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Question on rolling motion

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Question
    A disk is initially stationary on a platform. Force P is exerted on the platform. Mass of Disk is MD and Mass of Platform is MP.

    Determine the minimum coefficient of friction in order for the ball to not roll.
    What is the acceleration of the ball as seen from stationary frame?

    Attempt

    2. Relevant equations

    F=ma
    I(alpha)=Summation of Moment

    3. The attempt at a solution

    Acceleration of Ball has to be equal to the acceleration of the Platform since both starts from stationary.

    F (friction on ball)= MD x a
    P - F (friction on platform) = MP x a

    Since F can obtained, the angular acceleration of the ball is to be determined next.

    I x (alpha = angular acceleration) = (moment about center of the ball = Fr)

    With this "alpha" can be determined.

    Now for the last step,

    Acceleration of ball from neutral frame = acceleration of cart - (alpha x radius of ball)
    Hence, negative acceleration is obtained, which infer that the ball roll backwards.

    Please do correct me if there is any mistake.

    Thanks in advance.
     
  2. jcsd
  3. Nov 23, 2009 #2

    Doc Al

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    Your problem description is not quite clear. I assume that the problem begins with a ball at rest on a platform (on top of a frictionless surface)? And that you need the minimum coefficient of friction so that the ball rolls without slipping?
     
  4. Nov 23, 2009 #3
    Yes. Initially the disk and the platform are stationary. A force P is then applied to the platform alone. What is the minimum coefficient of friction so that the ball rolls without slipping.
     
  5. Nov 23, 2009 #4

    tiny-tim

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    Welcome to PF!

    Hi estalas! Welcome to PF! :smile:

    Like Doc Al, I don't understand either. :confused:

    You talk about a disk and a ball.

    Are they the same thing, or two different things?

    If it's a disk (like a coin), in which direction is it pointing, compared with the force, P?
     
  6. Nov 23, 2009 #5

    Doc Al

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    I assume you mean the ball and the platform are initially stationary. (There's no disk, right?)
     
  7. Nov 24, 2009 #6
    Sorry, my mistake for the ambiquity. Thanks for pointing it out !

    Its an infinitely thin disk, with I= 1/2 mr^2.
    So the question is on what is the coefficient of friction to keep the disk from slipping.
     
  8. Nov 24, 2009 #7

    tiny-tim

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    (ambiguity! :wink:)

    Just one more point …

    originally you said "rolling" … should we forget about that?

    is this a disk, lying flat on the platform, and the only thing it can do is to slip (or slide)?

    ok, do Ftotal = ma for the disk …

    what do you get? :smile:
     
  9. Nov 24, 2009 #8

    Doc Al

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    OK, so if I understand you correctly, there is no ball in this problem only a disk. You have a disk on its edge resting on a platform. (I'll assume that the axis of the disk is perpendicular to the acceleration of the platform.) The problem is to figure out the minimum coefficient of friction that would have the disk roll without slipping. Right?

    Assuming that is correct, let's return to your original post.
    The acceleration of platform and disk cannot be the same--otherwise they would move together and the disk would never roll.

    Hint: How do the accelerations of the disk and the platform and the angular acceleration of disk relate to each other, considering that the disk is rolling along the platform without slipping?
     
  10. Nov 24, 2009 #9
    By mental visualization,
    The acceleration of the disk should be lower than that of the platform itself. And the angular acceleration of the disk should be resulting in acceleration that is in opposite direction to the platform. Am I right?

    However, i am having a difficulty in coming up with the relation to all these.
    Please do enlighten me with the relationship and how it come about... thanks !
     
  11. Nov 24, 2009 #10

    Doc Al

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    Good.
    The translational acceleration of the disk (its center of mass) will be in the same direction as the force on it. That same force produces the angular acceleration of the disk, causing it to roll back (with respect to the platform).

    For the moment, view things from the frame of the platform. What's the condition for rolling without slipping? That will allow you to relate the angular acceleration with the translational acceleration of the disk with respect to the platform.
     
  12. Nov 24, 2009 #11
    Okay

    In the case whereby "force P aact to the right", causing platform to accelerate to the right,

    For rolling without slipping, acceleration of the disk w.r.t platform should be equal to the product of angular acceleration x radius

    Hence,
    translational acceleration = angular acceleration x radius

    With this, Is it correct for me to say the frictional force is acting to the left, because the disk is rotating to the left wrt platform. Hence, the disk is actually acceleration in opposite direction if i were to look at it at the platform?

    And, is the translational acceleration of the disk = acceleration of the platform due to force, but in opposite direction?

    Thanks for the fast reply !
     
  13. Nov 24, 2009 #12
    Oh yeah to add on,

    Just like you said, translational acceleration should be in the same direction as angular acceleration, this means frictional force on the disk will be to the left.

    Hence, the frictional force acting on the platform will be to the right, hence this will cause the platform to move even faster ? i dont really get this part.
     
  14. Nov 24, 2009 #13

    Doc Al

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    Good. Note that this translational acceleration is with respect to the platform.

    No. Friction acts to prevent slipping. If there were no friction, the disk would slip to the left (with respect to the platform); friction acts to prevent that slipping, thus it acts to the right. Think of someone pulling the platform to the right and the platform dragging the disk along via friction.
    That's true. The disk rolls to the left with respect to the platform.

    No.
     
  15. Nov 24, 2009 #14

    Doc Al

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    No. The translational acceleration with respect to the platform will be in the same direction as the disk rolls with respect to the platform. But friction acts to the right. (Careful about analyzing the dynamics in an accelerating frame. You need to add inertial forces.)
    That wouldn't make much sense. Luckily, it's not true. :smile:
     
  16. Nov 24, 2009 #15
    Ok.

    So

    1) Friction on cylinder is to the right.
    Hence, using F=(M disk) a, Translational acceleration of Cylinder is to the Right

    2) The risk rolls without slipping to the left wrt platform.
    r(alpha)=a (nonslip condition)

    3. With the alpha obtained, taking moment about centre of circle
    I(alpha)=Fr
    This will allow me to find F, the frictional force, hence the coefficient of friction.

    Anything wrong with the above equation?

    And finally, i still need to find the Acceleration of the Circular Disk. I could not relate it to the acceleration of the platform..
     
  17. Nov 24, 2009 #16
    OK. i Think i got the relation.

    F= Frictional force between disk and platform.
    m= mass of disk
    M= mass of platform
    a= acceleration of disk, wrt to neutral frame
    A= acceleration of platform, wrt to neutral frame.
    r= radius of disk

    Hence,

    A- (alpha) r = a

    In this case, lets consider the disk also move at A due to platform. However, it rolls in opposite direction, hence, the actual acceleration wrt neutral frame is A - (alpha) r

    Is my relation right?
     
  18. Nov 24, 2009 #17

    Doc Al

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    Make sure you realize that the acceleration in (1) is not the same acceleration as in (2).

    Yes, that looks good to me. (What you are calling a 'neutral' frame is usually called an inertial frame.)
     
  19. Nov 24, 2009 #18
    Alright! Thank you for the guidance along the way =)
     
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