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Rolling cylinder

  1. Mar 16, 2004 #1
    A cylinder is rolling on an inclined plane. We now that friction force does no work. But when we consider the pure rotational motion, friction force is responsible for the rotational acceleration, and 1/2Iù^2 = FRè ( FRè = work )
    What does realy happen here;
     
  2. jcsd
  3. Mar 16, 2004 #2
    The friction force does no work because it is normal to the displacement. On a frictionless plane the cilinder would 'slide' down. Because of the conservation of angular momentum, the angular momentum of the cilinder stays zero when there is no torque present.

    [tex]\frac{dL}{dt}=\Gamma[/tex]

    So I guess the friction force acts as a torque which causes the angular momentum to change, according to the equation above. And so the friction force is responsible for the rotational acceleration.
     
  4. Mar 16, 2004 #3
    The friction force does work. As you (nkehagias) say correctly, it causes the cylinder to rotate. Friction ("sticky friction"; sorry, I don't know the correct word) is in this case simply the way how potential energy is transformed into rot. en., it permits the gravitation to do work an the cylinder.

    What confuses you is probably simply that if we talk about "friction doing work", we usually mean that there is dissipation, "loss of energy".
     
  5. Mar 16, 2004 #4

    Doc Al

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    Staff: Mentor

    For rolling without slipping, the (static) friction force does no work since there is no displacement at the point of contact. It is actually gravity doing the work and providing the energy for rotation and translation.

    Newton's laws still hold, and it is true that it is the static friction which allows the cylinder to rotate. And it is also true that Τθ = FfRθ = 1/2Iω2, where torque is calculated about the center of mass and θ is the angular displacement of the cylinder. But this should not be interpreted as a real "work-energy" relationship: it is just an integration of Newton's 2nd law.

    You can show that the loss of gravitational PE as the cylinder falls will exactly equal the gain in translational plus rotational KE.
     
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