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Homework Help: Physical Pendulum

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data

    A metal rod of length 'L' and mass 'm' is pivoted at one end. A thin disk of mass 'M' and radius 'R' (<L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached :

    Case A: The disc is not free to rotate about its centre of mass, the geometrical centre.
    Case B: The disc is free to rotate about its centre of mass, the geometrical centre.

    The rod-disc system performs simple harmonic motion in vertical plane after being released from the same displaced position. Which of the following statements is/are true ?

    1. Restoring torque in case A is greater than that in case B.
    2.Restoring torque in case A is equal to that in case B.
    3.Angular frequency in case A is more that that of B.
    4.Angular frequency in case A is less that that of B.

    (Visualize the diagram yourself.)
    2. Relevant equations

    3. The attempt at a solution

    Option 2 should be correct because free body diagram of both the cases should be same.

    I don't know how moment of inertia of a body is affected by its motion. I think that angular frequency should also be same (because moment of inertia of both cases of physical pendulums should be same) but answer key says its 2 and 4. How ?

    Please help !

    Thanks in advance..:)
  2. jcsd
  3. May 8, 2015 #2
    The torque is the same, as you have the same force and the same radius to the center of mass, so I am glad you got that part. 4 is a bit trickier, if I have understood the set up correctly, then case B will have a greater frequency, as the disk will spin as well. Basically the disk will spin clockwise and counterclockwise absorbing the angular momentum experienced by the pendulum, causing it to change directions faster and thus have a shorter period meaning a faster frequency.
  4. May 8, 2015 #3
    Thanks. But can you please explain it mathematically. Angular frequency of a physical pendulum is inversely proportional to square root of moment of inertia of system about pivot, other factors being same. So case A and B should have different moment of inertia if their angular frequency is different. How? I am confused... How will that be different?
  5. May 9, 2015 #4


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    The set up is not entirely clear, but I gather the rod and disc are coplanar.
    Yes the moments of inertia will be different. In case B, the disc is not rotating around the pivot in the usual sense of rigid body rotation.
    Think about how the parallel axis theorem works. It adds the moment for turning about the mass centre to the effect of a displacement from there. But in case B, the disc will not rotate about its centre.
    I think we intend the same, but I would express it differently. In case A the disc rotates, in case B it doesn't (or may do so, but in a constant manner unrelated to the pendulum movement).
  6. May 9, 2015 #5


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    In a system of point masses, the sum of forces F=∑(miai) and the sum of torques with respect to an axis is τ=∑(miρi2αi), where ρi is the distance from the axis of rotation of the point i and αi is its angular acceleration.

    In case of a rigid body, the angular acceleration is the same for all points, and you get the torque equation τ=Iα where I is the moment of inertia I = Σ(miρi2).

    In case B, the pendulum is not a rigid body. The rod is, and the disk is, but the two together are not. The disk can rotate freely with respect to the rod.
    You can write the torque equation with respect to the pivot for the rod, and the weight of the disk contributes to that torque.

    There is no torque rotating the disk about its centre. So the angular acceleration of the disk is zero. In the original equation, the points of the disk cancel from the torque equation. In the sum τ=∑(miρi2αi) only the points of the rod appear.
  7. May 9, 2015 #6
    Yeah that is what I meant, the disk wil rotate with respect to the point mass of the pendulum as the rest frame. But from a 3rd frame looking at the swinging pendulum the disk will not rotate at all
  8. May 9, 2015 #7
    Thanks for all the replies!!!

    I think that I am starting to get it.

    I just need to clarify the following two points :

    1. I have read in textbook that moment of inertia only depends upon "mass distribution" in the body, it has nothing to do with motion of the rigid body. If moment of inertia in two cases be different then won't that be contradicting the point in textbook ? (Yes as per questions answer moment of inertia changes.)

    2. Brage mentioned that " case B will have a greater frequency, as the disk will spin as well. Basically the disk will spin clockwise and counter-clockwise absorbing the angular momentum experienced by the pendulum......" Brage can you please explain "absorbing the angular momentum experienced by the pendulum"....?
  9. May 9, 2015 #8
    Hi yes I reaslize that post was quite a bit confusing in retrospect. Firstly, you are right moment of intertia is independent of motion, and you can calculate it for any object independent of time (assuming the object itself doesnt change with time).
    Right so onto the "spinning wheel". The confusion comes from how you look at the wheel, if you were to "sit" on the rod of the pendulum looking down you would see the wheel indeed rotate forwards and backwards in a simple harmonic motion, however, if you were to look at the pendulum such that your body and the pendulum were parallel, you would see the wheel not rotating at all, staying in the same orientation as the pendulum swings. This is because the pendulum is connected to the wheel at its centre, and as such does not provide any torque on the wheel, which would mean the wheel would remain in the same orientation throughout the experiment.
    In case A, however, the wheel cannot rotate freely, and as such the wheel must experience a change in angular momentum in order to rotate forwards and backwards. As the wheel and pendulum will rotate in the same direction when fixed together (i.e. clockwise-clockwise and counterclockwise-counterclockwise) this means the wheel effectively "absorbes" angular momentum from the pendulum (note this is not strictly correct, as the wheel and pendulum are now one system rotating together, but it is a way of thinking about it that can help with undersanding the problem). Thus as the wheel in A absorbes some of the angular momentum of the pendulum the pendulum in case B must be greater, this translates to a greater angular frequency.

    Appollogies again for the poor wording of my previous responce, I hope what I was trying to say makes more sense now.
  10. May 9, 2015 #9
    Alternatively you can consider case B as all the mass of the disk fixed at one point, i.e. the centre of the circle. In case A the wheel is fixed so this is not the case and you must consider the mass as spread out around the disk. Both wheels must have the same angular momentum, as they experience the same torque. The moment of inertia of case A is larger than case B (as can be calculated by comparing moments of inertia of a disk and a point rotated about some fixed length L from their geometrical centres), then the angular angular velocity of case B must be larger such that the angular momentum of case B and case A are the same.
    This is the mathematicl interpretation.
  11. May 9, 2015 #10
    Ahh yes thanks. Now I have only one query to clear :
    Thanks brage. But....
    I still need some clarity regarding this point.
  12. May 9, 2015 #11
    "it has nothing to do with motion of the rigid body" that is true, but in one case the body is rigid, in the other it is not as the wheel is free to rotate. You can therefore not say that the moment of inertia of the two bodies are the same
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