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Given the situation shown in the figure below. A thin disk with mass m and radius r is rigidly connected to one end of a massless rod with length l at its centre, with the other end of the rod connected to a fixed point O and able to pivot freely around the fixed point, without friction. The disk is lifted to a certain $\theta_0$, and then released from rest.

I'm wondering how, given this setup, one might calculate the kinetic energy of the disk when the centre of the disk is at its lowest point (or perhaps how to retrieve velocities using laws of energy). Namely, I was taught the kinetic energy of a simple system or body may be described as

$$E_k=E_{k,translational}+E_{k,rotational}=\frac{1}{2}m||\vec v||^2+\frac{1}{2}I_O||\vec\omega||^2$$

where:

$m$ is the mass of the body,

$\vec v$ is the linear velocity vector of the centre of mass,

$I_O$ is the moment of inertia around the axis of rotation,

$\vec \omega$ is the angular velocity vector.

One could perhaps use some simple trigonometry to express the initial potential gravitational energy (with the lowest height of the centre of mass set to a height of 0, for example), and use conservation of mechanical energy to then equate said amount of potential energy to the amount of kinetic energy when at lowest height.

All that seems to make sense when the disk is rotating around its centre, and translating at the same time. However, I'm having a hard time deciding whether or not this system would be describing "linear motion" with a linear velocity, when the centre of mass is at its lowest point. By the parallel axis theorem, we can say that for this disk, $I_O=I_C+m\:d^2=\frac{1}{2}m\:r^2+m\:d^2$, where d is the perpendicular distance between the axis of rotation through the centre of mass parallel to the axis of rotation, and the actual axis of rotation through O.

Thus, from what I understand, the disk is just rotating around an axis through O, but ... wouldn't that mean this is pure rotation, without translation? This seems counter-intuitive to me, but at the same time, if this isn't pure rotation, then I would say it is pure translation, as if the centre of mass were moving along a circular and frictionless rail. Both interpretations give different results, which cannot be correct. For all I know, the disk's kinetic energy could be a combination of both!

As a follow-up: say the disk were able to spin freely around its centre as well, and it were given a certain angular velocity $\omega_0$ before being released. How would the rotational kinetic energy differ, in that case? Both the disk's rotation and revolution would have to be taken into account, but here also, it doesn't seem obvious how exactly.

Setup kinetic energy

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marked as duplicate by sammy gerbil, John Rennie energy Jan 17 at 6:44

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You can view it both ways. You can consider it a pure rotation and use the moment of inertia that you calculated, or you can consider it an instant traslation of the CM plus a rotation around the CM. You need to add the rotation aroung the CM because in your set up the disk is rotating around the CM: imagine that the disk is a clock with the numbers painted on it. As the system rotates the numbers will change in position relative to a vertical line (say the 12 is at the top then it is on the side). You should not include a rotation around the CM if the 12 were always pointing up.

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  • $\begingroup$ That seems to make sense. In the second case, though, I'm wondering whether or not an extra independent variable is introduced, i.e. the $\omega$ of the disk around its CM, of which I don't know if it's the same as the $\omega$ of the rod-disk system. Am I wrong in assuming one situation requires more information than the other as to calculate the kinetic energy? $\endgroup$ – Mew Jan 16 at 10:28
  • $\begingroup$ in your case you know $\omega$, but if the disk could itself rotate independently, you would need extra information. Although in such a case the two angular speeds are independent, because now the rod does not force the disk to rotate. In fact, if it were free to rotate and you start moving the rod, the angular speed around the cM would be zero, the clock would remain with the 12 up as it moves. $\endgroup$ – Wolphram jonny Jan 16 at 11:33
  • $\begingroup$ Alright, so then, in the situation I originally posed (with the rod fixed to the disk) and when utilising linear velocity of the CM and the rotation of the disk around the CM, the $\omega$ around the CM would be the same as the $\omega$ around the axis when I'd use adjusted moment of inertia without linear velocity? $\endgroup$ – Mew Jan 17 at 20:31

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