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 Cart Rolling Down Incline

  1. Jan 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A cart is rolling down an incline (angle to horizontal is theta). It has two wheels, each wheel has mass m and moment of inertia mR^2/2. The total mass of the cart and the two wheels is M. Using conservation of energy (assuming no friction between cart and axles), show that the acceleration of the cart along the incline is a = [M/(M + 2m)] g sin theta.

    2. Relevant equations
    K = Iw^2/2
    K = mv^2/2
    v = wR (w is angular velocity).
    v^2 = 2as

    3. The attempt at a solution

    E0 = E1
    Ug = 2(K_rolling) + K_trans

    Letting d be the distance traveled on the incline, this gives:

    Mgd sin(theta) = (Iw^2/2)2 + Mv^2/2 = Iw^2 + Mv^2/2

    Using the given moment of inertia I = mR^2/2:

    Mgd sin(theta) = (mR^2/2)w^2 + Mv^2/2

    Mgd sin(theta) = mv^2/2 + Mv^2/2

    v^2 = 2Mgd sin(theta)/(M + m)

    v^2 = 2ad:

    a = Mg sin(theta)/(M + m)


    I can't figure out where they got the 2m from.

    I've tried adjusting the height by the radius of the wheels, but that gets complicated, because there are two of them, at two different heights, and I can't believe that would lead to the formula they give.

    When you all recover from New Years, I'd appreciate any suggestions :wink:

    (This is from Serway and Jewett, btw, 10.89)
    Last edited: Jan 1, 2007
  2. jcsd
  3. Jan 1, 2007 #2
    Consider motion starting from rest over distance [tex]x[/tex] along the incline:

    [tex]Mgx\sin\theta = \frac{1}{2}Mv^2 + 2(\frac{1}{2}mR^2)(\frac{v}{R})^2[/tex]

    [tex]2Mgx\sin\theta = (M+2m)v^2[/tex]

    On the other hand

    [tex]v^2 = 2ax[/tex]

    [tex]2Mgx\sin\theta = (M+2m)2ax[/tex]

    [tex]a = \frac{Mg}{M+2m}\sin\theta[/tex]
    Last edited: Jan 1, 2007
  4. Jan 1, 2007 #3
    Ok, but I don't understand how you got that rolling friction term.

    Why isn't it


    Isn't the KE of rolling friction

    [tex] (\frac{1}{2}){I\omega^2}[/tex]

    Why would it be different in this case?

  5. Jan 1, 2007 #4
    The condition states that the cart is rolling, which I suppose means that it's rolling without slipping. The condition for rolling without slipping is:

  6. Jan 1, 2007 #5
    Dorothy, I agree with your solution. Is there a possibility that the book is wrong?
  7. Jan 1, 2007 #6
    Hi Quirk... Yes, I understand that bit. What I don't understand is what happened to the 1/2 in Iw^2/2, or if you prefer, Iv^2/2R.
  8. Jan 1, 2007 #7
    Hi Hawire,

    That was my thought, yes, that the book was wrong. But I've thought that before (smile) and it wasn't. Just want to make sure I understand all of this.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook