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Ball on a Turntable Simple Harmonic Motion

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    ball turntable 2.png
    A spherical ball of mass “m”, moment of inertia “I” about any axis through its center, and radius “a”, rolls without slipping and without dissipation on a horizontal turntable (of radius “r”) describe the balls motion in terms of (x,y) for a function of time.

    **The turntable is rotating about the vertical z-axis at a constant unspecified angular velocity.

    **Radius and mass of the turntable and the ball are unspecified.

    Can someone please show me step by step how to solve this problem?

    2. Relevant equations
    (angular momentum at Center of Mass)=(moment of inertia of a ball)*(angular velocity) or
    hcm=IW

    3. The attempt at a solution
    page 1.png
    page 2.png

    This is my attempt at solving the problem. Can you please show me step by step how you would solve it and what answer you get. Thank you :)
     
  2. jcsd
  3. Dec 10, 2016 #2

    berkeman

    User Avatar

    Staff: Mentor

    Thread closed for Moderation... Thread re-opened so the OP can show more work. :smile:
     
    Last edited: Dec 10, 2016
  4. Dec 10, 2016 #3
    Here's my work in text format:

    In part A
    of my work I determined that angular momentum is conserved about the moving point P. (Point P being the point on the ball in contact with the turntable.) If I'm not mistaken that would mean that the angular momentum at moving point P is coupled to that of the turntable.

    In part B I broke the angular momentum into its components to examine how the ball spins:
    (angular momentum at Center of Mass)=(moment of inertia of a ball)*(angular velocity) or
    hcm=IW

    ν= (2/5)ma^2*<wxi + wyj + wzk>


    (angular momentum at Point P)=(angular momentum at Center of Mass)+(Radius “a”) cross product (linear momentum)

    hp= hcm+ a x p

    = hcm + a x mv

    = hcm + a x m<vxi + vyj>

    = hcm + <-mavyi + mavxj>


    Therefore,

    (2/5)ma^2*wx - mavy = C1 C=constant

    (2/5)ma^2*wy + mavx = C2

    Wz=C3


    In part C I acknowledged that because the ball doesn’t slip “The velocity of point P on the turntable” and the “velocity of point P on the ball” will be equal to each other. I used this knowledge to break up the equation and eliminate the angular velocity components so that we can solve for x and y positions

    Ωk x r = v + w (cross product) (-ak) Ωk= constant rotation about the z axis

    My final answer was:
    x=xo + RCos((2/7)Ωt)
    y=yo + RSin((2/7)Ωt)

    This is my attempt at solving the problem, but my professor has told me it's wrong. Can you please show me how you would solve it and what answer you get. Thank you :)
     
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