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Angular acceleration of a cylinder rotating around another

  1. Dec 14, 2016 #1
    Screen Shot 2016-12-15 at 7.29.03 AM.png

    For Q11(b), what is the relation between the angular acceleration ##\alpha## of the bottom right cylinder and its horizontal acceleration ##a_x##?

    I get ##\alpha=\frac{a_x}{\sqrt{3}R}##, which is half the given answer (7.84) below.

    After the bottom right cylinder rolls around the top cylinder by an angle of rotation ##\theta## about its (the bottom right cylinder's) own center of mass, the cylinders will look as follows:
    image.jpeg
    (The bottom left cylinder is omitted in the drawing.)

    Let the axis of symmetry of the figure of 3 cylinders at ##t=0## be ##x=0##. (In other words, the point of contact of the two bottom cylinders has an ##x## coordinate of 0).

    Then the ##x## coordinate of the center of the bottom right cylinder, ##x = 2R\sin\phi##, where ##\phi=30^\circ+\theta##.

    ##\dot{x}=2R\cos\phi\dot{\phi}##

    ##\ddot{x}=2R(\cos\phi\ddot{\phi}-\sin\phi\dot{\phi}^2)=2R(\cos\phi\alpha-\sin\phi\omega^2)##, where ##\alpha=\ddot{\theta}## is the angular acceleration of the bottom right cylinder and ##\omega=\dot{\theta}## is its angular velocity.

    At ##t=0##, ##\phi=30^\circ## and ##\omega=0##. Thus ##\ddot{x}=2R(\cos 30^\circ\alpha)##. And we get ##\alpha=\frac{a_x}{\sqrt{3}R}##.

    The given answer:
    Screen Shot 2016-12-15 at 7.29.31 AM.png

    I believe the given answer is wrong because I suspect there is an approximation involved when it uses an infinitesimal distance ##d## in deriving (7.84).
     
    Last edited: Dec 14, 2016
  2. jcsd
  3. Dec 15, 2016 #2

    TSny

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    I don't believe that ##\ddot{\theta}## equals the angular acceleration ##\alpha## of the bottom right cylinder.
    https://en.wikipedia.org/wiki/Coin_rotation_paradox
     
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