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Normal force in rotational motion

  1. Dec 8, 2012 #1
    Hi there,
    I've got a conceptual question about the normal force as applied to rotational motion. Suppose you have an object like a uniform disk. If the disk were set up so that its axis of rotation were about its centre of mass, it would just sit there and the normal force would be equal to +mg.

    What happens if the disk is instead hinged so that its axis of rotation is at its rim (see attached image)? Here if the disk is held horizontally and then released, it'll experience an angular acceleration and start to rotate. Ultimately were it frictionless, it would oscillate back and forth.

    My question is what happens to the normal force in this situation? Does it remain equal to mg since ultimately the hinge isn't accelerating? Or, does the normal force decrease since the centre of mass of the disk is accelerating downwards. Or am I totally confused?



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  3. Dec 8, 2012 #2

    Jano L.

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    The normal force is the force due to axis rod which supports the disk? If so,
    this will be initially lower than mg, since the disk accelerates downwards. Continually, as the disk rotates and falls down, the acceleration decreases and so the normal force increases. When the disk is at its lowest position, the normal force will be at its maximum value, which is greater than mg (due to rotation of the disk).
  4. Dec 8, 2012 #3
    Thanks, Jano. This helps. Just to clarify, if the disk we just suspended at its lowest point without rotating, the hinge force would be equal to mg? But, since it's rotating and a radially directed centripetal force is required, the hinge force must exceed mg? Do I have that correct? If I interpret your response correctly, then, the force at the hinge will have a magnitude that varies in quite a complicated manner as the disk undergoes its rotational motion?

    Thanks for your help, again.

  5. Dec 9, 2012 #4


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    Not too complicated: just the force needed to balance the radial component of the gravitational force, + centripetal force. What will make it complicated for large perturbations is the 'circular error', i.e. the extent to which it is not SHM. For small perturbations, taking the SHM approximation, the hinge force will be something like A-Bθ2.
  6. Dec 10, 2012 #5

    Jano L.

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    That's right. Exact variation can be found as a function of the angle, but probably it won't be the simplest function possible.
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