Solving Toroid: Monte-Carlo Method Homework

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The discussion focuses on using the Monte Carlo method to calculate the mass, center of mass, and moment of inertia of a toroid. The user outlines their approach, which involves sampling points within the toroid and determining valid coordinates based on the toroidal equation. They successfully calculate the mass and center of mass, using the average of the valid coordinates. The user also discusses their method for calculating the moment of inertia, indicating that it depends on the number of partitions of the volume. Ultimately, they confirm that they have resolved their problem.
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Homework Statement



Hi everyone, I'm supposed to find the mass, centre of mass and moment of inertia of a toroid. The basic idea I have is:

a)Take a sample of N set of points (x,y,z)
b)Assign a value of 1 to each correct set of (x,y,z) that lie in the toroid (by seeing whether it satisfies the equation)
c)Sum up all the number of correct sets
d)Ratio R is defined as the sum/N
e)Volume of shape is R*(volume of region, which is usually a box)

Assuming p = 1 as in the question, I've managed to find the mass and centre of mass.

2im4553.png


The Attempt at a Solution



The centre of mass Ixx is defined as mixx. I basically found the mass by simply finding the volume of the shape * 1 as ρ = constant = 1. For the centre of mass, I took the Ʃ(xcorrect * 1)/(number of correct sets) which simply gives the average x..

But for mixx i have this idea:
1. The number of sets of Ʃ(y^2 + z^2) you can take before you finish summing up the entire volume depends on how many partitions the volume is split up into.
2. Thus the number of sets taken is simply the number of partitions = Volume/number of correct sets

This gives a value for mixx to be around 82, miyy 140, mizz, 210

Code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()

{

	int i=1;
	double x;
	double y;
	double z;
	int s;
	double p;
	double v;
	double cmx1;
	double cmx2;
	double cmy1;
	double cmy2;
	double cmz1;
	double cmz2;
	double mixx1;
	double mixx2;
	
	srand48(time(NULL));
	
	while (i<=100000)
		
	{
		x = 1 + 3 * (drand48());
		
		y = 4 - 7 * (drand48());
		
		z = 1 - 2 * (drand48());
		
		if ( x*x + y*y + z*z - 6*sqrt(x*x + y*y) <= -8)
		{s = 1;}
		else 
		{s = 0;}
		
		
		
		p = p + s;
		
		v = (p/i)*42;
		
		cmx1 = cmx1 + x*s;
		cmy1 = cmy1 + y*s;
		cmz1 = cmz1 + z*s;
		
		cmx2 = cmx1/p;
		cmy2 = cmy1/p;
		cmz2 = cmz1/p;
		
		mixx1 = mixx1 + (pow(y,2) + pow(z,2))*s;
		mixx2 = (mixx1/p)*v;
		
		i++;
		printf("%d \t %.0f \t %f \t %f \t %f \t %f \t %f \t %f\n", s, p, v, cmx2, cmy2, cmz2, mixx1, mixx2);
		
	}
	
}
 
Last edited:
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bumpp?
 
Nevermind, problem solved!
 

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