# Find axis to minimize moment of inertia

1. Nov 14, 2006

### vu10758

Two masses (M1 and M2) are connected by a thin massless rod of length L.

a) Determine the location of an axis perpendicular to the rod) which minimizes the moment of inertia of the system.

The answer is "with M1 at the origin x = M2*L/(M1+M2)

I know that I = Sum of ( M1*r1^2)
However, how do we know that M1 is at origin? I know that M1 + M2 is the total mass of the system. I know L is the length between the two points. But I don't know why the answer is above. I think I am missing something in understanding this but I don't know what.

b) Verify that the moment of inertia about this axis can be obtained by determine I about an axis through M2 and then using the parallel-axis theorem.

I know that I = I_cm + Mh^2 , where h is the distance between the axis, M is the total mass (M1+M2). Is the distance between the axis L? How do I use this to verify the moment of inertia.

2. Nov 14, 2006

### physics girl phd

for part a)
Hint --
How would you calculate the inertia of the system if the system was rotated about some known position -- say a distance "x" down the rod from the mass M1?

3. Nov 14, 2006

### vu10758

Wouldn't I use the formula I = Mx^2 then where x is the distance down from the rod?

4. Nov 14, 2006

### physics girl phd

Not quite -- what do you mean by "M"... are both masses contributing?

5. Nov 14, 2006

### FunkyDwarf

Ok theres two ways you can do this. First is intuitively. The moment of intertia is effectively a measure of mass for rotational motion. So if you have your axis straight through the middle of the rod, you can picture just through everyday experience that its going to require a fair bit of effort or torque in order to spin it. If you have the rod through one of the masses, the torque required to spin that object is virtually nil (t = fd and if d is some infintesimle radius...) And the only major force drain is the other mass at the max length. As im sure you'll see from the answer you have there, if the masses are equal its best to have it in the middle, if ones larger its best to have the axis closer to that one, or through it.

This can also be solved mathematically. You have written one definition of I but theres another, more useful (for me anyway) term.

$$I = \int r^2dm$$
Now this more general form is more for complex objects and so your sumation one is actually better for this point particle problem, but its a good idea to keep the above one in mind. Your moment of inertia is going to be

$$I = r^2M1 + (L-r)^2M2$$

right? From your calculus classes you know that if you want to minimize a function you can differentiate it and equate to zero. Bobs your uncle :)

-G

6. Nov 14, 2006

### vu10758

For part 2,

I know that I = I_cm + mh^2, where h is the distance between the axes. So is h, L-r in this case? How can I get I about an axis through M2.

Is it

I = M2*L^2 + M1*(L-r)^2 ?

7. Nov 15, 2006

### civil_dude

It is my recollection that the min. I is that axis which is the same as the centor of mass. Am I right?