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Wheel rolling uphill

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A wheel with mass m, inner radius r and outer radius R is placed on a slope. The slope has angle θ related to horizontal. When the wheel is affected by a constant, known torque Tau in the centre of mass, the wheel begins rolling uphill.

    http://sveskekat.dk/files/uploads/phys3.PNG [Broken]

    a) I need to draw a force diagram (free body diagram?) and find the acceleration of the wheel

    b) The wheel starts at rest and rolls uphill. What is the angular momentum when the wheel has rolled one revolution?

    2. Relevant equations
    sum(Tau) = I * alpha


    3. The attempt at a solution
    I did not attempt b) yet, so please disregard that one so far.

    For a) I am thinking I can use sum(Tau) = I * alpha.
    I did a calculation for I of the wheel, but doing so I assumed that the wheel has even density, and is a solid disk. I dont know if it is correct to assume this, but I tried it, and I then replaced sum(Tau) for Tau and alpha for alpha = a / R, and then solve for a.

    But I think I need to use the angle θ of the slope somehow. I need suggestions.



    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 25, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    Your idea is right with respect to using τ to determine the angular acceleration α.

    τ = I*α

    You will need however to calculate a proper I based on the given that the wheel, while evenly distributed, is only distributed over the radial range of r to R.

    ∫ ρ*2πr*r² dr from r to R

    (Solving the integral requires recognizing the total mass m is given by the mass area density ρ times the area of the ring which is π(R² - r²). )

    For b) you know that

    L = I*ω

    and you should know from linear kinematics analogy that

    Final-ω² = Initial-ω² + 2*α * x

    where α is your angular acceleration and x is the distance in radians. (1 revolution is how many radians again?)
     
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