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M2 Work done question: where did I go wrong? Watch

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    (Original post by Zacken)
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    I meant what exam board
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    (Original post by Hjyu1)
    I meant what exam board
    Edexcel.
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    (Original post by Zacken)
    Like I said, your attachments aren't showing.
    Oh my bad

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    Attachment 526191526193526200



    And this one if you don't mind

    [attach]5.261915261935262e+23[/attach]

    [attach]5.261915261935262e+29[/attach]
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    (Original post by Zacken)
    Look at the diagram properly. The reaction force does come out diagonally/perpendicular to the cylinder, as you've rightly said; but in their diagram, they've drawn two arrows resolving that perpendicular/diagonal force horizontally and vertically. I've zoomed it up for you:



    As you can see, they've split the normal perpendicular force into the horizontal and the vertical component.
    But isn't friction acting between the rod and the cylinder supposed to be (imagine in they didn't split the normal force) acting perpendicular to the normal?
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    (Original post by creativebuzz)
    But isn't friction acting between the rod and the cylinder supposed to be (imagine in they didn't split the normal force) acting perpendicular to the normal?
    Yes, but they have also split the friction into horizontal and vertical, look at the first diagram of the model answers.
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    (Original post by Zacken)
    Yes, but they have also split the friction into horizontal and vertical, look at the first diagram of the model answers.
    Have they made friction act in the direction which acts "into" the cylinder? (I find their diagram confusing, hence I'm here :/ )
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    (Original post by creativebuzz)
    Have they made friction act in the direction which acts "into" the cylinder? (I find their diagram confusing, hence I'm here :/ )
    The rod wants to slip off the cylinder in the obvious intuitive direction/way. Hence friction will oppose this and will act parallel to the rod (perpendicular to the normal, as you said up the rod.) When you resolve the parallel force up the rod, the force acts vertically upwards and horizontally rightwards (or into the cylinder). Sketch it for yourself and see.
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    (Original post by ghostwalker)
    Why not do it that way then?



    You have the MR of the triangles incorrect - should be half what you've got.

    Not at home, so can't post again for a bit.
    For part B, why is it wrong to just find the centre of mass of the whole thing?
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    (Original post by creativebuzz)
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    As for the (d) of the projectiles question, the speed is the square root of the sum of the squares of the the horizontal and vertical speed. Your horizontal speed seems spot on! But your vertical speed is totally off.

    So we have (resolving horizontally) a constant horizontal speed of \sqrt{27ag}\cos \theta and a vertical speed (here we need to use SUVAT in the vertical direction:

    v^2 = u^2 + 2as = 27ag\sin^2 \theta - 2g \times 6a = 27ag\sin^2\theta - 12ag.

    Now we know that \cos \theta = \frac{1}{\sqrt{26}} so sketching a little triangle tells us that \sin \theta = \frac{5}{\sqrt{26}}

    Hence the resultant speed is \displaystyle \sqrt{27ag \left(\frac{5}{\sqrt{26}} \right)^2  - 12ag + 27ag\frac{1}{26}}
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    (Original post by Zacken)
    The rod wants to slip off the cylinder in the obvious intuitive direction/way. Hence friction will oppose this and will act parallel to the rod (perpendicular to the normal, as you said up the rod.) When you resolve the parallel force up the rod, the force acts vertically upwards and horizontally rightwards (or into the cylinder). Sketch it for yourself and see.
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    Oh I see! I managed to get up to taking moments about A but I was unsure about how to take moments of (root3/3)(R) because it acts up and along the rod
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    (Original post by creativebuzz)
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    Oh I see! I managed to get up to taking moments about A but I was unsure about how to take moments of (root3/3)(R) because it acts up and along the rod
    Sorry, I'm crap at this moments stuff! I'll let ghostwalker answer this one. :lol:
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    (Original post by Zacken)
    Sorry, I'm crap at this moments stuff! I'll let ghostwalker answer this one. :lol:
    Haha, that's cool! As you can tell this M2 reserved paper really did kill me :ashamed2:
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    (Original post by Zacken)
    As for the (d) of the projectiles question, the speed is the square root of the sum of the squares of the the horizontal and vertical speed. Your horizontal speed seems spot on! But your vertical speed is totally off.

    So we have (resolving horizontally) a constant horizontal speed of \sqrt{27ag}\cos \theta and a vertical speed (here we need to use SUVAT in the vertical direction:

    v^2 = u^2 + 2as = 27ag\sin^2 \theta - 2g \times 6a = 27ag\sin^2\theta - 12ag.

    Now we know that \cos \theta = \frac{1}{\sqrt{26}} so sketching a little triangle tells us that \sin \theta = \frac{5}{\sqrt{26}}

    Hence the resultant speed is \displaystyle \sqrt{27ag \left(\frac{5}{\sqrt{26}} \right)^2  - 12ag + 27ag\frac{1}{26}}
    Ah yes, I've got the right answer now! Thanks Did you figure out the q8 part c I posted (it was in the same post as this speed question)
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    (Original post by creativebuzz)
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    Oh I see! I managed to get up to taking moments about A but I was unsure about how to take moments of (root3/3)(R) because it acts up and along the rod
    If the line of action of the force acts along the rod, what's the perpendicular distance of that line of action from A?

    (not checked details of your working).
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    (Original post by ghostwalker)
    If the line of action of the force acts along the rod, what's the perpendicular distance of that line of action from A?

    (not checked details of your working).
    Is it the perpendicular distance the yellow line?
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    (Original post by creativebuzz)
    Is it the perpendicular distance the yellow line?
    Here's some examples. In each case the perpendicular distance is from the origin to the line of action (red), and is repressented by the line in green. You can see (last diagram) that when our line of action goes through the origin, the perpendicular distance is zero, and so the moment of that force is zero.

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    (Original post by ghostwalker)
    Here's some examples. In each case the perpendicular distance is from the origin to the line of action (red), and is repressented by the line in green. You can see (last diagram) that when our line of action goes through the origin, the perpendicular distance is zero, and so the moment of that force is zero.

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    I've got the answer now, thanks

    Would you mind giving me hand on part B? I managed to get part a correct ?(3cm)

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    (Original post by Zacken)
    Sorry, I'm crap at this moments stuff! I'll let ghostwalker answer this one. :lol:
    Sorry to be such a bother but would you mind giving me a hand on the Q8c I posted in the 23rd message?
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    (Original post by creativebuzz)
    Sorry to be such a bother but would you mind giving me a hand on the Q8c I posted in the 23rd message?
    I agree with you that y = -\frac{7}{c}, this is your velocity when P is at Q (it's v). Your initial velocity is 7c. (it's u)

    Now, using v = u + at we have: -\frac{7}{c} = 7c - gt.

    Now solve for t in terms of c and plug this value of t into your equation for horizontal displacement, which is x = 7t. (horizontal speed * time)
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    (Original post by creativebuzz)
    I've got the answer now, thanks

    Would you mind giving me hand on part B? I managed to get part a correct ?(3cm)

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    Knowing the angle with the vertical, you should be able to find the vertical position of G, the CofM of L.

    Then consider the centre of mass of, L, the isosceles triangle, and the square you should be able to work out the latter.

    triangle = L plus square, so to speak.
 
 
 
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