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Cylinder with point mass angular momentum

  1. Aug 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A uniform cylinder of mass M and radius R can be rotated about a perpendicular axle through its centre. A particle of mass m is attached to the cylinder's rim. The system is rotated with angular velocity w about the axle, which is held in a fixed direction during the motion.

    Discuss the angular momentum L of the system about the centre of the cylinder. Find the angle between w and L, and explain why torque must be applied to maintain the axle's direction.

    2. Relevant equations

    L = Iw
    L
    = r x p = r x mv = rmv sin(theta)
    v = wxR

    3. The attempt at a solution
    After drawing a diagram, I am aware that there must be a torque to counteract the gravitational force of the point mass on the cylinder. Hence, I am suspecting the angle between w and L to be non-zero. I know that if we assume that the cylinder is rotating counter-clockwise when viewed from upwards (above the plane of the cylinder) then the w vector is directed upward along the axis around which it rotates.

    My thoughts are that we can simply take the vector sum of angular momentum contributions from the point mass and the cylinder as this is appropriate for rigid bodies and particles. Where I am stuck is developing a formula for the angular momentum of the cylinder.

    The direction of the velocity of the cylinder while rotating is what is really getting me. I keep imagine it as being perpendicular to the axis through the centre of the cylinder (i.e. tangential to the circular path that the cylinder rotates in). In which case, there is a moment arm of l/2 where l is the length of the cylinder and some velocity v, which could be rewritten as v = wxR in which case the moment of inertia of the cylinder would be (l/2)xM(wxR). However I am almost uncertain that I am not doing that right.

    In terms of the point mass, I intended to (as previously mentioned) summate it vectorially with the angular momentum of the cylinder to determine the angle between w and L. The angular momentum of the point mass about the axis would simply be (l/2)xmv.
     
  2. jcsd
  3. Aug 27, 2015 #2

    BvU

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    Hello Dee, welcome to PF :smile: !

    I need to ask a bit more information first :

    is not sufficient to uniquely determine the direction (perpendicular to what, exactly?) of the rotation axis. Please clarify. OR is this the litteral problem text ?

    can you post the diagram ?
     
  4. Aug 27, 2015 #3
  5. Aug 27, 2015 #4

    BvU

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    I could be corny and claim that still isn't unique
    1 it could be a side view for a very short cylinder ( why don't they call it a disk then :smile: ? ) and
    2 it could be a horizontal thin cylinder rotating about a vertical axis.

    The statement that m is attached to the rim makes me suspect that 1 is the correct interpretation.
    But your using l/2 (what is l ? confusing!) makes me think your idea differs.

    Your exercise probably doesn't want you to worry about gravitational forces. So focus on rotation only. Which way is the L of the mass m pointing ?

    Is that a 2.5 negation ?
    This is indeed not the right way to find a moment of inertia.
    There is no ##\omega## in a moment of inertia. For a cylinder rotating like this one there is no single velocity.

    You want to re-hash that part of your textbook. My approach is based on the equivalence between ##E_{\rm kin} = {1\over 2} m v^2## for translational kinetic energy and ##I = {1\over 2} I\omega^2## for rotational kinetic energy. If a lot of masses have different radii then $$\Sigma \ E_{{\rm kin}, i} = {1\over 2} I \omega^2 = \Sigma \ m_i {1\over 2} (\omega r_i)^2\; ,$$so ##I = \Sigma \ m_i {1\over 2} r_i^2## .

    For a solid you get ##I \equiv \int r^2 dm## and for a cylinder rotating about its (:wink:) axis ## I = {1\over 2} MR^2##.
     
  6. Aug 27, 2015 #5
    Thanks for your help. I understand how to derive the moment of inertia of a cylinder, however given that the question wants the angle between the angular velocity and L, a cross product would need to be used. As far as I know, L = Iw is not a cross product so one would have to use L = r x mv in order to get an angle.
     
  7. Aug 27, 2015 #6

    BvU

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    You are absolutely right. ##\vec L = I\;\vec \omega## is the expression...and written as such it's not a cross product.
    It also spells trouble, because ##\vec L## and ##\vec \omega\ ## appear to be along the same axis ...

    About ##\vec L= \vec r \times \vec {m v} \;##:

    With ##\vec v = \vec \omega \times \vec r ## you are back to square 1 again: ##\vec L## and ##\vec \omega## are coaxial !


    Still not clear to me if we have a disk-like cylinder or a broomstick-like cylinder !

    And if ##\vec L## and ##\vec \omega## are coaxial, then what about
    :rolleyes:

    --
     
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