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Thread: ball rolling

  1. #1
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    ball rolling

    Iíve come across a physics problem and attempted to solve it, but Iíve forgotten a lot of it and havenít arrived at a solution. See the attached pic.

    ball rolling-1479402387036.png

    Can someone guide me in the right direction or solve it?

    Iíve tried

    sin 30ļ = h / PQ
    1/2 = 2 m / PQ
    PQ = 4 m

    Fg * cos 30ļ * 4m * μ = Fg * x * μ
    cos 30ļ * 4m = x
    sqrt(3)/2 * 4m = x
    x = 2 * sqrt(3)

    Anyhow, I fail to obtain μ and I feel like my approach is wrong and x is thus also wrong.
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  2. #2
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    Re: ball rolling

    Actually you are doing OK so far. The energy change of the system is equal to the decrease in PE as the ball descends from height h:

     \Delta PE = mg h

    The work done by friction comes in two parts. First is friction force on the incline of  \mu mg \cos \theta operating over the distance of the incline, which is  \frac h {\sin \theta}. The second component of work is the friction force along the horizontal section times the distance it slides until it stops, or  \mu mg x. So the full equation is:

     mgh =  \mu mg \cos \theta \frac h {\sin \theta} +  \mu mg x

    You already have the value for x, so substitute that value into this equation and solve for  \mu
    Last edited by ChipB; Nov 17th 2016 at 01:06 PM.
    Thanks from topsquark and metlx
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  3. #3
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    Re: ball rolling

    Thanks, it makes sense now.

    I’m just a little puzzled about x.
    If I try μ = 0.2 and x = 6.5 m (answer (a)), your equation works.
    It looks like I did something wrong when I tried to work out the horizontal section (x).
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  4. #4
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    Re: ball rolling

    Sure those values satisfy one of the equations, but do they satisfy the fact that work done by friction on the incline = the work done by friction on the horizontal section? There is nothing wrong with your answer for 'x'.
    Thanks from metlx
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  5. #5
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    Re: ball rolling

    If I use my ’x’ then the result for μ is:

    h = 2
    x = 2 * sqrt(3) ≈ 3.464
    --
    mgh = μgh cosΦ * (h/sinΦ) + μmgx
    2 = 2μ * (sqrt(3)/2) * 4 + μ * 2 * sqrt(3)
    μ = sqrt(3)/9 = 0.192

    I’m guessing x = 3.464 m and μ = 0.192 are good enough for me to choose answer (2) then - 3.5m and 0.2 friction coefficient.
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  6. #6
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    Re: ball rolling

    Quote Originally Posted by metlx View Post
    If I use my íxí then the result for μ is:

    h = 2
    x = 2 * sqrt(3) ≈ 3.464
    --
    mgh = μgh cosΦ * (h/sinΦ) + μmgx
    You have included 'h' twice in the first part of the right hand side, instead of 'm'. This should be:

     mgh = \mu mg \cos \theta  \left( \frac {h}{\sin \theta} \right) + \mu mgx

    Try it again from here.
    Thanks from metlx
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  7. #7
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    Re: ball rolling

    Ooops. I hope itís right this time.

     \mu = \frac{\sqrt{3}}{6} \approx 0.29

    (3) then.
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  8. #8
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    Re: ball rolling

    Quote Originally Posted by metlx View Post
     \mu = \frac{\sqrt{3}}{6} \approx 0.29

    (3) then.
    Bingo!
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