Instantaneous axis of rotation

[Yoonique]

Homework Statement

Two hoops are fastened together as shown below. Smaller hoop mass m, and larger hoop mass 3m. The system is now placed on the table and the system is released from rest in the position shown below. There is sufficient friction between the large hoop and the table so that it rolls without slipping. How fast relative to the table is the centre of the large hoop moving when the centres of the two hoops like in a vertical line?

Homework Equations

The Attempt at a Solution

There is two methods to solving this question. The first method is considering the instantaneous axis of rotation the moment the centres of the two hoops lies in a vertical line which is the point on contact between the table and the hoop system.
So Δmgh = 0.5Iω², where I is the moment of inertia of the hoop system with respect to the instantaneous axis of rotation.

There is a second method where I can consider Δmgh = 0.5I_cmω² + 0.5M_cmV_cm², where I_cm is the moment of inertia of the hoop system with respect to the centre of mass.

Here is the part where I do not really understand. The first method uses v=ωr to find the total K.E of the hoop system. But I cannot use v=wr to find the total K.E of the hoop system for the second method as the answer said the centre of mass is rotating with respect to some other points. If I want to use the second method, I need to consider the rotational energy of the moving centre of mass. Can someone explain this to me? I don't really get it.

 

[haruspex]

If v is the velocity of the centre of the larger hoop, and r is the radius of the larger hoop, and omega is the rotation rate then ##v=\omega r## in all approaches. But I wouldn't use the mass centre of the hoop pair at all. Either use the instantaneous centre of rotation or the centre of the larger hoop. Anything else is more complex.

 

[Yoonique]

If v is the velocity of the centre of the larger hoop, and r is the radius of the larger hoop, and omega is the rotation rate then ##v=\omega r## in all approaches. But I wouldn't use the mass centre of the hoop pair at all. Either use the instantaneous centre of rotation or the centre of the larger hoop. Anything else is more complex.

How do I use the centre of large hoop as a frame of rotation? Δmgh = rotational kinetic energy with respect to the centre of large hoop + translation kinetic energy of the centre of mass with respect to the ground?

 

[haruspex]

How do I use the centre of large hoop as a frame of rotation? Δmgh = rotational kinetic energy with respect to the centre of large hoop + translation kinetic energy of the centre of mass with respect to the ground?

As I said, I would avoid having to determine the mass centre. Express the motion of each hoop, separately, as a rotation about the centre of the large hoop, plus a horizontal linear velocity.