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Finding the angular velocity and rotational kinetic energy

  1. Nov 30, 2013 #1
    1. The problem statement, all variables and given/known data
    3MenJG3.gif

    A solid cylinder is rolling along a horizontal plane and is friction less around its symmetric axis. The cylinder is pulled by a constant force, F and travels the distance, d. The cylinder does not glide and has a friction force, f on the ground.

    Known values:
    Mass: M
    Radius: R
    Moment of inertia: ##I=\frac{1}{2}MR^2##
    Force: F
    Angle of the force: θ
    Friction force: f
    Distance traveled: d
    Angular acceleration: α

    Find the angular velocity and rotational kinetic energy for the cylinder when it starts at rest and travels the distance d.

    2. Relevant equations
    [itex]\frac{d\omega}{dt}=\alpha[/itex]

    [itex]K=\frac{1}{2}I\omega^2[/itex]


    3. The attempt at a solution

    From a previous question, I found the angular acceleration:
    o44rN1P.gif

    I know the relation between angular velocity and angular acceleration is:
    [itex]\frac{d\omega}{dt}=\alpha[/itex]

    However what got me confused is the time, t. If I can integrate my angular acceleration in terms of t then I can find my angular velocity but I don't know the time?
     
  2. jcsd
  3. Dec 1, 2013 #2

    haruspex

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    Any conservation law you might be able to use?
     
  4. Dec 1, 2013 #3
    I really have no clue. Maybe energy conservation between the state at rest and when it has traveled d distance? I'm not sure...
     
  5. Dec 1, 2013 #4

    haruspex

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    Sounds appropriate. What work will F have done on the system?
     
  6. Dec 1, 2013 #5
    It would have done:
    [itex]W=F\cdot \cos(\theta)\cdot d[/itex]

    Right? I'm not sure how to use this in my energy conservation equation.
     
  7. Dec 1, 2013 #6
    Why would it even have angular velocity if the force is being applied through its cm?
     
  8. Dec 1, 2013 #7
    Because I also have a friction force.
     
  9. Dec 1, 2013 #8
  10. Dec 1, 2013 #9

    haruspex

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    Yes.
    Where has that work gone? It's rolling contact with the ground, and there's no friction at the axle, so no work has been done overcoming friction.
     
  11. Dec 1, 2013 #10
    sorry I thought it was a frictionless surface.
     
  12. Dec 1, 2013 #11
    Has it gone to the friction between the ground and the cylinder?
     
  13. Dec 1, 2013 #12
    I'm still not sure about this.
     
  14. Dec 1, 2013 #13

    haruspex

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    No, as I posted before, that's rolling contact, so the friction does no net work. The work energy is still in the system. What form is it taking? You have a choice of KE and/or PE.
     
  15. Dec 1, 2013 #14
    I'd say there is KE. We have KE because the cylinder has a velocity but we have zero PE because our plane is horizontal.
     
  16. Dec 1, 2013 #15
    Since I have no PE and the energy is conserved I have:
    [itex]K_1=K_2[/itex]
    [itex]\frac{1}{2}m\cdot 0=\frac{1}{2}mv^2[/itex]

    Because it starts at rest so the velocity is zero.

    I also know that the angular velocity is:
    [itex]v=R\omega[/itex]

    So if I can find an expression for the velocity after it has traveled distance d then I can find my angular velocity as well. Problem is how do I find the velocity....
     
  17. Dec 1, 2013 #16

    haruspex

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    Suppose it has velocity v. Write out expressions for the angular velocity, the linear KE, the rotational KE and the total KE.
     
  18. Dec 1, 2013 #17
    [itex]\omega=\frac{v}{R}[/itex]

    [itex]KE_{linear}=\frac{1}{2}mv^2[/itex]

    [itex]KE_{rotational}=\frac{1}{2}I\omega^2[/itex]

    [itex]KE_{total}=KE_{linear}+KE_{rotational}=\frac{1}{2}mv^2+\frac{1}{2}I \omega^2[/itex]
     
  19. Dec 1, 2013 #18
    I'm not sure how it will help me.
     
  20. Dec 1, 2013 #19

    haruspex

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    Good. You know how to write ω in terms of v and R, and you know how to write I in terms of M and R. So you can get KE in terms of M, v and R.
    You previously wrote
    but that's wrong. That would be true if no work had been done on the system. The force F has done work on the system, and you already calculated that value. So what equation should you write for work conservation?
     
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