By the parallel axis theorem, a pendulum that rotates around a point $P$ and a distance $l$ from it's center, has kinetic energy $E_{kin}= \frac{\omega^2}{2}(\frac{2mR^2}{5}+ml^2)$. Where R is the radius of the spherical pendulum bop (we assume the string or what not to be massless). Let's say, however, that in the middle of the pendulum's motion, the string is broken. Now the pendulum moves linearly with Kinetic energy $E_{kin} = \frac{mv^2}{2}$. My question is: why is this not equal to the rotational kinetic energy? Where does the small bit of energy $\frac{2mR^2}{5}$ (small if the bop is small) go? Something is not right here since energy should be conserved, but I'm not sure what.
[Physics] Rotational Kinetic Energy of a Pendulum
classical-mechanicsnewtonian-mechanicsoscillators
Best Answer
Consider your equation
$\Rightarrow $ kinetic energy $=\frac 12 (\frac 2 5 mR^2)\omega^2 + \frac 12 (ml^2)\omega^2$
The first term is the rotational kinetic energy of a bob which is rotating about its centre of mass with an angular speed $\omega$, which in this case it is not.
As you have stated the second term $(= \frac 12 m v^2)$ is the kinetic energy of the bob.
The terms which are often used in this context are spin which relates to the rotation about the centre of mass and orbital which relates to the rotation of the centre of mass about a point.
For example if the Earth is rotating about the Sun with an angular speed $\omega$ and spinning about its axis the kinetic energy is $=\frac 12 (\frac 2 5 mr^2_{\rm Earth})\Omega^2 + \frac 12 (mr^2_{\rm orbit})\omega^2$.
If in some way the Sun "disappeared" the Earth would move off at a tangent to its orbit, $\frac 12 (mr^2_{\rm orbit})\omega^2 = \frac 12 m v^2$, and still have the rotational kinetic energy component, $\frac 12 (\frac 2 5 mr^2_{\rm Earth})\Omega^2$.