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    Questions and solutions from the book:
    https://gyazo.com/5ae429683a09828f267b585560a1b09e
    https://gyazo.com/7e102090cc62cd2a68b37bf658b1e96c
    My approach:
    Image attached.
    All I did was have the hollow cone at the top of the cylinder instead of at the bottom, and my centre of masses were adjusted accordingly due to their distance from its origin, but how come they answers are different?
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    (Original post by Nerrad)
    Questions and solutions from the book:
    https://gyazo.com/5ae429683a09828f267b585560a1b09e
    https://gyazo.com/7e102090cc62cd2a68b37bf658b1e96c
    My approach:
    Image attached.
    All I did was have the hollow cone at the top of the cylinder instead of at the bottom, and my centre of masses were adjusted accordingly due to their distance from its origin, but how come they answers are different?
    The centre of mass of your cone seems iffy; I'm not sure why you would measure it that way? We have to measure the distance of the centre of mass from AB, so it'd simply be \frac{h}{4}.

    Two other things that could help lay your work out easier:
    1. State from which point of the body you're taking moments from - this makes life a little easier for the examiner
    2. Start using mass ratios! The algebra becomes a lot simpler.
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    (Original post by Nerrad)
    Questions and solutions from the book:
    https://gyazo.com/5ae429683a09828f267b585560a1b09e
    https://gyazo.com/7e102090cc62cd2a68b37bf658b1e96c
    My approach:
    Image attached.
    All I did was have the hollow cone at the top of the cylinder instead of at the bottom, and my centre of masses were adjusted accordingly due to their distance from its origin, but how come they answers are different?
    They're not. Your \bar{x} is the distance of the centre of mass from the bottom of the cyclinder. They want the distance of the centre of mass from the top of the cylinder. Hence the "distance from AB" part. This is easily remedied, just do 3r - \text{your answer} as such:

    \displaystyle 

\begin{equation*}\bar{x} = 3r - \frac{216r^2 - 12hr + h^2}{4(36r - h)} = \cdots = \frac{216r^2 - h^2}{4(36r - h)}\end{equation*}

    I'll let you sort out the intermediary steps and get to the final answer, sorry for the late reply, really had to crane my neck to see your working and verify it's correctness.
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    (Original post by aymanzayedmannan)
    ...
    Unanswered for an entire hour. *answers at the same time*... fml.
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    (Original post by Zacken)
    Unanswered for an entire hour. *answers at the same time*... fml.
    Yours was prettier either way.


    (Original post by Zacken)
    really had to crane my neck to see your working and verify it's correctness.
    I just downloaded the picture and rotated it. :laugh:
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    (Original post by aymanzayedmannan)
    Yours was prettier either way.
    Nah, but oh well - it was a different method, so it's okay. :lol:

    I just downloaded the picture and rotated it. :laugh:
    Cheeky, downloading pictures off the interwebz, eh?
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    (Original post by Zacken)
    They're not. Your \bar{x} is the distance of the centre of mass from the bottom of the cyclinder. They want the distance of the centre of mass from the top of the cylinder. Hence the "distance from AB" part. This is easily remedied, just do 3r - \text{your answer} as such:

    \displaystyle 

\begin{equation*}\bar{x} = 3r - \frac{216r^2 - 12hr + h^2}{4(36r - h)} = \cdots = \frac{216r^2 - h}{4(36r - h)}\end{equation*}

    I'll let you sort out the intermediary steps and get to the final answer, sorry for the late reply, really had to crane my neck to see your working and verify it's correctness.
    So my centre of mass is relative to my own graph's x-axis whilst theirs is relative to their x-axis, so that explains the 3r- (my CoM)?Is that correct?
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    (Original post by aymanzayedmannan)
    The centre of mass of your cone seems iffy; I'm not sure why you would measure it that way? We have to measure the distance of the centre of mass from AB, so it'd simply be \frac{h}{4}.

    Two other things that could help lay your work out easier:
    1. State from which point of the body you're taking moments from - this makes life a little easier for the examiner
    2. Start using mass ratios! The algebra becomes a lot simpler.
    Yea but I don't know in these cases what moments means because usually in M2 moment has a distance from a pivot and its perpendicular force, but when it comes to centre of mass I fail to see how you can visualise the moments and its direction if that makes sense?
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    Can I ask what UMS did the both of you get when you did M3? Do you think it's possible to get 100 UMS?
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    (Original post by Nerrad)
    So my centre of mass is relative to my own graph's x-axis whilst theirs is relative to their x-axis, so that explains the 3r- (my CoM)?Is that correct?
    You measured all your distances from the bottom of the cylinder, hence your COM distance will be the distance from the bottom. You implicitly took moments about the bottom and solved for the perpendicular distance from the bottom.

    They want the distance from the top, so you need to subtract your answer from 3r to get the distance from the top.
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    (Original post by Zacken)
    You measured all your distances from the bottom of the cylinder, hence your COM distance will be the distance from the bottom. You implicitly took moments about the bottom and solved for the perpendicular distance from the bottom.

    They want the distance from the top, so you need to subtract your answer from 3r to get the distance from the top.
    Thank you.
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    (Original post by Zacken)
    You measured all your distances from the bottom of the cylinder, hence your COM distance will be the distance from the bottom. You implicitly took moments about the bottom and solved for the perpendicular distance from the bottom.

    They want the distance from the top, so you need to subtract your answer from 3r to get the distance from the top.
    How well did you do in your M3 exam before?
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    (Original post by Nerrad)
    How well did you do in your M3 exam before?
    Bad day: horribly.
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    (Original post by Nerrad)
    Can I ask what UMS did the both of you get when you did M3? Do you think it's possible to get 100 UMS?
    I'm sorry, I have not sat the exam yet - I'll be sitting it this June with Edexcel International (exam's only a couple of weeks away, :dolphin::dolphin::dolphin::dolphin:). But it's definitely possible to get 100/100 (at least I certainly hope so!) if you have a good day and don't make any errors. You should be looking at 75/75 though, because that's usually what full is.


    (Original post by Nerrad)
    Yea but I don't know in these cases what moments means because usually in M2 moment has a distance from a pivot and its perpendicular force, but when it comes to centre of mass I fail to see how you can visualise the moments and its direction if that makes sense?
    The sum of moments caused by the composite body is equal to the sum of moments caused by its constituents is the best I can put it, basically \bar{x}\sum m_{i} = \sum m_{i}x_{i}.
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    (Original post by Zacken)
    They're not. Your \bar{x} is the distance of the centre of mass from the bottom of the cyclinder. They want the distance of the centre of mass from the top of the cylinder. Hence the "distance from AB" part. This is easily remedied, just do 3r - \text{your answer} as such:

    \displaystyle 

\begin{equation*}\bar{x} = 3r - \frac{216r^2 - 12hr + h^2}{4(36r - h)} = \cdots = \frac{216r^2 - h}{4(36r - h)}\end{equation*}

    I'll let you sort out the intermediary steps and get to the final answer, sorry for the late reply, really had to crane my neck to see your working and verify it's correctness.
    I haven't worked through this, but the dimensions of your answer are wrong. The last term on the top should either involve rh or h^2 (a useful check in Mechanics questions).
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    (Original post by tiny hobbit)
    I haven't worked through this, but the dimensions of your answer are wrong. The last term on the top should either involve rh or h^2.
    Yep, that h at the end should be an h^2, it's a given answer so no harm caused hopefully. I'll correct it when I get onto my computer! Thanks.
 
 
 
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