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Torque On A Unicycle

  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A woman is riding a unicycle. The mass of the woman is: m(w)=66.0 kg,
    the mass of the unicycle is: m(u)=5.40 kg, the diameter of the unicycle
    wheel is: d = 55.0 cm, and the length of the unicycle crank is L =
    11.0 cm

    What is the maximum torque that the woman is able to
    generate just by using her weight to push on the unicycle
    pedal?


    2. Relevant equations
    τ = r x F

    3. The attempt at a solution
    So, the maximum force that she can exert on the unicycle is her weight i.e. her mass times g. To find the torque all you have to do is multiply the length of the crank by the weight. I'm just wondering though why this is correct. I mean her weight acts vertically downwards, directly above the centre of the unicycle wheel. By the cross product formula r x F = |r||F|Sin(θ). But isn't theta in this case 0? Implying that the torque is 0, which can't be correct. I may just be visualizing this incorrectly though. If someone could explain this it'd be great. I'll attach a diagram of what I'm visualizing. Thanks.
     
  2. jcsd
  3. Apr 22, 2013 #2
  4. Apr 22, 2013 #3
    ImageUploadedByPhysics Forums1366653977.001940.jpg
     
  5. Apr 22, 2013 #4
    Am I visualising this incorrectly?
     
  6. Apr 22, 2013 #5
    Your diagram did not show the force acting on the pedal but other than that I think you are correct. I think the angle is between th radius and the force and this would be 90 deg and sin 90 = 1 so T = F X r
     
  7. Apr 22, 2013 #6
    Yeah, I was thinking that initially but our lecturer seemed pretty sure that it was the length of the crank times the applied force. This makes I sense to me as the rider is applying the force to the pedals and not the crank itself.
     
  8. Apr 22, 2013 #7
    It is the length of the crank times the force on the pedal. There is nothing magic about the pedal other than a way to get the force from the foot to the end of the crank.
     
  9. Apr 23, 2013 #8
    Ohhh, ok. I see. Thanks man.
     
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