Equivalence of Center of Inertia and Center of mass

Question

Why does center of gravity and center of mass correspond to the same point on the body only when gravitational field applied can be assumed to be constant?

Reference: IE Irodov, Fundamental laws of mechanics

Answer 1

The phrase "Center of Inertia" is here synonymous with "center of mass".

And the center of mass is only identical to the centre of gravity, if the gravitational influence is constant (uniform) throughout the object.

Think of a very tall skyscraper. Gravity decreases with altitude, so actually the gravitational pull is stronger in the lower half than in the top half of the building. While the centre of mass is sort of the midpoint - the point which all mass "averages down to" - then the centre of gravity will be skewed downwards and will be found in the lower half and thus does not coincide with the centre of mass.

But if the skyscrabe fell over and was placed horizontally, then the altitude differences would be neglegible and gravity would be basically constant throughout the entire structure, and so the centre of gravity would now coincide with the centre of mass.

Answer 2

The Center of gravity is a point at which the sum of net torques acting on a body gives a result of zero, that is $$\sum\vec\tau=0\tag1$$

If our body is continuous, for example like a baseball bat, then we need to consider integrals right from the start because we can't apply $\vec\tau=\vec r\times\vec F$ to the bat since there is not a single point at which gravity is acting upon. Thus we must divide our bat into small pieces, consider the torque on each piece, and then add all these torques to get the net torque. This means we have to take the integral of torques. Thus the condition becomes $$\int d\vec\tau=0\tag2$$ Where $d\vec\tau$ is the torque on each small piece of mass $dm$.

On the other hand, if we have noncontinuous bodies, like for example balls scattered across the pool table then we can't perform integration on the whole system directly. Instead, first, we'll calculate the net torque on each part of the system by the method of integration described above, this means that in the case of the pool balls we have to calculate the net torque on each ball first, then simply sum all these torques to get a final resultant torque

Cleared with this, first, consider a ball in a uniform gravitational field. The torque on some small piece of it having mass $dm$ is $$d\vec\tau=\vec r\times dm\vec g\tag3$$ Thus net torque on the whole body is simply the integral of this expression, hence $$\vec\tau=\int d\vec\tau=\int\vec r\times dm\vec g\tag4$$ Since we are assuming $\vec g$ to be a constant we get $$\vec \tau=\bigg(\int\vec r dm\bigg)\times\vec g\tag5$$ This integral is nothing but a definition of COM for a continuous body, thus our equation reduces to $$\tau=M\vec R\times\vec g\tag6$$ Where $\vec R$ is the position vector of COM from the point at which torque is being calculated and $M$ is the net mass of the body.

Now if our system had many other objects in it, then we'd have to add all these torques, that is

$$\vec T=\sum\vec\tau=\sum (M\vec R\times\vec g)\tag7$$ Again assuming $\vec g$ is constant we get $$\vec T=\sum(M\vec R)\times\vec g\tag8$$ This time the summation corresponds to the center of mass of all the balls in our system.

Notice that if $\vec g$ can't be assumed to be a constant then we couldn't pull it out of the integral in equation $4$ and our final equation for net torques would then be $$\vec T=\sum\bigg(\int\vec r\times dm\vec g\bigg)\tag9$$

In summary, if you have only one object in a uniform gravitational field then by virtue of equation $6$ the center of mass and center of gravity for the body becomes the same. If on another hand we have a system with many different continuous bodies in a uniform gravitational field then we have to use equation $8$. If the field is not uniform, then all bets are off and we need equation $9$.