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    (Original post by Star-girl)
    I think you, bananarama and I (once I've corrected it) should all post our solutions as they seem to be different. :cute:



    See above.



    No worries.
    I would, but I still have a stray minus Care to assist?

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    Guys, can't you do the question by considering v^2=0 ie. by considering the discriminant of a quadratic in x. In fact I tried and I got something very similar but a factor of g^2 off.
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    (Original post by bananarama2)
    I would, but I still have a stray minus Care to assist?

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    In line 6 you have inexplicably multiplied by -1!

    Oops, I misread your limits!
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    OK guys, here was how I did the question before posting it:

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    Let the distance travelled be x and the mass of the rope be m. Then:

    Gravitational force =  \dfrac{mg}{l}x

    Frictional force =  \dfrac{\mu mg}{l}(l-x)

    Putting this together, we get a second order differential equation:

     mx’’ = \dfrac{mg}{l}x - \dfrac{\mu mg}{l}(l-x)

     x’’ - \dfrac{1+\mu}{l}gx  = -\mu g

    Solving this differential equation yields the complimentary function:

     x = Ae^{kt}+Be^{-kt} , where  k = \sqrt{ \dfrac{(1+\mu)g}{l}}

    We can guess our particular integral as some constant:

      - \dfrac{1+\mu}{l}gx  =  -\mu g

     x = \dfrac{l\mu }{1+\mu}

    Which gives us the general solution:

     x = Ae^{kt}+Be^{-kt}+\dfrac{l\mu }{1+\mu}

    We can then differentiate this to get:

    x’ = kAe^{kt}-kBe^{-kt}

    Next, let’s apply boundary conditions. Firstly,  t = 0, x= 0 :

     \Rightarrow A+B = -\dfrac{l\mu }{1+\mu}

    Secondly,  t = 0, x’=v :

     \Rightarrow A-B = \dfrac{v}{k}

    We can then solve these simultaneously to get:

     A =  \frac{1}{2}(\dfrac{v}{k} -\dfrac{l\mu }{1+\mu})

     B =  \frac{1}{2} (-\dfrac{l\mu }{1+\mu} -\dfrac{v}{k})

    We want conditions on whether the rope falls off the table. If it does not, then there must be a solution to the equation:

     x’ = 0

     kAe^{kt}-kBe^{-kt} = 0

     Ae^{2kt} = B

    We can see very easily that B is always going to be negative. As such, for solutions to exist A must be negative:

     \Rightarrow \frac{1}{2}(\dfrac{v}{k} -\dfrac{l\mu }{1+\mu}) < 0

     \dfrac{v}{k} < \dfrac{l\mu }{1+\mu}

     \dfrac{v^2}{k^2} < \dfrac{l^2\mu ^2}{(1+\mu)^2}

     v^2 < \dfrac{(1+\mu)g}{l} \dfrac{l^2\mu ^2}{(1+\mu)^2}

     v^2 < \dfrac{lg\mu^2}{1+\mu}

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    (Original post by DJMayes)
    OK guys, here was how I did the question before posting it:

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    Let the distance travelled be x and the mass of the rope be m. Then:

    Gravitational force =  \dfrac{mg}{l}x

    Frictional force =  \dfrac{\mu mg}{l}(l-x)

    Putting this together, we get a second order differential equation:

     mx’’ = \dfrac{mg}{l}x - \dfrac{\mu mg}{l}(l-x)

     x’’ - \dfrac{1+\mu}{l}gx  = -\mu g

    Solving this differential equation yields the complimentary function:

     x = Ae^{kt}+Be^{-kt} , where  k = \sqrt{ \dfrac{(1+\mu)g}{l}}

    We can guess our particular integral as some constant:

      - \dfrac{1+\mu}{l}gx  =  -\mu g

     x = \dfrac{l\mu }{1+\mu}

    Which gives us the general solution:

     x = Ae^{kt}+Be^{-kt}+\dfrac{l\mu }{1+\mu}

    We can then differentiate this to get:

    x’ = kAe^{kt}-kBe^{-kt}

    Next, let’s apply boundary conditions. Firstly,  t = 0, x= 0 :

     \Rightarrow A+B = -\dfrac{l\mu }{1+\mu}

    Secondly,  t = 0, x’=v :

     \Rightarrow A-B = \dfrac{v}{k}

    We can then solve these simultaneously to get:

     A =  \frac{1}{2}(\dfrac{v}{k} -\dfrac{l\mu }{1+\mu})

     B =  \frac{1}{2} (-\dfrac{l\mu }{1+\mu} -\dfrac{v}{k})

