# Why can mass not be considered concentrated at CM (center of mass) for rotational motion?

Could anyone explain the following expression: Why can mass not be considered concentrated at CM (center of mass) for rotational motion?

The total linear momentum of a system of particle labeled with $i \in {1\dots n}$ can be found in the microscopic view just by summing the linear momentum of the constituents:

$$\vec{P} = \sum \vec{p} = \sum m_i \vec{v}_i$$

Now, writing $M = \sum m_i$ for the total mass, $\vec{X}$ as the position of the center of mass, and $V$ as the velocity of the center of mass, we can express the total linear momentum in the macroscopic view as

$$\vec{P}_{CM} = M \vec{V} = M \frac{\mathrm{d}}{\mathrm{d}t} \vec{X} = M \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{1}{M} \sum m_i \vec{x}_i \right) = \sum m_i \frac{\mathrm{d}}{\mathrm{d}t} \vec{x}_i = \sum m_i v_i = \vec{P}$$

That a system where it is OK to treat the mass as if it were concentrated at the CM.

Next consider angular momentum, and I'm going to proceed a little differently.

I will define $\Delta \vec{x}_i = \vec{x}_i - \vec{X}$ and the displacement of the $i$th particle from the center of mass, and note the consequence that $\Delta \vec{v}_i = \vec{v}_i - \vec{V}$.

Now write the microscopic angular momentum as

$$L = \sum m_i \vec{x}_i \times \vec{v}_i = \sum m_i \left( \Delta \vec{x}_i + \vec{X} \right) \times \left( \Delta \vec{v}_i + \vec{V} \right)$$

Computing all four partial products we get

$$L = \sum m_i \left( \Delta \vec{x}_i \times \Delta \vec{v}_i + \Delta \vec{x}_i \times \vec{V} + \vec{X} \times \Delta \vec{v}_i + \vec{X} \times \vec{V} \right)$$

and from the definition of the center of mass quantities you should be able to show that the two center terms are identically equal to zero [1]. That leaves us with

$$L = \left (\sum m_i \Delta \vec{x}_i \times \Delta \vec{v}_i \right) + \left( M \vec{X} \times \vec{V} \right) = L_{INT} + L_{CM}$$

where $L_{INT}$ is the "internal" angular momentum. The result is that you can not compute the angular momentum of the system from the macroscopic properties of the system without including an additional term that describes the rotation and mass distribution around the center of mass.

[1] As in

$$\sum m_i \Delta \vec{x}_i = \sum m_i \left( \vec{x}_i - \vec{X} \right) = \left( \sum m_i \vec{x}_i \right) - \left( \vec{X} \sum m_i \right) = \left( M \vec{X} \right) - \left( M \vec{X} \right) = 0$$

If it was, then the object would spin around its center of mass at infinite speed with 0 force. Clearly this does not happen in reality.

• Is the expression mean that as the car's wheel; there is no mass at the CM ?? Commented May 21, 2012 at 21:52
• There is negligible mass at any one point, since a point is a 0-dimensional construct. The mass is distributed across the full volume of the wheel. Commented May 21, 2012 at 22:01

This expression means that when considering rotational motion one cannot make the simplification of assuming that the rotating body in consideration is a point mass. In other words: shape and mass distribution matters.

This is due to the fact that spatial distribution of mass of the body around its rotation axis is a very important factor determining the rotational motion. The quantity which describes the distribution of mass relative to the rotation axis is called the tensor of inertia.

In practice, while calculating quantities describing the rotational motion one almost always uses moment of inertia with respect to the current rotation axis. Now, since this depends on the distribution of mass (and not only on the amount of mass), one ends up with very different results for different shapes and mass distributions.

Example: let's calculate rotational kinetic energy of two bodies of the same mass m, but with different moment of inertia. First, take a thin rod of length L rotating around one of its ends with angular velocity ω:

$$E_k = \frac{I \omega^2}{2}= \frac{mL^2}{3}\frac{\omega^2}{2}=\frac{mL^2\omega^2}{6}$$

Then, replace the rod with a point mass positioned at the center of mass of the rod, i.e. point mass m placed L/2 away from the rotation axis:

$$E_k = \frac{I\omega^2}{2}=\frac{mL^2}{4}\frac{\omega^2}{2}=\frac{mL^2\omega^2}{8}$$

Because rotational inertia depends on the distance (radius) of mass from the center of rotation. That's one simple reason.

Consider rotating ball. It's center of mass can be at rest, but it still has rotational kinetic energy. Given the angular velocity, this energy is dependent on mass distribution inside the ball. If all the mass will be concentrated in the center of the ball, the rotational kinetic energy will be zero.

Please note that this very simple argument is totally ignored when elementary particles are considered :).

The total angular momentum of a rigid body about an arbitrary point A can be thought of as the sum of 2 parts: 1) the angular momentum about A of the total mass concentrated at the center of mass (point G), and 2) the angular momentum of the body with respect to G.

If you want to calculate the angular momentum, you can only treat the body as a point mass when the angular momentum about its mass center is small compared to the total angular momentum of the body about A.

For example, if you were to treat a large body which is rotating rapidly about its center of mass as a point mass, and then calculate the angular momentum about its center of mass, you would get 0, even though the angular momentum should obviously be high!

Now if you calculate the angular momentum of a slow-spinning yoyo that is swinging at the end of its string about your finger, you could treat it as a point mass and still observe almost the exact dynamics as if you were to do all of the extra math to include the angular momentum of the yoyo spinning about its bearing.