# Is it correct to choose Rod's COM as origin of an inertial cordinate system?

I was trying to apply Chasles' theorem to the question attached below. To study the Dynamics I need to apply Chasles' theorem to convert the motion into a combination of rotation about the center of mass and translation of the center of mass.( Problem may be solved without it also)

I have studied in Kleppner and Kolenkow mechanics that when applying Chasles' theorem origin (with respect to which we calculate angular momentum and other origin dependent quantities) must be in an inertial frame. (Attached below.)

I applied Chasles' theorem considering origin to be the center of mass of the rod which is basically the center itself and the answer comes to be correct.

Now, the center of the rod cannot be the origin of an inertial frame because initially it was at rest and later it starts to move after collision with the ball. This has happened in several other problems also related to rigid body dynamics. What is the problem? This is An Introduction to Mechanics, Kleppner and Kolenkow. Clearly written "wrt origin of inertial cordinate system" Am I interpreting this wrong?

The question was Example 35

• I think you should perhaps rewrite the whole question , it's difficult to read as of rn Dec 23, 2020 at 17:40
• Should I attach a page from Mechanics Kleppner instead Dec 23, 2020 at 17:42
• No no, just like the wording of it. It's not really 'what the information is' but rather how you structured it Dec 23, 2020 at 17:42

To be more specific, the center of mass is chose as the origin for motion of an unconstrained body in space. If the body is constrained by some external forces to rotate about some fixed point $$O$$ (fixed in the inertial frame), that point is used for the origin. In both these cases, the forces and torques are those in the inertial frame. [If the fixed point for rotation $$O$$ (not the center of mass) is accelerating, then fictitious forces need to be considered.]
Choosing that origin has the advantage that $$\mathbf r_{com} \times \mathbf p_{com} = 0$$, so the angular momentum of the rod is the value calculated from its COM.