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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:

    $\displaystyle \Delta PE = mg h$

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

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

    You already have the value for x, so substitute that value into this equation and solve for $\displaystyle \mu$
    Last edited by ChipB; Nov 17th 2016 at 12:06 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:

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

    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.

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

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

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

    (3) then.
    Bingo!
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