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    (Original post by matt5822)
    7.58 isn't right for the last question because there is work done against friction
    I included work against friction in my solution and got 7.5m as my answer
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    Lol I posted this on the exam discussion thread too.
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    (Original post by TheLifelessRobot)
    Lol I posted this on the exam discussion thread too.
    Again... this question dosen't account for tension in the string...
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    Ohhh i see
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    (Original post by IrrationalRoot)
    Because if v<gl then the tension is negative which is impossible.
    But why would the energy method give you an incorrect answer? Surely tension could be 0, and the resultant centripetal force could be equal to the weight?
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    (Original post by aoxa)
    Yeah, but if you go off the basis that to make revolutions, v>0 this gives min u = root (4gl). This question has come up before, and the answer was definitely 2root(gl) then.


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    Nope. Alright I'll explain this because it seems half the people here don't understand it.

    This question has come up before. And the answer was definitely \sqrt{5ag}.

    Something similar to this has come up before, with a BEAD so that there is a REACTION force instead of tension. Now a reaction force can act upwards unlike tension, hence the difference.

    For the tension problem, what you are saying is that v>0 is a necessary and sufficient condition for circular motion. It is not. It is just a necessary condition. What is sufficient is that v \geq gl. This is because if v<gl, then the particle is going so slow that the centripetal force is so small that we would actually need the tension to go upwards for this to happen. But this is impossible.
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    (Original post by matt5822)
    7.58 isn't right for the last question because there is work done against friction
    Yeah you'd loose energy cause of that...
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    (Original post by matt5822)
    7.58 isn't right for the last question because there is work done against friction
    I didn't get this either and a lot of others seemed to, what value did you get?
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    That answer does include work done against friction, or at least I hope so, as that's what I got and I definitely included work done against friction!
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    (Original post by IrrationalRoot)
    Nope. Alright I'll explain this because it seems half the people here don't understand it.

    This question has come up before. And the answer was definitely \sqrt{5ag}.

    Something similar to this has come up before, with a BEAD so that there is a REACTION force instead of tension. Now a reaction force can act upwards unlike tension, hence the difference.

    For the tension problem, what you are saying is that v>0 is a necessary and sufficient condition for circular motion. It is not. It is just a necessary condition. What is sufficient is that v \geq gl. This is because if v<gl, then the particle is going so slow that the centripetal force is so small that we would actually need the tension to go upwards for this to happen. But this is impossible.
    In the bead questions the reaction force always acts towards the centre of the circle if I remember correctly.
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    (Original post by jjsnyder)
    Here are my answers from todays exam, not all of them will necessarily be correct so please correct me if you think they are wrong!
    1. a) 9.6
    b) i) 24.3
    ii) 12.7
    iii) No Air Resistance
    2.
    a)  a = (8-4x^3)i - 18e^{-3x}j
    b) i)  F = (16-8x^3)i - 36e^{-3x}j
    ii) 8.20
    c)  r = (4x^2- \frac{x^5}{5} + 3)i - (3+ 2e^{-3x})j
    3. a) By symmetry
    b)10.4 cm
    c) 16.5 degrees
    d) Explanation Q
    4. a)78.4
    b) 41.4 degrees
    c) r=2.89
    5. Lots of very long answers, will upload later if I have time. Part d) was  sqrt(5gl)
    6. a) Show that Q
    b)  \frac{g-(g-\lambda\mu)e^{-t\lambda} }{\lambda}
    7. a) Diagram
    b)  tanx = \frac{1-2\mu^2}{4\mu}
    8. I got 6.05m but many others got 7.58m so this seems like the correct answer, can anyone confirm this? There have been lots of proposed answers for this question.
    On 6b is there suppose to be a constant by e^-t
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    (Original post by steph-carys)
    I didn't get this either and a lot of others seemed to, what value did you get?
    7.58 because I didn't take that into account
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    Guys, its root(5lg).
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    2ii was 1.26 (cube root of 2), the question was "At what time does F act towards south?" so i component = 0, 2 = t^3, t = (2)^1/3
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    (Original post by matt5822)
    7.58 because I didn't take that into account
    Ahhh okay, cause I got something like 5.05m as I calculated that the mass moved around 1.05m down the slope
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    (Original post by MrMagic12345)
    It was sqr(5gl) not sqr(5gl/2) because it was 1/2mu^2=1/2mv2+2mgl
    I wrote that the minimum U occurs when initial ke =mg2l
    So u=2 root(gl)
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    (Original post by matt5822)
    7.58 because I didn't take that into account
    I took it into account and got 7.58 maybe u got lucky
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    How many marks do you reckon would be lost for using the method that gets you 2root(lg)? 3/4?
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    Can someone post their solution to question 8?
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    Had a go at both ways of working out u. In the exam I got u is 2root(ag)

    I think the problem is there is actually two conditions for complete revolutions, v>0 AND T>=0 as many others have already said, so you have to choose the most restricive.

    The energy method gives 2root(ag), but if you put this into the equation for centripetal force at the top of the circle, you have 4ag on the LHS and T(a/m) + 5ag on the RHS, which implies a negative tension,w hich isn't possible, unlike the bead on the ring question, where the reaction force from the ring can go either way. So as T>=0, u>=root(5ag)

    I got it wrong lol but that's the explanation as to why it's wrong
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