# Centre of mass - is this right?

1. Apr 19, 2005

### WY

Hello

I was given a question of 2 masses m_a and m_b whos centres of L distance apart and are connected by a massless rod (so it kind of looks like a dumb-bell). Each ball through its centre has an axis perpendicular to the rod (axiz a and b respectively) and each has a moment of inertia, I_a and I_b respectively. And it is also fiven that the ratio of I_a to I_b equals three.

Okay I worked out the ratio of the masses of the two balls which was the first part of the question and that came out okay, and then the second part looked way to simple and I can't seem to dig up another answer:
b) Find d_a, the distance from ball A to the system's center of mass (Express in terms of L, length of the rod)
is it just L/2 or is it acutally more complicated than this?

Can someone help me out ? Thank you!!!

2. Apr 19, 2005

### HallsofIvy

Staff Emeritus
Since the two masses are not the same- yes, it's more complicated than that! Imagine trying to support this "dumbbell" by holding at its center of mass- you will need to grip it closer to the heavier weight.

One way to calculate the center of mass is to use the "torques"- twisting force- around that point: Take xm to be the distance from A to the center of mass. Mass A causes a "twisting" equal to its weight times that distance: maxm. Mass B is L- xm from the center of mass and so causes a "twisting" of mb/sub]xm.
Since those masses are on opposite sides of the center of mass, those twists are in opposite directions- and since, by definition, there is no "twist" about the center of mass, they must be equal so that they offset: maxm= mb(L- xm).
Moments of inertia are proportional to mass so "ratio of I_a to I_b equals three" tells you that mb= 3ma That means your equation is maxm= 3ma(L- xm). Notice that the ma terms cancel out. Solve that equation for xm.

3. Apr 19, 2005

### WY

Hey thanks for that!
That makes much more sense now!