translational kinetic energy vs. rotational kinetic energyby mm2424 Tags: energy, kinetic, rotational, translational 

#1
Nov1311, 10:32 PM

P: 44

1. The problem statement, all variables and given/known data
My question relates to a physics problem dealing with the orbit of a moon. In the problem, the orbit of the moon shrinks such that its radius to the center of the planet in question (here, Earth) is smaller. The question gives a great deal of information, but at one point, it asks how much work was done in shifting the orbit. 2. Relevant equations Conservation of energy 3. The attempt at a solution I set this problem up such that the Total Initial Energy = Kinetic Energy at Orbit 1  Potential Energy at Orbit 1. Similarly, I said that Total Final Energy = Kinetic Energy at Orbit 2  Potential Energy at Orbit 2. I also reasoned that Total Final Energy  Total Initial Energy = Work done. My question is how best to conceive of kinetic energy for a problem like this. I'm having trouble distinguishing between translational kinetic energy and rotational kinetic energy...I know that a rolling cylinder has both types of energy...so I tried to reason by analogy and figured that the moon does as well due to its rotation about its own axis (rotational kinetic energy) and orbital movement (which I figured was translational kinetic energy). So, I set the problem up such that the kinetic energy at beginning and end were found using 1/2mv^2. The periods at both positions were given, so I found the velocities using v = ωR (with omega equaling 2∏/period). The answer key I have used the rotational kinetic energy instead of the translational kinetic energy for the kinetic energy components in the expressions above. My answer was close but just far off enough to make me wonder whether it was rounding error or an error in my strategy. My two questions, I suppose, are: 1) Is it equally acceptable to conceive of the kinetic energy of moving planets as rotational kinetic energy or translational kinetic energy? In other words, are the expressions 1/2mv^2 and 1/2Iω^2 equal provided you correctly solve for v using ω in the first place? And, 2) Why do we conceive of the motion of moons in an orbit to be rotational kinetic energy? Shouldn't the term be reserved only for the moon's rotation around its own axis? Thanks!! 



#2
Nov1311, 11:17 PM

HW Helper
P: 3,394

I would just say there are two kinds of rotational energy going on. And ignore the rotation of the moon on its own axis in this question unless there is something to indicate that rotation changes.
It should work out the same way whether you use Ek = ½m⋅v² or Ek = ½m⋅(rω)² 


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