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Rolling without slipping and friction

  1. Jul 7, 2016 #1
    I am sorry for drawing up very old and closed posting https://www.physicsforums.com/threads/rolling-friction.150891/
    but I saw this question many times and I can not understand why do not people simply use well-known theorem which says that power of the forces which are applied to a rigid body is calculated by formula ##(\boldsymbol F,\boldsymbol v_A)+(\boldsymbol \omega,\boldsymbol M_A)##. Here ##A## is a body fixed point; ##M_A## is a net torque abput the point ##A##; and ##\boldsymbol F## is a net force.
    From this formula it is clear that the work of friction is equal to zero
  2. jcsd
  3. Jul 7, 2016 #2
    It is zero because the first term is negative, while the second term is equal but positive. And that's if its rolling without slipping. Am I right?
  4. Jul 7, 2016 #3
  5. Jul 7, 2016 #4
    Well this is probably the most rigorous way to deal with why the work of friction is zero. It is also in agreement with my intuition cause what this theorem is saying ,in the case we apply it with point A being the c.o.m , is that friction takes translational kinetic energy and transforms it to rotational kinetic energy. This is the best way to understand things, cause just simply saying that the work of friction is zero, because the point of contact has zero relative velocity, is just not so appealing to me.
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