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Two ways to calculate the final speed of a rotating cylinder

  1. Sep 16, 2012 #1
    Situation:
    A uniform cylinder (Mass [itex]M[/itex], Radius [itex]R[/itex]) is rotating down a slope of incline [itex]θ[/itex] and distance [itex] s[/itex], there are two methods which I used to calculate the final speed, one of which considered forces acting on the cylinder and the other using energy, pure rolling assumed throughout, both give the same final answer.

    Force method:
    1) [itex]Mgsinθ * R = \frac{3}{2} MR^2 α[/itex]
    (Torque about point of contact * R = Angular acceleration * Moment of inertia)
    2) [itex] Mgsinθ - F = MRα [/itex]
    (Newton's Second Law, F is Friction).

    After finding a to be [itex] \frac{2}{3}Mgsinθ [/itex], I then used V2= 2as to find the final speed to be[itex]\sqrt{\frac{4}{3}sgsinθ}[/itex], which is the same as that given by the...

    Energy Method:

    Simply equated GPE and the final rotational kinetic energy.
    [itex] Mgh = \frac{1}{2}( \frac{3}{2} MR^2 )ω^2 [/itex]

    Problem: The similarity of the answers assumes that somehow the cylinder does no force against static friction, which must be present for rolling.
     
  2. jcsd
  3. Sep 16, 2012 #2
    The cylinder does no work against static friction. The part of the cylinder in contact with the ground always has zero velocity.
     
  4. Sep 16, 2012 #3

    Simon Bridge

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    Not really, you have assumed that there is no work against static friction.

    [edit] Hah! I was too sloow!!!
     
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