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Confusion over Rolling with Friction and Rolling Without Slipping

  1. Dec 8, 2012 #1

    mhz

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    1. The problem statement, all variables and given/known data

    This isn't a specific question, more of a general one:

    Suppose there is a cylinder of mass [itex]m[/itex] that is rotating in the positive clockwise direction with initial rotational velocity [itex]\omega_0[/itex], and radius [itex]R[/itex].

    Then, suppose this rotating cylinder is placed on a surface with coefficient of kinetic friction [itex]\mu_k[/itex].

    How long after it is placed this surface does it begin to roll without slipping?

    2. Relevant equations

    [itex]\tau = I\alpha \\
    F = ma \\
    I_{cyl} = \frac{mR^2}{2} \\
    v = v_0 + at \\
    \omega = \omega_0 + \alpha t \\[/itex]

    3. The attempt at a solution

    I'm torn between two approaches.

    The first - couldn't I simply solve for a and alpha, set the two linear/angular velocity equations equal to each other for the case that v = r(omega) and solve for t?

    The second - using energy, I could say that the initial rotational kinetic energy equals the final rotational kinetic energy plus toe final kinetic energy plus the energy lost to friction, find d or final velocity when v = r(omega) and use kinematics to solve for t?
     
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  3. Dec 8, 2012 #2

    TSny

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    First approach sounds good (devil in the details).

    Second approach sounds complicated to me. The cylinder slips on the surface so the the distance that friction acts is not the same as the distance the cylinder travels. But if you take that into account, I guess this approach should work.
     
  4. Dec 8, 2012 #3

    mhz

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    So would this be correct?

    [itex] \tau = I\alpha = -fR \implies \alpha = \frac{-2\mu_kg}{R} \\
    -f=ma \implies a = - \mu_k g \\
    v = v_0 + at \implies v = at \text{ Since there is no initial velocity} \\
    \omega = \omega_o + \alpha t \implies at = r\omega_0 + r\alpha t \implies -\mu_k g t = r \omega_0 = 2\mu_k g t \implies t = \frac{r\omega_0}{\mu_k g}[/itex]
     
  5. Dec 8, 2012 #4

    TSny

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    I think it looks good except for one point. Why did you write ##-f = ma## with a negative sign for the force?
     
  6. Dec 8, 2012 #5

    mhz

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    To be consistent, considering that frictional force acts in the negative direction, if both were positive the final answer would be negative

    Also I have lowercase f as frictional force, F as net force
     
  7. Dec 8, 2012 #6

    TSny

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    What direction are you taking to be positive for the linear motion? If the cylinder is initially spinning clockwise, what direction will it the travel along the surface? Is the direction it travels also your positive direction?
     
  8. Dec 8, 2012 #7

    mhz

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    - <-----------> +

    Clockwise = Positive

    Actually, shouldn't the angular acceleration be positive? As the frictional force applies a positive torque.
     
  9. Dec 8, 2012 #8

    TSny

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    Think carefully about the direction of the friction force on the cylinder while it is slipping on the surface. Is it to the right or to the left?
     
  10. Dec 8, 2012 #9

    mhz

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    Assuming the ball is moving to the right, I would imagine the friction force is to the left.
     
  11. Dec 8, 2012 #10

    TSny

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    What propels the ball to move to the right?
     
  12. Dec 8, 2012 #11

    rcgldr

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    One possible source of confusion is that that the friction force could refer to the force the cylinder exerts onto the surface, or the force the surface exerts onto the cylinder (a newton third law pair).

    Assuming you want to consider the force the surface exerts on the cylinder, then if the cylinder is slidng and spinning clockwise, the direction of force from the surface onto the cylinder is to the right, resulting in linear acceleration to the right and angular deceleration of the cylinder (also keep in mind your convention of ω being positive for clockwise rotation). If you allow α to be negative (since there is angular deceleration), then you have:

    f = m a
    f R = τ = I α

    and you can continue with rest of the equations you've already determined.
     
  13. Dec 8, 2012 #12

    mhz

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    Here is a solution I found:

    lFGai.png

    However, why don't they account for energy lost due to friction?
     
  14. Dec 8, 2012 #13

    TSny

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    Everything in your solution was correct except the direction of the friction force. As the cylinder spins clockwise and slips on the surface, it scrapes against the surface and tries to push the surface to the left. So, the surface pushes on the cylinder to the right (+ direction). That force to the right on the cylinder produces a counterclockwise torque (your - direction). So, you had the sign of the torque correct. It is just the direction of ##f## in ##F = ma## that you need to correct.
     
  15. Dec 8, 2012 #14

    TSny

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    Right. That solution is bogus. They also have the friction on the cylinder pointing in the wrong direction as well as not taking into account heat produced by friction. Note that if the direction of the friction is as shown, then the cylinder would increase it's rate of spin!
     
  16. Dec 8, 2012 #15

    mhz

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    Haha oh man, it is quite challenge studying for my Physics final using the finals and solutions posted by my professors when half the answers are wrong (this has happened for maybe 6 or 7 questions since I've begun studying lol).

    Anyway that makes perfect sense guys, thanks a lot!
     
  17. Dec 8, 2012 #16

    haruspex

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    So, what do you get for the answer now? It's Rω0/(3gμk), right?
     
  18. Dec 8, 2012 #17

    mhz

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    Yes that's right
     
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