Simple Work-Energy Problem is stumping me

[Saladsamurai]

Homework Statement

So I am going over some old problem as review. I have the solution to this problem and something is bothering me about it.

Here is the problem and solution:

what bothers me is this: When they are finding the expression for the kinetic energy of the disk, they use I about its center 'A' and not about the hinge-point of the system 'O'.

The statement says that the "bar is welded at A"... and hence it cannot rotate about A. Angular velocity \omega that they are using is of the bar not the disk about its center.

Shouldn't the expression for the KE of disk use I about A ?EDIT:

I get this...the disk does not rotate, hence no KE (angular) ... but in the first version i think angular KE should be WRT point A
Here is the solution to the version where the disk is "pinned at A with a SMOOTH pin"...

 

[KLoux]

Take a look at their formulation of the disk's kinetic energy. It has two terms, one describing the rotation of the disk about it's center, and one describing the rotation of the disk about O. If you like, you can use the parallel axis theorem to compute the inertia of the disk about O. You should get the same answer.

The difference is that when you consider the rotation of the disk about it's own center, you need to account for the translation of the disk separately. Using the inertia of the object about it's instantaneous center of rotation has this translation "built in" to the inertia.

-Kerry

 

[Saladsamurai]

Take a look at their formulation of the disk's kinetic energy. It has two terms

Yes I understand this...

one describing the rotation of the disk about it's center, and one describing the rotation of the disk about O.

...I do not understand this. It is not rotating about its center, so why does it make sense to talk about its KE like this?

Thanks,
Casey

 

[KLoux]

The disk is rotating around it's center. If you painted a horizontal mark on the right side of the disk, and let the whole system swing down 90 degrees, where will the mark be? Will the disk have rotated? I think what you're getting at is that the disk did not rotate relative to the bar. This is correct. When you consider the system, you need to consider the rotation of the disk relative to some fixed set of axis. The disk will rotate relative to a set of axis fixed to the ground.

If you're still having trouble accepting this, I recommend working through the problem using the inertia of the disk about point O, just so you can see that the math is the same.

-Kerry