I = sym(600);
%W = sym(3000);
%p = sym(7850);
X = (1:1:10).' ;
syms I W p x positive
Pi = sym(pi)
Pi =
π
thickness =
mass =
energy =
D = mass;
G = energy;
y = subs(Y, x, X)
y =
d = subs(D, x, X)
d =
g = subs(G, x, X)
g =
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.

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