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An object on an inclined plane

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A homogenous thin cylinder is put on an inclined plane (angle of elevation is α). The coefficient of friction is [tex] \mu [/tex] . What is the speed of the centre of the mass of the cylinder and it's angular velocity at distance l from the start of motion.


    2. Relevant equations
    [tex] F=ma [/tex]
    [tex] I=mR^2/2 [/tex]
    [tex] v=at [/tex]



    3. The attempt at a solution
    I drew a free body diagram and found out that the acceleration is
    [tex] a=g(sinα-\mu\cosα) [/tex] and found the velocity
    [tex] v=\sqrt{2lg(sinα-\mu\cosα)} [/tex]

    The problem is, i can't find the angular velocity in any way.
     
  2. jcsd
  3. Aug 10, 2011 #2

    HallsofIvy

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    I presume this is pure "rolling", no sliding, motion. In one complete rotation, then, the cylinder will have moved a distance equal to the circumference of the cylinder. Since you know the linear speed of the center axis, you know the time it will take the cylinder to move that far. The angular speed is [itex]2\pi[/itex] radians divide by that time.
     
  4. Aug 10, 2011 #3

    Doc Al

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    Staff: Mentor

    How did you derive this result?
    Same question here.

    The translational and rotational velocities are related by the constraint that the cylinder rolls without slipping.
     
  5. Aug 10, 2011 #4
    [tex]N=mgcos\alpha[/tex]
    [tex]ma=mgsin\alpha-F[/tex]
    [tex]F=\mu N[/tex]

    and from here
    [tex]a=g(sinα−μcosα)[/tex]
    and the distance l is
    [tex]l=at^2/2[/tex]
    from here [tex]t=\sqrt{2l/a}[/tex]
    and velocity is [tex]v=at=\sqrt{2lg(sinα−μcosα)}[/tex]

    But that would mean the cylinder is sliding i guess. I'm completely lost now.
     
  6. Aug 10, 2011 #5

    Doc Al

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    Staff: Mentor

    Do not assume this. F = μN would be the maximum possible value of static friction for the given surface; the actual friction needed to prevent slipping will be less. Just call the friction F and solve for it.

    (Don't forget Newton's 2nd law for rotation.)
     
  7. Aug 10, 2011 #6
    So
    [tex] mg sin \alpha - F = ma [/tex]
    [tex] FR=mR^2\beta/2 [/tex]
    [tex]a=\beta R[/tex]

    and from here [tex]a=2gsin\alpha /3[/tex]

    and i can find the translational velocity, but what about the angular velocity? it seems i will need the radius, which is not given
     
  8. Aug 10, 2011 #7

    Doc Al

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    Staff: Mentor

    Yes, you are correct. It seems you do not have enough information to obtain the angular velocity.

    (If this is a textbook problem, what book is it from?)
     
  9. Aug 10, 2011 #8
    It's not a textbook problem. However, thanks a lot for your help ! :)
     
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