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Bead on a Helix, angular velocity
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[QUOTE="vela, post: 6017758, member: 221963"] You have the right idea, but the distinction you're making between [i]rotational[/i] and [i]translational[/i] is not so clearcut. Consider a point mass ##m## rotating about a point at a distance ##r## with angular speed ##\omega##. On the one hand, you could say the speed of the mass is ##v = \omega r##, so its translational kinetic energy is ##\frac 12 m (\omega r)^2##. On the other hand, you could say it has a rotational kinetic energy ##\frac 12 I\omega^2## where ##I = mr^2##. Either way, you get the same result for the kinetic energy. It depends on the way you want to look at the system. Now consider a uniform rod of mass ##m## and length ##L## rotating with angular speed ##\omega## about an axis that passes through the end of the rod and is perpendicular to the rod. The moment of inertia of the rod about this axis is ##I = \frac 13 mL^2##, and its kinetic energy is $$K = \frac 12 I \omega^2 = \frac 16 mL^2\omega^2.$$ We could also express its kinetic energy in terms of the motion of the center of mass and rotation about the center of mass. The center of mass of the rod moves with speed ##v_\text{cm} = \omega L/2##, and the moment of inertia about the center of mass is ##I_\text{cm} = \frac{1}{12}mL^2##. So the kinetic energy of the rod is $$K = \frac 12 I_\text{cm} \omega^2 + \frac 12 m v_\text{cm}^2 = \frac 12\left(\frac {1}{12}mL^2\right)\omega^2 + \frac 12 m \left(\frac{\omega L}{2}\right)^2 = \frac 16 mL^2 \omega^2.$$ Finally, we could calculate the translational kinetic energy of each piece of the rod and integrate: $$dK = \frac 12 v^2\,dm = \frac 12 v^2 \lambda\,dr \quad \Rightarrow \quad K = \frac 12 \int_0^L (\omega r)^2\ \lambda\,dr = \frac 16 mL^2 \omega^2,$$ where ##\lambda = m/L##. [/QUOTE]
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Bead on a Helix, angular velocity
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