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    (Original post by und)
    This didn't exactly inspire confidence:

    \displaystyle x=\left( \frac{\mu l}{2g(\mu +1)}+\frac{v}{2\sqrt{\frac{g}{l}  (\mu +1)}} \right) e^{\sqrt{\frac{g}{l}(\mu +1)}t} + \left( \frac{\mu l}{2g(\mu +1)}-\frac{v}{2\sqrt{\frac{g}{l}(\mu +1)}} \right) e^{-\sqrt{\frac{g}{l}(\mu +1)}t}-\frac{\mu l}{g(\mu +1)}
    :rofl:
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    (Original post by und)
    This didn't exactly inspire confidence:

    \displaystyle x=\left( \frac{\mu l}{2g(\mu +1)}+\frac{v}{2\sqrt{\frac{g}{l}  (\mu +1)}} \right) e^{\sqrt{\frac{g}{l}(\mu +1)}t} + \left( \frac{\mu l}{2g(\mu +1)}-\frac{v}{2\sqrt{\frac{g}{l}(\mu +1)}} \right) e^{-\sqrt{\frac{g}{l}(\mu +1)}t}-\frac{\mu l}{g(\mu +1)}
    That's similar to what I've got; not quite though.
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    Problem 54 *

    A triangle has sides of length at most 2, 3 and 4 respectively.
    Determine, with proof, the maximum possible area of the triangle.

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    Come on, it's a good question!
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    (Original post by shamika)
    Problem 54 *

    A triangle has sides of length at most 2, 3 and 4 respectively.
    Determine, with proof, the maximum possible area of the triangle.

    Spoiler:
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    Come on, it's a good question!
    It's the one question that I didn't even read when I was doing the paper. Someone I know who said he completed it got one mark. :lol:

    (Original post by DJMayes)
    That's similar to what I've got; not quite though.
    I was quite hoping you'd say it was completely wrong.
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    (Original post by around)
    In the Cambridge 4th year some people (tried) to run a Topos Theory reading group. I think in the end only about 5 or so people finished the course.

    I started it and then gave up after about 6-7 'lectures' because it was just utterly incomprehensible.
    Yeah it's on my list of "things to learn as I get older".
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    (Original post by DJMayes)
    Problem 50:

    An inextensible rope of length l and uniform mass per unit length lies on a rough table with one end on an edge. The co-efficient of friction between the table and the rope is μ. The rope receives an impulse which sets it moving off of the edge of the table at speed v. Prove that, if the rope does not fall off the table, then:

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

    By considering this as  \mu varies between zero and 1, find the maximum possible impulse that could potentially be given to the rope without it falling off of the table.
    Solution 50:

    Let the mass per unit length of the rope be  m . If the rope does not fall off the table, then the distance traveled by the rope off the table must be less than  l . Let this distance be  x .

    When the rope has just come to rest, a length x of the rope hangs off the table and length (l-x) is still on the table. Consider the case where friction here must be limiting in order to stop the rest of the rope from sliding off the table. For the section of the rope that's hanging, it's weight must be equal to the tension in the rope, which in turn is equal to the frictional force in the horizontal section of rope otherwise there would be an acceleration.

    \therefore F=T=\mu m(l-x)g = mxg \Rightarrow x= \dfrac{\mu l}{1+\mu } upon solving for  x .

    Now as the rope travels off the table, the frictional force slows it down from a speed v to  0 . As the frictional force is a function of x, then the work done by friction in stopping the rope, W, is:

    \displaystyle W= \int^x_0 F \ dx = \int^x_0 \mu m(l-x)g \ dx = \mu mg\int^x_0 (l-x) \ dx

    \displaystyle = \mu mg \left[ lx- \frac{x^2}{2} \right]_0^x = \mu mg\left( lx- \frac{x^2}{2} \right)

    The rope also falls, thus losing gravitational potential energy. By the principle of conservation of energy, the work done by friction is equal to sum of the loses in gravitational potential and kinetic energies:

    \displaystyle \Rightarrow \mu mg\left( lx- \frac{x^2}{2} \right) = \frac{1}{2} mlv^2  + \frac{mgx^2}{2}

    \Rightarrow 2\mu glx -\mu gx^2=lv^2 +gx^2

    \Rightarrow lv^2 = 2\mu lgx - gx^2(1+\mu )

    Substituting in the expression for  x that was derived earlier:

    lv^2 =  2\mu lg\dfrac{\mu l}{1+\mu } - g(1+\mu ) \left( \dfrac{\mu l}{1+\mu } \right)^2

    \Rightarrow lv^2 = \dfrac{2\mu ^2 l^2g - \mu ^2l^2g}{1+ \mu }

    \therefore v^2 = \dfrac{lg\mu ^2 }{1+\mu }

    This however gives the limiting value of the square of the velocity, since I assumed that conditions were limiting. Hence the range of values for which the rope does not fall must be given by

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

    As shown earlier, equality gives the limiting value of  v^2 and therefore of  v . Maximum impulse implies maximum change in velocity by the impulse-momentum principle, and so the maximum value for the impulse,  I_{max} , of the rope without it falling off the table will just be  mlv where  v is limiting:

    \displaystyle \therefore I_{max} = ml\mu \sqrt{ \frac{lg}{1+\mu } } .
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    (Original post by around)
    In the Cambridge 4th year some people (tried) to run a Topos Theory reading group. I think in the end only about 5 or so people finished the course.

