# Is newtons calculus approximate?

1. Nov 21, 2013

### nil1996

Hello PF

i was studying rotational motion
it is given in my textbook that the moment of inertia of a rod of mass M and length L about an axis passing through the center of rod and perpendicular to the rod is ML2/12

This formula was prepared using calculus.
So i experimented.....
a assumed a rod of length 101 atoms of copper.its atomic mass is 63.546. and Van der waals radius of 140pm.

so i made a computer program to calculate the moment of inertia of the rod by both methods
1] calculating the moment of inertia by each atom. (first method)
2] using that formula by newton. (second method)
(You can skip the code)
The code in c++ is below:
#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
int i=0;
int distance=280;
int diameter=280;
float mass=63.546;
float moment1=0.0;
float moment2=0.0;
int length=28000;
for(int i=0;i<50;i++)
{
moment1=mass*distance*distance; //calculates the moment of single atom
distance=distance+diameter; //increases the distance
moment2=moment2+moment1; //passes its value
}
cout<<"moment by nils method comes to be :"<<moment2*2<<endl;
cout<<"moment by newtons method comes to be :"<<101*mass*length*length/12<<endl;
getch();
}

the results come
1]4.27705 (by first method)
2]4.19319 (second method)

the result of the above code is in moment.zip ( in the attachment)

also check if i have gone wrong anywhere:tongue2:

#### Attached Files:

• ###### moment of inertia.zip
File size:
4.4 KB
Views:
60
Last edited: Nov 21, 2013
2. Nov 21, 2013

### MikeGomez

Try making the initial distance 140 instead of 280, since that will be the distance from the center of the rod to the center of the first atom.

Also, you loop 50 times. Doesn't that give you 100 atoms if there are 1 each on oppossing side from the center? If that is so, then multiply by 100 instead of 101 for the Newton equation.

3. Nov 21, 2013

### nil1996

Got it right :)

i added 280 to the length of the rod as the radius of the end atoms was not considered while calculating the moment of inertia according to newtons formula. Yet the last 2 digits are different,don't know why?

4. Nov 21, 2013

### phinds

Cumulative rounding error, perhaps?

5. Nov 21, 2013

### nil1996

yes it must rounding error

thanks

6. Nov 21, 2013

### Staff: Mentor

Try declaring the the two variables as double instead of float, see if you get an interestingly different result.

7. Nov 21, 2013

### Staff: Mentor

The integration assumes that the mass distribution is continuous, and you are assuming that the mass distribution is discrete. This will account for the slight difference in results.

Last edited: Nov 21, 2013
8. Nov 21, 2013

### nil1996

So integration does a small assumption,which isn't true in reality isn't it?

9. Nov 21, 2013

### Staff: Mentor

In this case, yes. In general, it depends on what you are integrating.

10. Nov 21, 2013

### voko

Frankly, I fail to see the point of the entire exercise. First of all, this has nothing to do with Newton. The concept "moment of inertia" was introduced by Euler, about half a century after Newton published his Principles. Second, obviously an integral and a finite sum approximating that integral will be different somewhat depending on the quality of the approximation. This has nothing to do with Newton or Euler, either, because none probably even bothered thinking on convergence and stability of numerical schemes, or even the soundness of integral's definition as a limit of integral sums to begin with, which was first done by Riemann another century later.

11. Nov 21, 2013

### Staff: Mentor

Well, if you had 1023 atoms, it wouldn't matter. And, what happens in the discrete calculation when you have 1023, and one atom is sticking out a little more on one end than on the other, or, if you don't know whether you have 1023 atoms or 1023+1 atoms? Or, how would you handle the discrete calculation if the atoms were vibrating. In practice, if you had 1023, you couldn't do the discrete calculation anyway. There is some assumption made in every calculation you do.