Rotation about the Center of Mass

[ninevolt]

Why is it that when an object moves both transitionally and rotationally the center of mass follows the path as if there was no rotational energy?

My theory is that objects naturally rotate around the center of mass and since the center of mass is a point, rotation doesn't affect its location. If this is true then I have another question.

Why do objects naturally rotate around the center of mass?

 

[K^2]

Simple example. Consider just two masses, m₁ and m₂ joined by a weightless rod of length L.

Imagine some forces F₁ and F₂ are applied to the two masses. In addition, there may be some tension in the rod. It applies force T to m₁ and -T to m₂. (Newton's 3rd Law)

Acceleration of m₁:

$$m_1 a_1 = F_1 + T$$

$$a_1 = \frac{F_1 + T}{m_1}$$

Acceleration of m₂:

$$m_2 a_2 = F_1 - T$$

$$a_2 = \frac{F_2 - T}{m_2}$$

Lets say that m₁ is located at some point r₁ and m₂ at r₂. Then the center of mass, r, is located at:

$$r = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$$

From this, it is easy to convince yourself that acceleration of center of mass, a, is given by the following.

$$a = \frac{m_1 a_1 + m_2 r_2}{m_1 + m_2}$$

Substituting earlier equations for a₁ and a₂ into the above, we get this.

$$a = \frac{F_1 + T + F_2 - T}{m_1 + m_2}$$

Tensions cancel. If I now call the total mass of the system M = m₁ + m₂, the result is exactly what you expect.

$$a = \frac{F_1 + F_2}{M}$$

$$a M = F_1 + F_2$$

Or in other words, the Newton's 2nd Law is valid for center of mass motion.

It is easy enough to extend the above for any collection of point masses. If you know a bit of calculus, you can derive it for continuous rigid bodies.

As far as the object rotating "around center of mass", any kind of motion can be separated into linear and rotational, with center of rotation being an arbitrary point. The above equations show that in absence of an external force, center of mass moves with no acceleration. That is, uniformly, along a straight line. Therefore, it is a natural choice for origin of rotational motion. You can pick any other point, but that point is going to experience a net force from its neighbors, and therefore, its motion is much more difficult to describe.