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Rolling solid disk

  1. Apr 15, 2008 #1

    lx2

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    I have attempted to solve this question and I got stuck at the last part. Here you go.

    A uniform solid disk is set into rotation with an angular speed X about an axis through its center. While still rotating at this speed, the disk is placed into contact with a horizontal surface and released.

    (a) What is the angular speed of the disk once pure rolling takes place?
    (b) Find the fractional loss in kinetic energy from the time the disk is released until pure rolling occurs.
    (c) Furthermore, assume that the coefficient of friction between disk and surface is u. Show that the time interval before pure rolling motion occurs is RX/3ug
    (d) Show that the distance the disk travels before pure rolling occurs is (RX)^2 / (18ug)

    For part (a), I assumed a constant frictional force acting at the point of contact until pure rolling occurs. Say the force = F, final angular velocity = Y, and t = duration of the force. Moment of inertia of the disk = 0.5MR^2

    Using Newton's Second Law (both linear and rotational),

    Ft = Mv = MRY ................ (1)
    FRt = (0.5MR^2)(X-Y)

    Equating the two equations,
    0.5X - 0.5Y = Y
    1.5Y = 0.5X
    Y = 1/3 X

    (b) Have calculated the fractional loss as 2/3.

    (c) Using equation (1), uMgt = MR(1/3 X)
    t = RX/3ug

    (d) I couldn't solve this part. I tried to apply conservation of energy for the loss of energy due to friction.

    Let the distance traveled be s and F be the frictional force.
    Fs = uMgs = loss of energy = (2/3)(0.5 x 0.5MR^2 x X^2) = (1/6 MR^2X^2)
    s = (1/6 MR^2X^2) / uMg = (1/6 R^2X^2) / ug

    My answers are correct for part a,b,c
    Couldn't get the coefficient as 1/18 as required by question :(
    Someone please tell me what's going wrong :(
     
  2. jcsd
  3. Apr 16, 2008 #2

    Doc Al

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    Staff: Mentor

    You cannot assume that the distance the center of the disk moves is the same as the distance the surfaces slip. But there's an easier way: What's the acceleration of the disk?
     
  4. Apr 16, 2008 #3

    lx2

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    Acceleration of the center of mass? Do u mean I'll have to solve it using kinematics equations?

    Before pure rolling occurs, what happens to the disk actually? slipping+rolling? i couldn't imagine how smth slip and roll at the same time. anything to help me picture that?
     
  5. Apr 16, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes.
    Solve what? For the distance, sure. To get the acceleration, consider the force that acts. Hint: You already have the time.

    Yes, rolling and slipping until the translational speed is increased enough for rolling without slipping.
     
  6. Apr 16, 2008 #5

    lx2

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    I got u. Thanks!!
     
  7. Oct 13, 2011 #6
    Hi. I'm sorry for bumping this old thread. I was wondering about something in his workings. For part B, the fractional loss is not 2/3 right? 2/3 is just the loss in rotational energy.

    The total loss in K.E will be Initial rot. energy - final rot. energy - final translational energy?

    Thanks.



    EDIT:

    Is the fractional loss in K.E 1/6?
     
    Last edited: Oct 13, 2011
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