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I have a sphere of a mass $m$ and radius $r$, attached to a massless string with a length $l$. The pendulum moves with an angular velocity $\omega$.

I know that axes of rotation can be relative. If I chose that one going through a pivot, then I would write the toal kinetic energy as

$$E_k=\frac 12[\frac 25 mr^2+m(r+l)^2]{\omega}^2 .$$

But I saw another way to calculate it: the axis went through the centre of mass of the sphere and I somehow demeed the movement translational. Then the centre of mass had a velocity $v$ and the energy was then

$E_k=\frac 12 \frac 25 mr^2 {\omega}^2+\frac 12 mv^2$,

which is the same since $v=\omega (r+l)$.

But I do not undertstand how is it possible to have such axis of rotation? How can it go through the centre of mass? It does not rotate around it. Or yes from a specific point of view?

And how can I consider the motion to be translational when all point making the sphere do not have the same trajectory?

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  • $\begingroup$ Presumably you mean maximum kinetic energy rather than total KE, since the KE oscillates between zero at the extremes of each swing and some maximum value at the centre point. $\endgroup$
    – gandalf61
    Jul 7, 2023 at 17:44
  • $\begingroup$ When I think about, why is the energy in my first equation constant? The only possible way it can change is by changing the angular velocity, but that remains constant, doesn't it? So how is that possible when the kinetic energy is zero at the maximum point? $\endgroup$
    – Lootr
    Jul 7, 2023 at 18:14
  • $\begingroup$ $\omega$ is the angular velocity of the pendulum bob, which clearly changes over the course of a swing. The frequency of the pendulum's oscillation is constant (assuming there is no friction or drag), but that is a different quantity. $\endgroup$
    – gandalf61
    Jul 8, 2023 at 8:05

2 Answers 2

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It is the same as with a rolling ball, you can add the translational energy to the rotation about its center, or you calculate the rotational energy from the point of contact. And yes your mass rotates consider the lowest point when it goes through the deepest point and this point at the maximum elongation.

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  • $\begingroup$ But translational motion is when all points move with same velocities. And in case of pendulum, that is not truth since all points aren't in the same distance from the pivot, so their trajectories aren't the same either. So why is that the translational motion? This is what confuses me most. $\endgroup$
    – Lootr
    Jul 9, 2023 at 11:44
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You are considering the instantaneous translation of the centre of mass and the instantaneous rotation about the centre of mass as shown below.

enter image description here

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