# Centre of mass and Rotational kinetic energy

A massless rod of length $l$ is pivoted at the upper end and two equal point masses of mass $m$ are attached to it, one at the centre of rod and one at its lower end. Then how much horizontal velocity must be provided to the lower end so that the rod just becomes horizontal?(consider only gravitational force is acting).

My question is - How to solve above question by applying mechanical energy conservation law on Centre of mass of these 2 masses rather than applying it for individual masses and equating. If we can't do it by centre of mass approach, then why so?

I asked my teacher about it and he said that you can't find out velocity for COM just like that because it's performing rotational motion.

Why is it so? What's the reason behind it?

I have taken the reference plane which passes through pivoted end.

• -1. Not clear what you are asking. Try googling these concepts, or searching on this website. Commented Nov 2, 2017 at 17:01
• @sammygerbil -i hope it's somewhat clear what am i trying to ask... Commented Nov 2, 2017 at 18:43
• Still pretty unclear what you're asking. It sounds like you're describing a pendulum with two masses on it? If so you can definitely apply mechanical energy conservation to it.
– Chris
Commented Nov 2, 2017 at 21:37
• @Chris -I'm asking how to solve above question by applying mechanical energy conservation law on Centre of mass of these 2 masses rather than applying it for individual masses and equating. If we can't do it by centre of mass approach Then why so? Commented Nov 3, 2017 at 2:47
• Yes, now it is clear to me what you are asking. Commented Nov 3, 2017 at 11:17

You also have to take into account rotation around the center of mass, $T_{\rm rot}=I\omega^2$. Consider an object just rotating around it's center of mass. It should be pretty intuitive that it has kinetic energy, and yet its center of mass is not moving. So, just the translational kinetic energy you get from considering the motion of the center of mass is not the full story- you also need rotational kinetic energy whenever rotation is involved to fully account for mechanical energy conservation.