    I started it and then gave up after about 6-7 'lectures' because it was just utterly incomprehensible.
    Does anyone understand it?
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    (Original post by Star-girl)

    Since the rope also falls, thus losing gravitational potential energy, the kinetic energy lost is less than the work done by friction:

    \displaystyle \therefore \frac{1}{2} mlv^2 - 0 < \mu mg\left( lx- \frac{x^2}{2} \right)

    \Rightarrow lv^2 < 2\mu lgx - \mu mx^2g

    I think putting in the GPE here (As KE plus GPE gained = Work Done) gets rid of the two and solves your co-efficient issue. I think you then need to sub this in to the other expression for x and work with possible values of x to get an inequality.
    I've put my comment up there in bold. Your method isn't the way I was using when I did it but it looks solid.
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    (Original post by shamika)
    Does anyone understand it?
    Peter Johnstone.
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    (Original post by DJMayes)
    I've put my comment up there in bold. Your method isn't the way I was using when I did it but it looks solid.
    Thanks - I thought of putting it in, but I didn't think it was necessary. I'll try that. :cute:
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    (Original post by ukdragon37)
    Peter Johnstone.
    Does he do undergraduate supervisions? I'm scared now.
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    (Original post by Star-girl)
    Thanks - I thought of putting it in, but I didn't think it was necessary. I'll try that. :cute:
    I'm making no promises it will work - I did it the same way I think und has approached it which means that I don't necessarily have the best bearing on what you're trying to do.
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    (Original post by DJMayes)
    I'm making no promises it will work - I did it the same way I think und has approached it which means that I don't necessarily have the best bearing on what you're trying to do.
    I don't know why my x is different to what you got but I'll leave it now anyway since it's been solved using a much nicer method.
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    (Original post by DJMayes)
    I'm making no promises it will work - I did it the same way I think und has approached it which means that I don't necessarily have the best bearing on what you're trying to do.
    I'll chug on and see if I can solve it.
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    (Original post by und)
    Does he do undergraduate supervisions? I'm scared now.
    I'm not actually sure, but John's is certain to have more than enough supervisors for undergrads to not have to bother him if they don't want to. :laugh:

    His Elephant, currently standing at 1278 pages over two volumes, is one of the few books I might own in my life that is able to serve as an excellent murder weapon (along with TAOCP).
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    (Original post by und)
    I don't know why my x is different to what you got but I'll leave it now anyway since it's been solved using a much nicer method.
    It's sign differences in a couple of places, but otherwise essentially identical.

    (And it hasn't been fully solved yet)
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    (Original post by DJMayes)
    ...
    Is 0 \leq \mu \leq 1 ?
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    (Original post by Star-girl)
    Is 0 \leq \mu \leq 1 ?
    That's the general convention in Mechanics questions I've seen although strictly speaking I don't think it's true.
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    (Original post by shamika)
    Problem 54 *

    A triangle has sides of length at most 2, 3 and 4 respectively.
    Determine, with proof, the maximum possible area of the triangle.

    Spoiler:
    Show
    Come on, it's a good question!
    From the formula for the area of a triangle:

     \frac{1}{2}absin\theta we can see the area will be maximised by maximising two sides and making the angle between them a right-angle for two known sides N. This is because sin theta is greatest when theta = 90 degrees when sintheta is one. Since the third side is not involved in the area formula, the area will be maximised by maximising the two shortest sides.

    If you sides of length 2 and 3... If you increase the side length of 3, then the other side can becomes at most 3. This means that you have this new side of length 3 and the old one of length 2. If you increase the one of length 2 the other side must become at most 2, hence you still have sides of at most length 3 and 2. If you increase both then you have sides lengths of at most 2, just over 3 and just over 3. But you can have side lengths of 2 and 3 and one larger so this still gives the most area as the area depends only on the two sides 2 and 3 and the angle between them. Therefore, the lengths of the shorter two sides are 2 and 3, so from above the angle between them is 90 degrees.

    0.5 x 3 x 2 = 3 units^2.
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    Can you guys help me find the stray minus?

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    Neck injury warning.
 
 
 
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