1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)

    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)
    gh = ad
    Therefore, a = gh/d
    And height/distance is equal to sin(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


    User Avatar
    Staff Emeritus
    Science Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook