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What is the instantaneous axis of rotation?

  1. Nov 2, 2009 #1
    What is the instantaneous axis of rotation? Can you give some examples and a few situations in helping me understand this better. I can across this term while doing a problem in my physics textbook and I did not understand what it meant.
  2. jcsd
  3. Nov 2, 2009 #2

    Doc Al

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    Imagine a rod of length 2L that moves within the x-y plane. Say it's rotating about its center with angular speed ω. Say its center is moving along the +y axis with speed ωL. At some instant, let the rod be parallel to the x-axis.

    At that instant, one end of the rod is momentarily at rest and the other is moving at speed 2ωL. So you can view the rod as if it were instantaneously in pure rotation about that one end.

    Another common example is a wheel rolling without slipping. At any given moment, the point of the wheel touching the ground is at rest, so that contact point can be considered the instantaneous axis of rotation of the wheel.

    Make sense?
  4. Nov 2, 2009 #3

    D H

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    It is akin to the concept of instantaneous velocity. Suppose a particle is accelerating. Its position x(t) and velocity dx(t)/dt change as a function of time. The values of x(t) and dx(t)/dt at some specific time t are the particle's instantaneous position and velocity at that point in time.

    Now consider the motion of a physical object (as opposed to a point mass) through space. This object can be viewed as a collection of point masses. Each point mass will follow some time-dependent trajectory through space characterized by xi(t) and dxi(t)/dt. Here the subscript i denotes that this is the state description for the ith point mass in the collection.

    In the special case of rigid body, the velocity of some element of the rigid body can always be described in terms of the velocity of the body's center of mass plus a rotational velocity [itex]\boldsymbol{\omega}(t)\times \mathbf r_i[/itex], where [itex]\mathbf r_i[/itex] is displacement vector from the center of mass to the element in question and [itex]\boldsymbol{\omega}(t)[/itex] is the instantaneous angular velocity. The direction in which [itex]\boldsymbol{\omega}(t)[/itex] points is the instantaneous axis of rotation.
  5. Nov 2, 2009 #4
    Thanks both of you. It took some time to understand but finally i think i got it.

    Consider this situation.
    A rod is placed along two walls x axis and y-axis such that one end is on x-axzis while the other is on the y-axis. It is released from rest with the rod being vertical. I need to calculate the angular velocity when the angle made by rod with the horizontal is @.

    I devised two methods to solve this:
    As the rod can be viewed to be in pure rotation about IAOR (inst. axis of rotation), can I conserve the energy using Work Energy Theorem? But I have a doubt in this case. The IAOR continuously varies. So is conservation of energy applied here?

    Strangely though the answer comes out to be correct

    Method II:
    I can assume that the angular velocity of the rod be w. Can I calculate KE of the rod using Integral Calculus and then using this in conservation of energy?

    I am also thinking of a third method using constraint relations between the various points on the rod. Can the answer be found through this method as well?

    Thanks again for your replies
  6. Nov 4, 2009 #5

    Doc Al

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    OK. Are you assuming no friction and that the rod is still in contact with the wall at the given angle?

    Please show what you did. I don't understand how you are using the IAOR to solve this problem. As long as there's no friction you can certainly use energy conservation.
  7. Nov 4, 2009 #6
    I meant that the position of the IAOR varies.. So how can we conserve energy about a varying point ?
  8. Nov 5, 2009 #7

    Doc Al

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    I don't understand what you mean by conserving energy "about a point", varying or otherwise. Can you show how you solved the problem? (Or at least explain what you mean.)
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