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Homework Help: Rolling balls angular rotation

  1. Apr 17, 2010 #1
    If a cylinder is rolling without slipping, C is the centre of zero velocity for a moment and O is the centre

    Does the angular rotation about O equal to the angular rotation about C, or is there only one angular rotation when a cylinder is rolling, that is the rotation about the point of contact?

    3. The attempt at a solution

    This is a question that came to me, not a assignment question or anything, but anyway

    I think in the ordinary frame of reference a rolling cylinder only has an angular rotation about the point of contact, not about the centre

    Where in a frame of reference where you are following the cylinder, you should see the cylinder rotating about the centre and the surface is moving linearly without sliding.

    I know angular rotation is always about an a line, so a single motion can have many angular velocities with respect to many axis's.

    Is it possible to evaluate the angular velocity with respect to O in the normal frame of reference when the ground is stationary?

    i am not really sure i guess the angular velocity for each point on the shape would vary if you measure it from O since the whole object is kind of translating... and since C has a zero velocity
  2. jcsd
  3. Apr 18, 2010 #2


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    Hi SpartanG345! :smile:

    (btw, better to say "instant" rather than "moment", so as not to confuse with other types of moment :wink:)

    Angular velocity (unlike angular momentum) is the same about any point.

    Angular velocity is a "free" vector (strictly, a "free" pseudovector), so (unlike force) it has a direction, but not a specific line in that direction.

    Changing to a different inertial frame will, of course, alter the velocity, but will not alter the angular velocity.

    Formulas that combine I and ω use the same ω, no matter whether I (the moment of inertia) is about the centre of rotation or the centre of mass (btw, they don't generally work about any other point).
    Yes, you get τ = IOω, instead of 0 = ICω - rmv, which is the same since IC = IO + mr2.

    (I've used your notation, but usually we use C for centre of mass, and O for centre of rotation :wink:)
  4. Apr 18, 2010 #3

    Doc Al

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    Just to add to what tiny-tim has already explained...
    Since the cylinder rolls without slipping, its instantaneous axis of pure rotation is the point of contact. So you can describe the motion in two ways:
    (1) As a pure rotation about the point of contact.
    (2) As a combination of rotation about the center of mass plus translation of the center of mass.

    (Same ω in both cases, of course.)
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