    We want conditions on whether the rope falls off the table. If it does not, then there must be a solution to the equation:

     x’ = 0

     kAe^{kt}-kBe^{-kt} = 0

     Ae^{2kt} = B

    We can see very easily that B is always going to be negative. As such, for solutions to exist A must be negative:

     \Rightarrow \frac{1}{2}(\dfrac{v}{k} -\dfrac{l\mu }{1+\mu}) < 0

     \dfrac{v}{k} < \dfrac{l\mu }{1+\mu}

     \dfrac{v^2}{k^2} < \dfrac{l^2\mu ^2}{(1+\mu)^2}

     v^2 < \dfrac{(1+\mu)g}{l} \dfrac{l^2\mu ^2}{(1+\mu)^2}

     v^2 < \dfrac{lg\mu^2}{1+\mu}

    You've got less than and I have equals... uh oh... :lol:
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    (Original post by Star-girl)
    You've got less than and I have equals... uh oh... :lol:
    The question has to by nature be an inequality, as there will be more than one value for v which is sufficiently slow to prevent it from falling.
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    (Original post by DJMayes)
    x
    Why is your PI negative?
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    Problem 55*

    The problem is concerned with constructing lines of length root  n , where  n is not a perfect square. Suppose you only have a ruler which can only measure integer lengths and assume you can draw perfect right-angled triangles. Show that unless  n is of the form \displaystyle 4k-2 , then you only ever need one right-angled triangle to construct a line of length root  n , e.g. To construct  \sqrt 2 , you create a  \displaystyle \left \{ 1, 1, \sqrt2 \right \} triangle, to construct  \sqrt3 you create a  \displaystyle \left \{ 1, \sqrt3, 2 \right \} triangle.
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    (Original post by DJMayes)
    The question has to by nature be an inequality, as there will be more than one value for v which is sufficiently slow to prevent it from falling.
    Ah yes, I suppose. So equality denotes the maximum value of v^2, then?
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    (Original post by und)
    Why is your PI negative?
    Could you elaborate? As far as I can see it isn't...
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    (Original post by bananarama2)
    I would, but I still have a stray minus Care to assist?

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    I will indeed, I just need to first finish all the modifications on the first part of my solution.
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    When I come to the student room Mr M always turns up, like :

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    Solution 50

    Using u as the initial velocity.

    Since the function of v is continuous, if v \to 0^{+} for some x, then at that instant the frictional force must be greater than the weight, thus the rope will remain stationery. It remains to find a range of values of  u for which v will become  0 .

    Creating and solving the differential equation, we obtain v^2=\frac{g(\mu +1)x^2}{l}-2\mu g x+u^2, so if v is to be  0 for some positive x, then there must exist a positive root x=\frac{\mu l \pm \sqrt{(\mu gl)^2-u^{2}gl(\mu +1)}}{g(\mu +1)}, so a necessary and sufficient condition (assuming all the constant terms are positive) is u^2 \leq \frac{{\mu}^{2}gl}{(\mu +1)} as required.

    If the rope has mass m, then the impulse applied to accelerate the rope to a speed u is given by the change of momentum mu. Hence \text{Impulse}^2 \leq  \frac{{m^{2}\mu}^{2}gl}{\mu +1}, giving the maximum impulse as m\mu \sqrt{\frac{gl}{\mu+1}}.
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    (Original post by Star-girl)
    Ah yes, I suppose. So equality denotes the maximum value of v^2, then?
    No, I don't think so. If you look at my DE equations solution, equality implies that B = 0, which means that there aren't solutions to x' = 0 as it implies A =0, which is impossible.
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    (Original post by DJMayes)
    Could you elaborate? As far as I can see it isn't...
    Sorry, I meant why is it positive? When you plug x into the DE with a positive complimentary function, the constant term has the wrong sign.
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    (Original post by DJMayes)
    No, I don't think so. If you look at my DE equations solution, equality implies that B = 0, which means that there aren't solutions to x' = 0 as it implies A =0, which is impossible.
    Hmmm... :holmes:
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    (Original post by und)
    Sorry, I meant why is it positive? When you plug x into the DE with a positive complimentary function, the constant term has the wrong sign.
    No it doesn't, because the co-efficient of x is negative.
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    (Original post by bananarama2)
    When I come to the student room Mr M always turns up, like :

    :rofl::rofl:

    Right - I'll have a look now at your solution. Could you check over mine (I've modified it now) as well as it seems to be some discrepancies with DJ's...
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    (Original post by DJMayes)
    No it doesn't, because the co-efficient of x is negative.
    Oh wait nvm... :lol:
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    (Original post by bananarama2)
    When I come to the student room Mr M always turns up, like :

    Why run when you can teleport?
 
 
 
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