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Compare Velocity of Center of Mass - Two Hollow Cylinders

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=59497&stc=1&d=1371091306.png


    2. Relevant equations
    [itex] mgh=\frac{1}{2}mv^2+\frac{1}{2}Iω^2 [/itex]


    3. The attempt at a solution
    I used conservation of energy, and found that the final velocity of the center of mass is independent of the object's mass, so in both cases: [itex] v=\sqrt{gh} [/itex]. Therefore, the ratio is 1.

    However, the answer is 0.5. I think I need to take into account the friction force, but I'm not sure how to proceed, the question does not give me the static friction coefficient.

    Any help would be greatly appreciated.
    LovePhys
     

    Attached Files:

  2. jcsd
  3. Jun 12, 2013 #2

    gneill

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    Staff: Mentor

    Cylinder A rolls, it does not slip. Where can gravitational PE end up? (Hint: rolling implies rotation...)
     
  4. Jun 13, 2013 #3
    @gneill

    Thank you. I came up with a different answer now, but it is not 0.5

    For cylinder A, the final velocity of the center of mass is still [itex] v=\sqrt{gh} [/itex]
    For cylinder B, I think that since the surface is frictionless, and all the forces that are acting on it (gravity and normal force from the incline) are going through the center of mass, so the net torque is 0 and it will not roll but slide. So using conservation of energy:

    [itex] mgh=\frac{1}{2}mv^2 [/itex] and [itex] v=\sqrt{2gh} [/itex]

    So the ratio is [itex] \frac{1}{\sqrt{2}} [/itex]

    Please correct me if I am wrong. Thank you.
     
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