Consider a case where a ball or object of mass $m$ is attached to a spring that is hinged at one point. If the ball is given a velocity perpendicular to the spring, it does spiral motion due to elongation of spring. When applying conserving energy, why do we not consider rotational kinetic energy? Doesn't the angular velocity possess rotational kinetic energy?
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$\begingroup$ rotational kinetic energy doesn't exist $\endgroup$– basicsCommented Jun 18 at 8:04
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$\begingroup$ I have edited to make your question clearer. $\endgroup$– Vincent ThackerCommented Jun 18 at 16:33
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2 Answers
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If we assume that the spring is "light" i.e. the mass of the spring is much smaller than the mass of the ball then we can ignore the kinetic energy of the spring.
If the spring is not light then we cannot ignore its kinetic energy, and the motion will be much more complex and difficult to analyse, since the spring changes length during the motion and may bend as well, depending on its stiffness.
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This is an answer to version 2 before the OP was edited to make my answer somewhat invalid.
There is only kinetic energy. We sometimes break it into types to make analysis easier, but ultimately it's just a calculation trick.
In this case the kinetic energy only comes from the mass $m$ moving at speed $v$. Taking your circular motion example, the speed is constant, and the kinetic energy $K$ is just
$$K=\frac12mv^2$$
If we use the relationship between speed and angular speed $v=R\omega$, where $R$ is the radius of the circular path the mass travels on, we have
$$K=\frac12mR^2\omega^2$$
Since we know the moment of inertia of the point mass is $I=mR^2$, we can rewrite the kinetic energy as
$$K=\frac12I\omega^2$$
which is just your "rotational kinetic energy".
So then, in this case, since the mass is undergoing only rotational motion, feel free to treat the rotational kinetic energy as the total kinetic energy and vice versa.
answered Jun 18 at 2:08
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$\begingroup$ This is incorrect. The length of the spring is changing so the mass cannot be modeled as a rigid body around the origin. It has radial velocity as well. The kinetic energy for a point mass is strictly $mv^2/2$. The formula $I\omega^2/2$ only applies to a rigid body, i.e. if the spring were replaced by a massless inextensible rod. See my answer here. $\endgroup$ Commented Jun 18 at 16:29
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$\begingroup$ @VincentThacker The OP has specified circular motion. You are correct in general it does not have to be circular motion though $\endgroup$ Commented Jun 18 at 19:35
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$\begingroup$ @VincentThacker Ah, I see you have edited the OP to make my answer invalid. Seems kind of rude to do that; at least be honest with me that you did that instead of doing it then criticizing my answer to a version I didn't make an answer to $\endgroup$ Commented Jun 18 at 19:36
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$\begingroup$ I don't think that I edited the OP to make your answer invalid. I simply rephrased the question more clearly. I don't think I changed the meaning of the question. The OP was talking about the case of a mass attached to a spring all the while, which isn't circular in general. It took me a long time to make out what OP was trying to ask as the original question was extremely unclear. Feel free to make further edits. $\endgroup$ Commented Jun 18 at 19:49
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$\begingroup$ This discrepancy itself is proof of how unclear the original question was. $\endgroup$ Commented Jun 18 at 19:56