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Is the tangential acceleration component always zero(wheel?)

  1. Apr 28, 2015 #1
    If a wheel is rolling down a hill without slipping then we know the contact point has zero velocity.

    Is it also true that the tangential acceleration of this point is zero too?
  2. jcsd
  3. Apr 28, 2015 #2


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    Why dont you try computing it?
  4. Apr 28, 2015 #3
    Ok well,

    The velocity at point A is zero, V_A = 0

    So d(V_A)/dt = (a_t)_A = 0

    Therefore the tangential acceleration is zero at this instant.

    Is this correct? It is kind of weird, because then how would the wheel continue turning?
  5. Apr 28, 2015 #4


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    No, it is not correct. The velocity is on momentarily zero and the derivative of a function which is zero at a given point may still be non-zero (take the derivative of x with respect to x at x = 0 for example).

    Can you describe how the position of a given point on the wheel (in a reasonable coordinate system) depends on time?
  6. Apr 28, 2015 #5
    I don't know any physics theory. I'm taking a dynamics course in engineering and its really applied, but this concept is one that I'm curious about because on a previous exam there was a question about a wheel rolling down a hill without slipping, and there was a point P that connects the ground and wheel. The question was what was the acceleration of the point P, and the answer only used the normal component of the wheel, and said the tangent component was 0.

    Altough, I just wrote my calculus 2 final exam so I'm going to try and do some calculus and to give your question a try using what I know, I don't think the n-t coordinate system will be helpful because it has nothing to do with position.

    But I think I can describe the position with parametric equations. If we start from the top of the wheel , then we know its coordinates are x = rsinwt, y = rcoswt (w is angular velocity)

    So taking the derivatives

    x' = wrcoswt
    y' = -wrsinwt

    x'' = -w^2rsinwt
    y'' = -w^2rcoswt

    When the top point of the wheel touches the ground, then it has rotated 180 degrees, so (wt) = pi

    Then x'' = 0 and y'' = rw^2

    Meaning that this point P is only moving up at this exact point

    Since in n-t coordinate system, "up" is y which is n, "left/right" is x which is t

    But our "left/right" is 0. So the only acceleration is "up" So this has to mean there is no tangential acceleration, only normal

    This conclusion agrees with the exam solution too, and I'm pretty confident with it.

    Even though we'd have some new term inside our sin/cos wave changing the angular velocity (due to a gravity component), I'm going to say its the same even when we have a hill, just because the exam answer sort of implied this.

    EDIT: But to do some further calculus, if its rolling down a hill and accelerating without slipping

    r is radius, p is the angular acceleration due to gravity

    x = rsin[ (w + pt)t ] = rsin[ wt + pt^2 ]
    y = rcos[ (w + pt)t ] = rcos[ wt + pt^2 ]

    x' = [ wt + pt^2 ]' rcos[ wt + pt^2 ] = wrcos[ wt + pt^2 ] + 2ptrcos[ wt + pt^2 ]
    y' = [ wt + pt^2 ]' -rsin[ wt + pt^2 ] = -wtrsin[ wt + pt^2 ] - 2ptrsin[ wt + pt^2 ]

    Wow, this is getting messy fast. I'm going to assume y'' is nonzero and only calculate x''

    x'' = -wr(w+2pt)sin[ wt + pt^2 ] + 2prcos[ wt + pt^2 ] - 2ptrsin[ wt + pt^2 ]

    When [ wt + pt^2 ] = pi

    x'' = 0 + 2pr - 0

    Ok, x is nonzero, it shouldn't be... because the normal component is tangent to the hill.

    So I'm just going to assume the tangent component is zero based off of the exam answer
    Last edited: Apr 28, 2015
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