M2 centre of mass Watch

vinsta
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#1
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http://www.thestudentroom.co.uk/show...06#post4182001
The paper is on this thread
I'm stuck on question 5b)
I don't know what to do.
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thenewromance1234
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M2 centre of mass always messes me up :hmmm:
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vinsta
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(Original post by Clements-)
M2 centre of mass always messes me up :hmmm:
I just hate the ones with a uniform lamina and right angled triangles :sigh:
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thenewromance1234
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(Original post by vinsta)
I just hate the ones with a uniform lamina and right angled triangles :sigh:
I don't mind those. It's ones that are made up of bars I can never do.
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vinsta
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(Original post by Clements-)
I don't mind those. It's ones that are made up of bars I can never do.
Bars, which ones are they?
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Unbounded
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a standard rule:  \bar{x} \displaystyle\sum_{i=1}^{n} m_i = \sum_{i=1}^{n} m_ix_i

for point masses, this turns into:

 \displaystyle \bar{x} = \frac{m_1x_1 + m_2x_2 + m_3x_3 \cdots m_nx_n}{m_1 + m_2 + m_3 \cdots m_n}

you can do a similar thing for the y coordinates...

Spoiler:
Show
or you can take a column vector approach and do them both together:

\displaystyle m_1\binom{x_1}{y_1} + m_2\binom{x_2}{y_2} +\cdots + m_n\binom{x_n}{y_n} =  \binom{\bar{x}}{\bar{y}} \displaystyle\sum_{i=1}^{n} m_i
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3thr3e
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"[Unparseable or potentially dangerous latex formula. Error 6 ]"

^ lol, dangerous maths.
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Unbounded
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(Original post by 3thr3e)
"[Unparseable or potentially dangerous latex formula. Error 6 ]"

^ lol, dangerous maths.
just missed out a bracket in the latex :p:
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vinsta
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(Original post by GHOSH-5)
a standard rule:  \bar{x} \displaystyle\sum_{i=1}^{n} m_i = \sum_{i=1}^{n} m_ix_i

for point masses, this turns into:

 \displaystyle \bar{x} = \frac{m_1x_1 + m_2x_2 + m_3x_3 \cdots m_nx_n}{m_1 + m_2 + m_3 \cdots m_n}

you can do a similar thing for the y coordinates...

Spoiler:
Show
or you can take a column vector approach and do them both together:

\displaystyle m_1\binom{x_1}{y_1} + m_2\binom{x_2}{y_2} +\cdots + m_n\binom{x_n}{y_n} =  \binom{\bar{x}}{\bar{y}} \displaystyle\sum_{i=1}^{n} m_i
In the markscheme it says x bar is 4, without any working out, so are you supposed to see that?
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Unbounded
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(Original post by vinsta)
In the markscheme it says x bar is 4, without any working out, so are you supposed to see that?
it's just a bit of mental arithmetic in a sense. but it's best to put all the working. from what i've seen, markschemes have simplied working for the less challenging questions (usually). and examiners (who are the only ones meant to see this), would know how to solve the question, but just to check the answer themselves, the markscheme has it.

(always put down all your working, regardless of how little may be in the markscheme)
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vinsta
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(Original post by GHOSH-5)
it's just a bit of mental arithmetic in a sense. but it's best to put all the working. from what i've seen, markschemes have simplied working for the less challenging questions (usually). and examiners (who are the only ones meant to see this), would know how to solve the question, but just to check the answer themselves, the markscheme has it.

(always put down all your working, regardless of how little may be in the markscheme)
Ok Ghosh thanks.
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thenewromance1234
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(Original post by vinsta)
Bars, which ones are they?
Or rods, i just did a question from June 06, q.4 where a frame was made up of 4 bars, and weights were added to two corners.
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Unbounded
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(Original post by vinsta)
Ok Ghosh thanks.
just realised you wanted help with part b :>.<:

you need to find the centre of mass of the uniform triangle. this can be found by taking the average of the coordinates of the vertices:

uniform triangle ABC has coordinates A = (x_a,y_a), B = (x_b,y_b), C = (x_c,y_c)

\therefore \displaystyle (\bar{x}, \bar{y}) = \left(\frac{x_a+x_b+x_c}{3}, \frac{y_a + y_b + y_c}{3}\right)

then treat the centre of mass of the triangle as a point mass of km at  (\bar{x}, \bar{y}) and use the same prinicple in part A, with finding the centre of mass of a system of point masses, using:

