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Angular friction with ground

  1. Sep 21, 2010 #1
    I am trying to figure out how exactly Angular friction ("torque") is connected with friction force.

    Is there any connection between torque friction and force friction?
     
  2. jcsd
  3. Sep 21, 2010 #2

    rock.freak667

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    Well torque is related to force by T=rxF
     
  4. Sep 22, 2010 #3
    Yes i know for that relation but what if i have box on the ground which fully touches ground with all bottom area. Which i spin and i know that static friction is k_s and kinetic is k_k. How can i figure out when box will stop rotating?

    Box has some moment of inertia and some mass...

    I know that formula is.
    [tex]\omega = 0[/tex]

    [tex]\omega = \omega_0 + \alpha t[/tex]

    [tex]\tau = I \alpha[/tex]

    If i toss around elements
    [tex]\frac{(\omega - \omega_0)}{\alpha} = t [/tex]

    [tex]\alpha = \frac{\tau}{I}[/tex]

    and if i join equations together
    [tex]\frac{(0 - \omega_0) I}{\tau} = t[/tex]

    then what the heck is torque(T)? If i insert [tex]\tau = r \times F[/tex] Then i have missing variable r.
     
  5. Sep 22, 2010 #4

    rock.freak667

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    r would be the distance from the center to where the force is applied.
     
  6. Sep 22, 2010 #5
    Yes i know, that why i cant get it because box spins on ground and it has full contact on ground whole area is in contact. That why i cant figure out what i could do with r. because r is scalar and not contact area!
     
  7. Sep 22, 2010 #6

    rock.freak667

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    Friction is independent of area and given by Ffriction= μN where N is the normal reaction.
     
  8. Sep 22, 2010 #7
    I can use [tex]\tau = r \times F[/tex] for point contacts with ground and then calculate produced torque. Which gives me no problem.

    But what if i spin box around ground normal. how can i get then torque which will stop the box from spining on ground?
     
  9. Sep 22, 2010 #8

    rock.freak667

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    Multiply the frictional force by the distance from the center of rotation.
     
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