When do we only consider an object's rotational kinetic energy? My question is in regards to an object's translational and rotational kinetic energy.
Till now, my understanding was that, you must consider both translational and rotational kinetic energy in every situation. However, after solving a couple of problems, it has come to my attention that there are situations where you only have to consider rotational kinetic energy without having any translational kinetic energy.
For example, here is a question that I got stuck on:

Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0-kg Jane starts from rest at a height of 5.00 m in the trees and swings down to the ground using a thin, but very rigid, 30.0-kg vine 8.00 m long. She arrives just in time to snatch the 72.0-kg Tarzan from the jaws of an angry hippopotamus. What is Jane’s (and the vine’s) angular speed?

(the answer to this question is 1.28 rad/s)
In this problem, I first thought that you have to consider both translational and rotational kinetic energy for the vine and Jane, but the solutions show that you only have to take into account the rotational kinetic energy for both objects. We can see that the center of mass of the rod is clearly moving so I don't understand why is that the case?
How do we exactly know when to consider both translational and rotational kinetic energy or just one of them? How does this situation in the problem that I have given differ from that of a sphere that is rolling without slipping (in the latter, we need to consider both translational and rotational kinetic energy)?
Any kind of help would be greatly appreciated!
Edit:
I've added an image to show where I am struggling at. I don't understand why we don't include $\frac{1}{2}m_{J}v_{1}^2$ and $\frac{1}{2}m_{V}v_{1}^{2}$ when solving this particular question, even though we need them when we are dealing with a problem that involves, say, a ball that is rolling without slipping down a hill.

 A: The kinetic energy of an object is invariant to the location where it is measured, as long as the correct mass moment of inertia is used.
Consider a thin long rod of mass $m$ and length $\ell$. This rod is pivoting about one end with rotational velocity $\omega$. Either of the two following ways of calculating KE yields the same result.

*

*Calculate the KE of the rod about one end

*

*The mass moment of inertia about the end is $I_{\rm end} = \frac{m}{3} \ell^2$

*The velocity at the end is $v_{\rm end} = 0$

*The kinetic energy is $$ \require{cancel} {\rm KE}_{\rm end} = \cancel{ \tfrac{1}{2} m v_{\rm end}^2} + \tfrac{1}{2} I_{\rm end} \omega^2 = \tfrac{m}{6} \ell^2 \omega^2 $$



*Calculate the KE of the rod about the center of mass

*

*The mass moment of inertia about the center is $I_{\rm cen} = \tfrac{m}{12} \ell^2$

*The velocity of the center is $v_{\rm cen} = \tfrac{\ell}{2} \omega$

*The kinetic energy is $${\rm KE}_{\rm cen} = \tfrac{1}{2} m v_{\rm cen}^2 + \tfrac{1}{2} I_{\rm cen} \omega^2 = \tfrac{m}{6} \ell^2 \omega^2 $$
In both cases, the result is the same, ${\rm KE}_{\rm cen} = {\rm KE}_{\rm end}$.
Specifically in your question, because the MMOI given is at the end of the vine you are at the first situation above, where the velocity of the end if the vine is not moving and thus translational KE is zero.
This is a point your teacher should make very clear, that KE has a point of summation (reference point) about which MMOI is to be measured and velocity calculated, but this point is completely arbitrary and so it is to your advantage of choosing one that simplifies the problem.
A: I dont get it clearly what you mean by rod is moving and is translational. The thing is the rod(vine here)(i.e. its centre of mass) and jane are rotating around the top point of vine which is fixed. Vine is not having translational motion upper end is fixed and lower is moving.
