Why do we only consider the monkey's translational energy and the pole's rotational energy? [closed]

Question

The problem states:

A 50-kg monkey sits at the top of a 5-m high 200-kg pole. A worker saws through the bottom of the pole so that it falls on the side. With what speed does the monkey hit the floor if she holds on to the pole?

Lets write:

$m = \rm 50 \ kg$

$M = \rm 200 \ kg$

$h = \rm 5 \ m$

So at first they have a combined potential energy of : $$W_p = mgh + Mg(h/2)$$ This part is fine with me.

But then, when you look at the moment when the monkey hits the floor, we wrote that this potential energy transforms to rotational energy of the pole $+$ translational energy of the monkey. I don't understand why this is so, why don't we consider the rotational and translational energy of both of them?

So we wrote:

$$W_k = \frac{1}{2} \cdot \frac{1}{3} \cdot M \cdot h^2 \cdot \omega^2 + \frac{1}{2} \cdot m \cdot v^2$$

Any help explaining this would be very welcome.

Answer 1

The monkey is treated as a point mass. To include its rotational energy you would need to know its moment of inertia. The monkey could have been included as part of the rotational inertia of the pole (m$h^2$) instead of as a separate entity.

Answer 2

In case of the pole, I think they have assumed, that when the pole falls, its centre of mass stays at the same place. Something like a see-saw. When the front of the pole falls, the back of it rises. So there is no translation, just pure rotation. That is why they have assumed only the rotational kinetic energy. As the monkey falls, its centre of mass does change, and so you consider the translational part.

However, I'd personally consider the translation and rotational energies of both the pole and the monkey, as both of their centre of masses would shift and rotate, when the pole falls down. A diagram along would the question would surely help remove such ambiguity. Hope this helps !