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A uniform solid cylindrical log begins rolling without slipping down a ramp that rise

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0 above the horizontal. After it has rolled 4.20 m along the ramp, the magnitude of its linear acceleration is closest to

    2. Relevant equations



    3. The attempt at a solution

    What I did was

    mgh=1/2mv^2+1/2Iω^2

    mgh=1/2mv^2+1/2[1/2mr^2]ω^2

    v=rω

    mgh=1/2mv^2+1/4mv^2
    √4/3gh=v

    solved for v and got 7.41m/s

    I then used the constant acceleration equations

    vx^2=vo+2aΔx

    solving for a and got 6.54 which is the incorrect answer.

    Where did I go wrong

    Thank you!
     
  2. jcsd
  3. Apr 15, 2012 #2
    Re: A uniform solid cylindrical log begins rolling without slipping down a ramp that

    Hi Mdhiggenz,

    The problem is in your first calculation, where you come up with 7.41m/s. Your manipulation of the algebra looks correct to me (double check!), but at the very last step you seem to have used h=4.20m. The problem says that the log moved 4.20m along the ramp, which means it would have descended 4.20m*sin(28deg), which is the value you should use for h. Does this fix your problem?

    Hope this helps,
    Bill Mills
     
    Last edited by a moderator: Apr 16, 2012
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