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I Pendulum with a sphere vs with a particle in its end

  1. Oct 25, 2017 #1
    So, I'm really bothered with something. Let's suppose there's a simple pendulum with a rigid sphere on it's end. In order to get the motion equations I thought we could use two approaches. One would be using rigid body dynamics (torque, moment of inertia ...), the other one would be using Newton's Second Law in the center of mass. The thing is, the results are slightly different depending on the approach. I suppose the second one is wrong, but why is that? Shouldn't the Law be applicable in any system's center of mass? What isn't right with this reasoning?
     
    Last edited: Oct 25, 2017
  2. jcsd
  3. Oct 25, 2017 #2

    mfb

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    Did you take the rotation of the sphere into account in the second approach? The force from the pendulum won’t act exactly on the center of mass.
     
  4. Oct 25, 2017 #3
    Oh, you mean something like a "spin"? I didn't tought of that. Where does it acts and why not in the center of mass? Is the tension in the string that provokes this rotation?
     
  5. Oct 25, 2017 #4

    mfb

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    If the sphere is connected to the string at a fixed place (not identical to its center of mass), it has to rotate while the pendulum swings. That makes the difference to a point mass.
    With an actual pendulum the string will be a tiny bit "ahead" of the sphere in the motion downwards, providing some off-axis force which (a) leads to torque too make it spin and (b) makes it accelerate slower. Upwards it is reversed, the spinning sphere pushes the string ahead and the off-axis force pushes the sphere up a bit while slowing its rotation.
     
  6. Oct 30, 2017 #5
    The classic equation for a pendulum period (T = 2 pi SQRT(l/g) is an approximation and is only valid for a point sized bob, and for small angles of swing (there are other assumptions). Angular momentum is ignored in the analysis.

    If you have an extended mass as the bob you now have to take account of the fact that the bob rotates a little on each swing.

    Note that I am assuming that the pendulum rod is a weightless, stiff rod attached to the top of the sphere and the bob cannot move relative to the rod. If I attach the pendulum to an axle running through the centre of the sphere which allows the sphere to rotate relative to the rod I get a different answer.

    This is the reason the two approaches give different results. If you set the size of the bob to zero in the torque/angular momentum approach, you will find you get the same answer for small angles of swing as with the point mass. So, in the equation below, if we set epsilon = 0, we get T = 2 pi SQRT(l/g) as expected.

    Incidentally, this is an excellent way to check to see if our equation is wrong - if it does not give the same answer we know it must be wrong.

    The effect is surprisingly large. Take a nominal one second pendulum of length about 40 inches. If you have a clock with a 40 inch pendulum with a spherical bob of radius 3 inches, the clock loses about 100 seconds per day compared with the idealised point mass pendulum. This is equivalent to lowering the bob by about 0.1 inch.

    My book on mechanics (An Introduction to the Theory of Mechanics by KE Bullen) gives the following analysis ... but you should always check something you read in a book to be sure it is correct! The last line is wrong because expanding the series actually gives the line shown below.

    Remember that Richard Feynman, probably the greatest 20th century physicist after Einstein had "If I cannot derive it, I don't understand it" written at the top of his blackboard.

    1.png
    The correct last line is
    2.png
    and we see that setting epsilon to zero gives the conventional answer.
     
    Last edited: Oct 30, 2017
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