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Minimum linear velocity attained by sphere

  1. Nov 8, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A sphere of mass M and radius R is moving on a rough fixed surface, having co-efficient of friction μ, with a velocity v towards right and angular velocity ω clockwise. It will attain a minimum linear velocity at time (take v>ωR)

    3. The attempt at a solution
    Since v>ωR the sphere rolls with slipping. So frictional force will act in the backward direction. Using the equation [itex]\int \tau dt = \int dL [/itex] where τ=μmgR.

    [itex]\mu mgRt= \frac{2}{5} mR^2 (\omega ' - \omega) \\

    \mu mg = m \frac{dv}{dt} \\

    \mu gt = (v' - v)
    [/itex]

    Using the relation v'=ω'R and solving the above two equations I get
    t= 2(v-ωR)/3μg. But the correct answer has 7 in the denominator.
     
  2. jcsd
  3. Nov 8, 2013 #2

    mfb

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    Staff: Mentor

    There is a minus sign missing.

    I don't see why this point is called "minimum linear velocity" - it is the point where the sphere stops slipping.
     
  4. Nov 8, 2013 #3

    utkarshakash

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    Why there should be a minus sign? Since friction acts backwards, so is the acceleration. Hence both minus cancels out. Are you trying to say that since velocity is decreasing that's why dv/dt should carry a '-' with it?
     
  5. Nov 8, 2013 #4

    mfb

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    Acceleration is backwards, but as you write your equation both sides are positive, so v increases. You can use a negative v everywhere, but then things get really confusing and the initial v>ωR does not work.
     
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