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

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In summary, the conversation discusses the calculation of the period for small oscillation of a pendulum made of two disks of equal mass and radius, connected by a massless rod and pivoted through one disk. The moment of inertia of the entire system is needed, including the moment of inertia of the top disk, which is rotating about its own axis. The concept of treating the entire configuration as one rigid body is also mentioned to simplify the analysis. The conversation also clarifies the moment of inertia when the bottom disk rotates.
  • #1
LCSphysicist
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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?

1588540343231.png
 
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  • #2
LCSphysicist said:
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.
 
  • #3
etotheipi said:
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.
 
  • #4
LCSphysicist said:
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.
 
  • #5
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?
 
  • #6
This is my crude representation of the configuration,

1588548245076.png


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.
 
  • #7
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?
 
  • #8
LCSphysicist said:
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.
LCSphysicist said:
If the bottom disk route, the moment of inertia would be just 1/2mr^2 + ml^2?
Yes. This is a physical pendulum.
 
Last edited:

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

1. What is a two-disk pendulum oscillation?

A two-disk pendulum oscillation is a type of simple harmonic motion where two disks of equal mass and radius are connected by a rigid rod and allowed to swing back and forth under the influence of gravity.

2. How is the period of a two-disk pendulum oscillation calculated?

The period of a two-disk pendulum oscillation can be calculated using the formula T = 2π√(I/mgd), where T is the period, I is the moment of inertia of the system, m is the mass of the disks, g is the acceleration due to gravity, and d is the distance between the disks.

3. What factors affect the period of a two-disk pendulum oscillation?

The period of a two-disk pendulum oscillation is affected by the mass and radius of the disks, the length of the connecting rod, and the acceleration due to gravity.

4. How does changing the mass and radius of the disks affect the period of a two-disk pendulum oscillation?

Increasing the mass and radius of the disks will result in a longer period for the oscillation, as the system will have a larger moment of inertia and will take longer to complete a full swing.

5. Can the period of a two-disk pendulum oscillation be changed by altering the length of the connecting rod?

Yes, the period of a two-disk pendulum oscillation can be changed by altering the length of the connecting rod. A longer rod will result in a longer period, while a shorter rod will result in a shorter period.

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