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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.

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  • $\begingroup$ -1. Not clear what you are asking. Try googling these concepts, or searching on this website. $\endgroup$ – sammy gerbil Nov 2 '17 at 17:01
  • $\begingroup$ @sammygerbil -i hope it's somewhat clear what am i trying to ask... $\endgroup$ – anamika Singh Nov 2 '17 at 18:43
  • $\begingroup$ 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. $\endgroup$ – Chris Nov 2 '17 at 21:37
  • $\begingroup$ @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? $\endgroup$ – anamika Singh Nov 3 '17 at 2:47
  • $\begingroup$ Yes, now it is clear to me what you are asking. $\endgroup$ – sammy gerbil Nov 3 '17 at 11:17
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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.

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  • $\begingroup$ Thanks! So,we'll have to apply the law of conservation for the individual masses only when rotational motion is present because the COM will not possess all of the energy possessed by whole system...? Am i interpreting it correctly? $\endgroup$ – anamika Singh Nov 3 '17 at 15:42
  • $\begingroup$ Or you can break it down into movement of the center of mass plus rotation around the center of mass. Both methods are equivalent. Though treating them as separate masses is probably a bit easier in this case. NB that the individual masses don't separately have conserved mechanical energy, though- it's the sum of the energies that is conserved. $\endgroup$ – Chris Nov 3 '17 at 20:00
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The CM can be used to calculate the change in the gravitational potential energy of the system in this example, but it does not always give the correct result. It works here because the gravitational field is assumed to be uniform. If the gravitational field were not uniform, using the CM would not give the correct result for the change in GPE. You would have to consider the change for each of the constituent masses separately.

The CM cannot be used to calculate the change in kinetic energy in this example, but it does work in some situations. For example, when the motion is in 1D even though the object is 3D. You can use the motion of the CM to calculate changes in kinetic energy of an object when there is no rotation of the object.

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