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Ball rotating on a ball

  1. Jun 13, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball of mass m and radius r starts off on top of a fixed ball of radius R. it rotates down without slipping.
    What's the ball's center of mass velocity as a function of θ.
    The angular velocity is ω.

    2. Relevant equations
    Moment of inertia: ##I=\frac{2}{5}mr^2##
    Potential energy: ##E_p=mgh##
    Kinetic energy due to rotation: ##E_ω=\frac{1}{2}I\dot\theta^2##

    3. The attempt at a solution
    Kinematics: the path the small ball makes is equal to the path on the big one, and so are the angular velocities
    $$R\dot\theta=r\omega\rightarrow\omega=\frac{r}{R}\dot\theta$$
    The height of the ball's center during the fall is ##(R+r)\cos\theta##
    Total energy at the top, relative to the center, is equal during the fall:
    $$mg(R+r)=\frac{1}{2}\left( \frac{2}{5}mr^2 \right)\omega^2+mg(R+r)\cos\theta$$
    $$mg(R+r)=\frac{mr^2}{5}\frac{R^2}{r^2}\dot\theta^2+mg(R+r)\cos\theta$$
    The center of the ball's velocity is ##v=(R+r)\dot\theta##
    I can't solve for v, and the answer should be an isolated v
     

    Attached Files:

  2. jcsd
  3. Jun 13, 2015 #2
    whats the meaning of isolated v ? If you can give me an example I can solve the question.
     
  4. Jun 13, 2015 #3
    I think ##v## is ##v_{cm}##. Check your 1st equation for energy consevation. Is it correct?
     
  5. Jun 13, 2015 #4
    $$mg(R+r)=\frac{1}{2}\left( \frac{2}{5}mr^2 \right)\omega^2+mg(R+r)\cos\theta+\frac{1}{2}mv_{cm}^2$$
    kinematics:
    $$v_{cm}=(R+r)\dot\theta\rightarrow\dot\theta^2=\frac{v_{cm}^2}{(R+r)^2}$$
    $$mg(R+r)=\frac{1}{2}\left( \frac{2}{5}mr^2 \right)\frac{R^2}{r^2}\dot\theta^2+mg(R+r)\cos\theta+\frac{1}{2}mv_{cm}^2$$
    $$mg(R+r)=\frac{1}{2}\left( \frac{2}{5}mr^2 \right)\frac{v_{cm}^2}{(R+r)^2}+mg(R+r)\cos\theta+\frac{1}{2}mv_{cm}^2$$
    $$v_{cm}^2=\frac{10g(R+r)^3(1-\cos\theta)}{2R^2+5(R+r)}^2$$
    It's not right
     
  6. Jun 13, 2015 #5
    ##\omega## is not ##\dot{\theta}##. The ratio is ##\frac{\omega}{\dot{\theta}} = \frac{R}{r}##
     
  7. Jun 13, 2015 #6
    If you know th exact answer can you tell me ? I solved it I guess.I think your equation is wrong why there is ##(R+r)mg## and the other side ##(R+r)mgcosθ##
     
    Last edited: Jun 13, 2015
  8. Jun 13, 2015 #7
    But i have done that in:
    $$mg(R+r)=\frac{1}{2}\left( \frac{2}{5}mr^2 \right)\omega^2+mg(R+r)\cos\theta+\frac{1}{2}mv_{cm}^2$$
    $$mg(R+r)=\frac{1}{2}\left( \frac{2}{5}mr^2 \right)\frac{R^2}{r^2}\dot\theta^2+mg(R+r)\cos\theta+\frac{1}{2}mv_{cm}^2$$
     
  9. Jun 13, 2015 #8
    Ok. You are make correct inside equation.
     
  10. Jun 13, 2015 #9
    Why theres two potantial energy ? Can somebody give me an answer ??
     
  11. Jun 13, 2015 #10
    The right hand is the potential energy to arbitrary angle ##\theta>0##.

    Your answer is correct. The angle that the small ball leave the big ball surface is another question.
     
    Last edited: Jun 13, 2015
  12. Jun 13, 2015 #11
    In yor perspective the potantial energy on "top" will be zero thats nonsense.The total potantial energy is ##mg(R+r)(1-cosθ)## or ##mg(R+r)(cosθ)##
     
    Last edited: Jun 13, 2015
  13. Jun 13, 2015 #12
    There is a problem with the unlucky point on the attached graph. With big ball centre as origin the potential energy is ##mg(R+r)\cos{\theta}##.
     
  14. Jun 13, 2015 #13
    If you consider left picture you need to realize the potantial energy is cannot be zero top but If you consider right picture you will consider that potantial energy iz zero.I am not wrong which one x and which one is x axis and which one is y axis ? I assumed left picture and If you assumed same thing then my equation is true.But If yıu assume right pic your equation is true.
     

    Attached Files:

  15. Jun 13, 2015 #14
    On the OP, the potential energy variation makes the move, indepented from origin. See a better image.
     
  16. Jun 13, 2015 #15

    ChrisVer

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    In your solution for [itex]v^2[/itex] why did you square the [itex]1-\cos \theta[/itex]?
    Was that a typo?
    Also a square should be missing from the denominator (where you add m^2 with m)
     
  17. Jun 13, 2015 #16

    ChrisVer

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    Also there is no problem with the potential energy as written.
    At first when the ball is on top you may say that you have potential energy [itex]mg (R+r)[/itex].
    Then the potential energy once the ball has taken some kinetic energies by slidding, the potential energy is [itex]mg (R+r) \cos \theta[/itex].
    The zero is taken at the height of the big ball's center.

    It doesn't matter where you take the zero in fact, since the potential energy of the system will have to be mg(R+r) at the beginning (at the end potential energy is not taken at a point but as the difference between two points). Either you take point A to be 0 and point B to be U, you can as well take point B to be 0 and point A will be U.
     
  18. Jun 13, 2015 #17

    ChrisVer

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    *deleted found the answer*
     
  19. Jun 14, 2015 #18

    haruspex

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    Not sure if this has been pointed out already, but that is not right. E.g. consider the case where r=R and it has rolled all the way from top to bottom. ##\theta = \pi##. Through what angle has the moving ball rotated?

    [Edit: there is quite a quick way. Pretend that at angle ##\theta## the moving ball encounters a ramp going down tangentially to the fixed ball. Will transition to the ramp immediately change the linear or angular velocities? What is its KE now?]
     
    Last edited: Jun 14, 2015
  20. Jun 14, 2015 #19
    Thank you
     
  21. Jun 14, 2015 #20

    ehild

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    As Haruspex pointed out, it is not right.
    See figure.
    ballonsphere2.JPG
    The contact point with the sphere is the red dot on the ball. The ball rolls a distance s on the ball , so the point A moves into A', and the length of arc P1A' on the sphere is s. The angle <P'O'A' = s/r, but the ball turned by the angle <P'O'P more. The angle of turn of the ball is Φ=s/r+θ. s=θR, so Φ=(R/r+1)θ. ω=dΦ/dt and ##V_{CM}=(R+r)\dot \theta##.
     
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