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    Edexcel M5 June 2006, question 7.

    I have done part (a), but I am stuck on part (b).

    The mark scheme says:
    "For Q:
    -I = 2maw - mu
    I = 6maw - 2maw = 4maw
    ... = (m/3)\sqrt{8ag}"

    Name:  Screen Shot 2016-04-27 at 13.08.52.png
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Size:  114.0 KB

    I tried using "moment of the impulse = gain of A.M.", so Ia = 6ma^2w - ma(2ag)^{1/2} but this leads to I = 6maw - m(2ag)^{1/2}, which looks similar to the mark scheme, but implies that w = g/2a.

    Please help!
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    Student403 what's going on :lolwut:
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    (Original post by ombtom)
    Student403 what's going on :lolwut:
    M5 has been put on hiatus until summer :cry2: I spent too long on M4 rel motion and by the time I got over it and decided to move on, things kept coming up so I couldn't continue
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    (Original post by Student403)
    M5 has been put on hiatus until summer :cry2: I spent too long on M4 rel motion and by the time I got over it and decided to move on, things kept coming up so I couldn't continue
    Oh no My exam's in under 2 months. Who can I tag?
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    (Original post by ombtom)
    Oh no My exam's in under 2 months. Who can I tag?
    I think Duke Glacia might be able to help a little if he remembers some stuff
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    (Original post by ombtom)
    Oh no My exam's in under 2 months. Who can I tag?
    Omb fam i ll get back to u soz
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    (Original post by ombtom)
    Edexcel M5 June 2006, question 7.

    I have done part (a), but I am stuck on part (b).

    The mark scheme says:
    "For Q:
    -I = 2maw - mu
    I = 6maw - 2maw = 4maw
    ... = (m/3)\sqrt{8ag}"

    Name:  Screen Shot 2016-04-27 at 13.08.52.png
Views: 105
Size:  114.0 KB

    I tried using "moment of the impulse = gain of A.M.", so Ia = 6ma^2w - ma(2ag)^{1/2} but this leads to I = 6maw - m(2ag)^{1/2}, which looks similar to the mark scheme, but implies that w = g/2a.

    Please help!
    Your 6maw suggests that you are looking at the whole system. If you do, then the impulse cancels out. You must therefore just look at the change in (angular) momentum of Q and R. The impulse in the string acts upwards on Q, hence the minus sign in the MS. They had a term mu, which is the same as your second term. Looking at part a, a couple of lines from the bottom, you can see that u = 6aw. That's where their 6 came from, whereas yours came from considering the whole system.
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    (Original post by tiny hobbit)
    Your 6maw suggests that you are looking at the whole system. If you do, then the impulse cancels out. You must therefore just look at the change in (angular) momentum of Q and R. The impulse in the string acts upwards on Q, hence the minus sign in the MS. They had a term mu, which is the same as your second term. Looking at part a, a couple of lines from the bottom, you can see that u = 6aw. That's where their 6 came from, whereas yours came from considering the whole system.
    I'm looking at \sqrt{2ag} = u = 6aw in the mark scheme for part (a), but I thought that it was found by considering the whole system? (The previous line sums the A.M. of each part.)

    A.M. before = moment of linear momentum of falling particle = "mrv" = ma\sqrt{2ag}.

    A.M. after = "Iw" of Q and R after coalescing = (m+m)*a^2 * w= 2mwa^2.

    Still quite confused about this topic
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    (Original post by ombtom)
    I'm looking at \sqrt{2ag} = u = 6aw in the mark scheme for part (a), but I thought that it was found by considering the whole system? (The previous line sums the A.M. of each part.)

    A.M. before = moment of linear momentum of falling particle = "mrv" = ma\sqrt{2ag}.

    A.M. after = "Iw" of Q and R after coalescing = (m+m)*a^2 * w= 2mwa^2.

    Still quite confused about this topic
    These give the desired result:

    Let "I" be the impulse on Q,R, acting in the opposite direction to the A.M of Q and R; and so we have:

    "A.M. of Q" = "A.M. of R" - Ia.

    So,

    Ia = ma\sqrt{2ag}-2mwa^2

    and

    I = m\sqrt{2ag}-2mwa

    Which is the linear equation of impulse/momentum (and where I would have started)

     = m\sqrt{2ag}-2ma\frac{1}{3}\sqrt{\frac{g}{2a}  }

     = m\sqrt{2ag}-m\frac{1}{3}\sqrt{2ag}

     = m\frac{2}{3}\sqrt{2ag}

     = \frac{m}{3}\sqrt{8ag}

    Alternatively, one can consider the mass P and the pulley, the impulse acting on that, and working with angular momentum, we have:

    Ia=3ma^2w + ma^2w

    Terms on the right due to P and the pulley, respectively.

    So,
    I=4maw

    I=4ma\frac{1}{3}\sqrt{\frac{g}{2  a}}

     = \frac{m}{3}\sqrt{8ag}

    as before.
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    (Original post by ghostwalker)
    These give the desired result:

    Since "I" is acting in the opposite direction to the A.M of Q and R, we have:

    "A.M. of Q" = "A.M. of R" - Ia.

    So,

    Ia = ma\sqrt{2ag}-2mwa^2

    and

    I = m\sqrt{2ag}-2mwa

    Which is the linear equation of impulse/momentum (and where I would have started)

     = m\sqrt{2ag}-2ma\frac{1}{3}\sqrt{\frac{g}{2a}  }

     = m\sqrt{2ag}-m\frac{1}{3}\sqrt{2ag}

     = m\frac{2}{3}\sqrt{2ag}

     = \frac{m}{3}\sqrt{8ag}

    Alternatively, one can consider the mass P and the pulley, the impulse acting on that, and working with angular momentum, we have:

    Ia=3ma^2w + ma^2w

    Terms on the right due to P and the pulley, respectively.

    So,
    I=4maw

    I=4ma\frac{1}{3}\sqrt{\frac{g}{2  a}}

     = \frac{m}{3}\sqrt{8ag}

    as before.
    Perfect. Thank you
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    (Original post by ghostwalker)
    These give the desired result:

    Since "I" is acting in the opposite direction to the A.M of Q and R, we have:

    "A.M. of Q" = "A.M. of R" - Ia.

    So,

    Ia = ma\sqrt{2ag}-2mwa^2

    and

    I = m\sqrt{2ag}-2mwa
    I was about to post an excoriating criticism of the mathematical nonsense that you posted, but then I realised that your I and that of the post that you quoted are different (your I = impulse, previous I = MOI). This baffled me a lot.

    The only other vaguely useful thing that I have to say is that I looked at this late yesterday evening, drew a picture, and then spent 1/2 an hour wondering why I couldn't get it to work via conservation of linear momentum, as it's an isolated system. It was only this morning that I realised that I hadn't drawn anything holding the pulley in place...
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    (Original post by atsruser)
    I was about to post an excoriating criticism of the mathematical nonsense that you posted, but then I realised that your I and that of the post that you quoted are different (your I = impulse, previous I = MOI). This baffled me a lot.
    Thanks - learnt a new word.

    I probably should have defined it explicitly - done.
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    (Original post by atsruser)
    I was about to post an excoriating criticism of the mathematical nonsense that you posted, but then I realised that your I and that of the post that you quoted are different (your I = impulse, previous I = MOI). This baffled me a lot.
    We need more letters. :lol:
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    (Original post by ombtom)
    We need more letters. :lol:
    We need to find a new language to borrow them from. Maybe one of those nice swirly far Eastern scripts? Or Akkadian? They won't mind.
 
 
 
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