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Acceleration of a solid cylinder on an incline

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the acceleration of a solid cylinder rolling down an incline that makes an angle (theta) with the horizontal is given by the equation:
    a=2/3 g sin(theta)


    2. Relevant equations
    Using the conservation of mechanical energy:
    1/2 Mv2 + 1/2 I w2 + Mgh = 1/2Mv2 + 1/2 Iw2 + Mgh
    ***Note: w = angular velocity, I'm just not sure how to insert the symbol
    v2 = v20 + 2a(x-x0)
    I=1/2MR2

    3. The attempt at a solution
    After simplifying the conservation of mechanical energy equation and substituting in the equation for moment of inertia, I got:
    1/4v2 + gh = 1/2v2 + 1/4v2

    In order to solve for acceleration, I substituted in the equation v2=2a(d) ... where d is equal to distance down the incline:
    1/4(2ad) + gh = 1/2(2ad) + 1/4(2ad)
    Simplified:
    gh = ad
    Therefore, a = gh/d
    And height/distance is equal to sin(theta):
    a=gsin(theta)

    The problem is that I don't know where the 2/3 comes in ...

    Thankyou in advance for your help :)
     
  2. jcsd
  3. Sep 13, 2009 #2
    Hello Chelseaalyssa
    Nice problem you have there. I see you found that if you flip the cilinder to the side the acceleration would fully go to sliding. The 2/3 factor is one to prove. Part of the potential energy comes into rolling the cilinder. I have tried to find if the moment of inertia is indeed
    I=MR^2/2. Have not found the page(s) in my mechanic book where this very nice way of using integrals is directed to the most common forms... Suppose that this is correct:
    The most important thing is to connect the rolling to the moving; so the omega w and the velocity v.
    Succes and greetings from Janm
     
  4. Sep 13, 2009 #3

    Astronuc

    User Avatar

    Staff: Mentor

    In this equation, 1/4v2 + gh = 1/2v2 + 1/4v2, the v on the left side is not the same as the v on the right side.

    If v on the left is initially zero, i.e. the cylinder is at rest, then the equation becomes

    gh = 1/2v2 + 1/4v2

    and then using v2=2a(d)

    gh = 1/2(2ad) + 1/4(2ad).
     
  5. Sep 14, 2009 #4
    Oh ok - I see where I went wrong...
    Thanks for the help!
     
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