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Why is it not necessary to take into account rotational kinetic energy when using the Law of Conservation of Mechanical Energy to solve vertical circular motion problems? After all, the particle is rotating about the centre of the circle and does have rotational KE, doesn't it? All the examples I have seen just use KE= 1/2 $mv^2$, e.g. here: http://www.physicsforums.com/showpost.php?p=2312566&postcount=4 |
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When dealing with point particles, rather than bulk (extended) matter, there is no need for the concept of angular kinetic energy, (regular) $\frac{1}{2}mv^2$ kinetic energy (in addition to potential energy) is the relevant conserved quantity. More accurately, when we deal with bulk matter in classical kinematics problems, we ignore the internal forces between the particles that constitute the matter, and consider their orientation fixed. This lets us separate the energy deriving from the motion of the center of mass from the rotation of that mass about it's center. In this case you have two terms for the energy because you have two parameters for it's "speed", velocity of the center of mass, and rate of angular rotation. |
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For a point particle, Translational KE is rotational KE: $$\frac12I\omega^2=\frac12mr^2\omega=\frac12mv^2$$ The formula for rotational KE ($\frac12I\omega^2$) is derived by adding up the KE of each particle in a rigid body in pure rotation. When body has both rotation and translation, we can derive that: $$KE=KE_R+KE_T=\frac12I_{com}\omega_{com}^2+\frac12mv_{com}^2$$ In this case, the point particle has no $\omega$ about its center of mass, so no problem. Though we can still apply the pure rotation formula. Just that we can't say it has both rotational and translational mortion. |
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