Two-Disk Pendulum Oscillation: Find Period w/ Mass M & Radius R

[LCSphysicist]

Homework Statement

Torque and angular moment

Relevant Equations

t = rf, l = rp

"A pendulum is made of two disks each of mass M and radius R
separated by a massless rod. One of the disks is pivoted through its
center by a small pin. The disks hang in the same plane and their
centers are a distance I apart. Find the period for small oscillation"

I don't understand why we consider the inertia moment of the topper disc,
he doesn't route, route?

 

[etotheipi]

I don't understand why we consider the inertia moment of the topper disc,
he doesn't route, route?

The whole configuration is rotating about the fixed axis, even the parts of the top disk on the other side of the pin. So you need the moment of inertia of the whole thing.

 

[LCSphysicist]

The whole configuration is rotating about the fixed axis, even the parts of the top disk on the other side of the pin. So you need the moment of inertia of the whole thing.

Which force produce the torque on the top disk?? The tension and the weight pass through the pivot, so i think that their torque is zero.

 

[etotheipi]

Which force produce the torque on the top disk?? The tension and the weight pass through the pivot, so i think that their torque is zero.

If you consider the top disk as its own system, there will be some torque applied by the massless rod in the tangential direction due to a shear component of the internal constraint force. Otherwise it would not go into rotation!

However it's best to treat the whole configuration as one rigid body. Where does the weight of this rigid body act through, so what is its torque?

For the sake of clarification, you could hypothetically split the rigid body into several different bodies and do the analysis on each. There's no physical law saying you have to use the whole rigid body. However, it's not feasible to consider arbitrary sub-bodies in this case, since we can't directly find the internal constraint forces that would result between these sub-bodies and consequently we can't apply $\tau = \frac{dL}{dt}$ to them.

But by considering the whole configuration, and by taking torques about a suitable (well chosen) point, you can do all of the analysis by considering only one torque.

 

[LCSphysicist]

I think my problem it is in not being able to view what is happen, i will try:

The bottom disk is rotating about the pivot [dont route about it axes], and the top disk is rotating about it own axes. Right?

So we sum the inertia moment of the top disk, and the bottom disk about the pivot?

 

[etotheipi]

This is my crude representation of the configuration,

The system is hinged at the cross. What is the moment of inertia of the system about the axis through the hinge? Remember that $I = \sum m_i r_i^2$ so it follows that the MOI of a system about an axis equals the sum of the MOIs of its individual parts about that axis.

You may need to use the parallel axis theorem.

 

[LCSphysicist]

I see, we sum the moment of inertia:

1/2mr^2 + 1/2mr^2 + ml^2. Is this a rigid body??

I think i understand, i just need to clarify a thing:

If the bottom disk route, the moment of inertia would be just 1/2mr^2 + ml^2?

 

[kuruman]

I see, we sum the moment of inertia:
1/2mr^2 + 1/2mr^2 + ml^2. Is this a rigid body??
[\quote]
Yes, this is a rigid body.

If the bottom disk route, the moment of inertia would be just 1/2mr^2 + ml^2?

Yes. This is a physical pendulum.