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Homework Help: Mechanics Qualifying Exam Question

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A wheel is pulled over a frictionless washboard surface with a constant horizontal velocity component. The equation of the surface is y=Acos(2πx/λ) where A is the height of the surface and x is the horizontal distance along the surface. At what value v_x does the wheel begin to lose contact with the surface, and at what point on the surface will this occur? (assume the radius of the wheel <<λ)

    2. Relevant equations
    not sure

    3. The attempt at a solution
    I wasn't quite sure where to start on this one. I played around with the chain rule (dy/dx) (dx/dt) but didn't get any promising results. If anybody can provide insight on this one, I'd certainly appreciate it.
  2. jcsd
  3. May 15, 2015 #2


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    Welcome to PF!
    Why don't you show what you tried? The chain rule is a good start.
    What is the direction of the velocity at a point with coordinate x? What are the x, y components?
    What are the components of the acceleration? And keep in mind Newton's second equation.
    What is he condition that the wheel loses contact with the surface?
    Last edited: May 16, 2015
  4. May 17, 2015 #3
    By statement of the problem, the velocity in the x-direction is constant, call it v. Since y=A cos(2πx/λ), The velocity in the y-direction (call it u) is given by u=dy/dt. By the chain rule, v=(dy/dx)(dx/dt)= -A(2π/λ)sin(2πx/λ)(dx/dt). But dx/dt = v. So, u = -A(2π/λ)v sin(2πx/λ).

    The sum of the forces in y, where vertically upward is taken to be positive, N-mg=ma_y. As long as the wheel is in contact with the surface, N = du/dt = du/dx dx/dt = -A(2π/λ)^2 v^2 cos(2πx/λ). So that a_y = -mA(2π/λ)^2 v^2 cos(2πx/λ)-mg. That is where I get stuck. The condition for the wheel to leave the surface is N=0, a_y = -g. Obvious N→0 as x→n(λ/4). But how does that depend on v_x? I feel like this is a good place to stop and see what comes back to me.
    Last edited: May 17, 2015
  5. May 17, 2015 #4


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    The equation N=du/dt is wrong. N is force, du/dt is acceleration.
    The normal force is perpendicular to the surface, it is not vertical. It has both horizontal and vertical components. And there is also the pulling force which ensures the constant horizontal velocity component. It is not clear if that force is horizontal, but we can assume it, as it is not stated otherwise.
    The acceleration multiplied by the mass is equal to the sum of forces. Collect all forces and write Newton's second law for both the horizontal and vertical components.
  6. May 20, 2015 #5
    In the x-direction:


    In the y-direction:


    where a=(du/dx)(dx/dt)=-A(2π/λ)^2 v^2 cos(2πx/λ).
  7. May 20, 2015 #6


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    Correct. What is the normal force when the wheel looses contact with the surface?
  8. May 20, 2015 #7
    The Normal force is 0 when the wheel loses contact with the surface
  9. May 20, 2015 #8
    You are correct. Any thoughts on how to proceed?
    Last edited: May 20, 2015
  10. May 21, 2015 #9
    My attempt,

    N2 for the y-direction reads,

    Ny-mg=ma_y or,

    Ny-mg=-mA(2π/λ)^2 v^2 cos(2πx/λ).

    When the wheel loses contact, Ny=0,

    -mg=-mA(2π/λ)^2 v^2 cos(2πx/λ)


    v^2=g/(A(2π/λ)^2 cos(2πx/λ)) (*)

    but when x=nλ, cos(2πx/λ)=1 and a this corresponds to local maxima on the cosine curve, it is the most likely place for the wheel to lose contact.

    So, v^2=g/(A(2π/λ)^2 ).

    When the wheel loses contact a_y = -g.


    a_y=-A(2π/λ)^2 g/(A(2π/λ)^2 ) cos(2πx/λ) = -gcos(2πx/λ)

    and for x=nλ, a_y=-g

    So that checks, but the general form of v^2 (*) bothers me. (*) violates the assumption that v is constant.
  11. May 21, 2015 #10


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    You are told the horizontal component is constant.
  12. May 21, 2015 #11


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    v is constant. Your starred equation tells you what v must be in order for Ny to equal zero when the wheel is at horizontal location x.

    An interesting approach to this problem is to go to the frame of reference moving horizontally with speed v so that the wheel only has vertical SHM motion in this frame. So, ay = -ω2y.
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