 \displaystyle \bar{x} = \frac{m_1x_1 + m_2x_2 + m_3x_3 \cdots m_nx_n}{m_1 + m_2 + m_3 \cdots m_n}

edit: also in the markscheme, for Q5, i saw none of it just having '4' as an answer on its own - it all had working - might take a while to load - or just run over the blank parts of it with your cursor to make it show up.
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vinsta
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(Original post by GHOSH-5)
just realised you wanted help with part b :>.<:

you need to find the centre of mass of the uniform triangle. this can be found by taking the average of the coordinates of the vertices:

uniform triangle ABC has coordinates A = (x_a,y_a), B = (x_b,y_b), C = (x_c,y_c)

\therefore \displaystyle (\bar{x}, \bar{y}) = (\frac{x_a+x_b+x_c}{3}, \frac{y_a + y_b + y_c}{3})

then treat the centre of mass of the triangle as a point mass of km at  (\bar{x}, \bar{y}) and use the same prinicple in part A, with finding the centre of mass of a system of point masses, using:

 \displaystyle \bar{x} = \frac{m_1x_1 + m_2x_2 + m_3x_3 \cdots m_nx_n}{m_1 + m_2 + m_3 \cdots m_n}

edit: also in the markscheme, for Q5, i saw none of it just having '4' as an answer on its own - it all had working - might take a while to load - or just run over the blank parts of it with your cursor to make it show up.
I did that and got 3 as xbar but the answer is 4.5.
I did 0+9+0/3=3
What have I done wrong?
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vinsta
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(Original post by GHOSH-5)
just realised you wanted help with part b :>.<:

you need to find the centre of mass of the uniform triangle. this can be found by taking the average of the coordinates of the vertices:

uniform triangle ABC has coordinates A = (x_a,y_a), B = (x_b,y_b), C = (x_c,y_c)

\therefore \displaystyle (\bar{x}, \bar{y}) = (\frac{x_a+x_b+x_c}{3}, \frac{y_a + y_b + y_c}{3})

then treat the centre of mass of the triangle as a point mass of km at  (\bar{x}, \bar{y}) and use the same prinicple in part A, with finding the centre of mass of a system of point masses, using:

 \displaystyle \bar{x} = \frac{m_1x_1 + m_2x_2 + m_3x_3 \cdots m_nx_n}{m_1 + m_2 + m_3 \cdots m_n}

edit: also in the markscheme, for Q5, i saw none of it just having '4' as an answer on its own - it all had working - might take a while to load - or just run over the blank parts of it with your cursor to make it show up.
Well not mark scheme, but examiners report it states "by intuition" some found x:
http://www.edexcel.com/migrationdocu...p_20060125.pdf
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Unbounded
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(Original post by vinsta)
I did that and got 3 as xbar but the answer is 4.5.
I did 0+9+0/3=3
What have I done wrong?
you forgot to multiply the coordinates by their respective masses.
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vinsta
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(Original post by Clements-)
Or rods, i just did a question from June 06, q.4 where a frame was made up of 4 bars, and weights were added to two corners.
I don't like it when they change the situation half way through.
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thenewromance1234
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(Original post by vinsta)
I don't like it when they change the situation half way through.
I prefer that cos if it follows on from the first bit. If I can't do the first bit I'm usually then screwed for the second bit.
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example.

3 particles A,B and C have masses of 1, 2 and 3kg respectively. they have coordinates (1,2), (1,1) and (0,3) respectively. find their centre of mass:

X-coordinate
 \displaystyle \bar{x} = \frac{m_1x_1 + m_2x_2 + m_3x_3 +\cdots +m_nx_n}{m_1 + m_2 + m_3 +\cdots +m_n}

 \bar{x} = \frac{1\times 1 + 2 \times 1 + 3\times 0}{1+2+3} = \frac{1+2+0}{6} = \frac{1}{2}

Y-coordinate
 \displaystyle \bar{y} = \frac{m_1y_1 + m_2y_2 + m_3y_3 +\cdots +m_ny_n}{m_1 + m_2 + m_3 +\cdots +m_n}

\bar{y} = \frac{1\times 2 + 2 \times 1 + 3\times 3}{1+2+3} = \frac{2+2+9}{6} = \frac{13}{6}

 \therefore (\bar{x},\bar{y}) = (\frac{1}{2},\frac{13}{6})

now try it with your question.
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thenewromance1234
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See, it's those questions I can do but they never seem to come up in the paper :puppyeyes: It's always some awkward rod question.
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