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KK Prob 1.14 - Angular Accl

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A drum of radius R rolls down a slope without slipping. Its axis has acceleration "a" parallel to the slope. What is the drum's angular acceleration A?

    2. Relevant equations

    x(t) = x_0 + v_0 + 1/2gt^2 (because of constant acceleration)
    w = S / 2pi

    3. The attempt at a solution

    The center of the circle moves according the kinematic equation for position with constant acceleration.
    If we consider its starting position as 0, then the position just gives the distance traveled by the center. I hope that I can also say that the initial velocity is 0 (can I?)

    So then the distance traveled is 1/2gt^2. Dividing by 2R * pi gets us the number of revolutions, and then multiplying by 2pi gives the number of radians.

    so (1/2R)gt^2 after simplification. If we divide by time, we should get w, or the angular velocity as gt/(2R).

    Upon taking the derivative to find the angular acceleration, I get g/(2R). But this definitely does not seem right.
     
  2. jcsd
  3. Sep 8, 2010 #2

    Delphi51

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  4. Sep 9, 2010 #3
    Yeah, I've already seen that page. However, I don't understand how that "formula" is derived. It should have some similar derivation as mine does (and I'm sure the authors Kleppner and Kolenkow would have wanted that).

    On that note, what's wrong with my derivation?
     
  5. Sep 9, 2010 #4

    Delphi51

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    The formula A = a/R for angular acceleration comes from the basic definition d = rθ for arc length on a circle of radius r.
    Differentiate with respect to time to get v = rdθ/dt = rω
    Differentiate again to get a = rdω/dt = rA.

    Your derivation makes considerable sense . . . but there are problems. For instance, the cylinder rolls down a ramp so it doesn't feel the full acceleration g of a vertical fall; you must begin with the given acceleration a rather than g. The step where you divide by time to get angular velocity assumes constant velocity which is not the case.
     
  6. Apr 3, 2011 #5

    jbunniii

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    My last physics class was in 1986, and although I have studied a lot of mathematics since then, my physics is pretty rusty. I recently started reading Kleppner and Kolenkow's "Introduction to Mechanics" to review the subject just for fun. I'm going to try to do all the exercises as part of this process.

    Most of the exercises don't have solutions (a few have answers or hints), so to ensure that I am not deluding myself, I thought I would post any here that I'm not 100% certain about. This one, 1.14, falls in the, hmm, 98% category, so here we go.

    Problem statement: A drum of radius [itex]R[/itex] rolls down a slope without slipping. Its axis has acceleration [itex]a[/itex] parallel to the slope. What is the drum's angular acceleration [itex]\alpha[/itex]?

    My answer: I think the slope is a red herring. Whatever effect gravity and the slope angle have must already be consolidated into the given acceleration [itex]a[/itex]. If I understand correctly, this is the distinction between kinematics and mechanics: the accelerations are simply given, as opposed to being calculated based on the physical forces that influence the object.

    The acceleration is [itex]a[/itex], parallel to the slope. But who cares what the slope is? I can simply rotate the problem so that the drum is rolling along a horizontal surface, with acceleration [itex]a[/itex] in the positive horizontal direction. With these coordinates, the position, velocity, and acceleration of the drum axis are all scalar functions of time.

    If [itex]v(t)[/itex] is the horizontal velocity of the drum's axis, then the angular velocity is

    [tex]\omega(t) = \frac{v(t)}{R}[/tex]

    (here I have used the fact that the drum is not slipping), and so the angular acceleration is

    [tex]\alpha(t) = \dot{\omega}(t) = \frac{\dot{v}(t)}{R} = \frac{a}{R}[/tex]

    quite independently of the slope.

    Is my reasoning sound? This seems deceptively simple for a book with a reputation for tricky problems.

    P.S. I recognize this is an old thread, but it covers the same problem as the one I am discussing. What's the better protocol in this case: replying to an old thread or starting a new one?
     
    Last edited: Apr 3, 2011
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