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Rotational/angular problem: the bike wheel with a valve stem

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data

    You have your bicycle upside-down for repairs. The front wheel is free to rotate and is perfectly balanced except for the 16 g valve stem. If the valve stem is 34 cm from the rotation axis and is located 32 degrees below the horizontal, what is the resulting torque about the wheel's axis? (torque = N*m)

    2. Relevant equations

    torque = rFsin(theta)
    torque = I(alpha)
    torque = MR^2(alpha)

    3. The attempt at a solution

    So I thought that torque = rFsin(theta) would be the most relevant since I think I know the radius, force, and the theta, giving me:

    torque = rFsin(theta)
    torque = r(mg)sin(theta)
    torque = (.34cm)(0.016kg * 9.8m/s^2)(sin32)
    = 0.02825 N*m

    But this is wrong. What did I do wrong?
     
  2. jcsd
  3. Mar 14, 2009 #2
    You did your cross product wrong.
     
  4. Mar 14, 2009 #3
    Thanks for the reply. But I am using the right equation, right? So the F part in the equation is where I'm doing it wrong...
     
  5. Mar 15, 2009 #4
    Ooh, so I took a break for a bit and I finally understand what you mean about the cross product. It all makes sense. I just had to redraw my diagram and draw the Force (mg) from the wheel but then draw the tangent line to the wheel where the Force is coming from. So it's torque = rFcos(theta).
     
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