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ganeshie8

  • one year ago

Find the center of mass of the following shape w/o using integrals

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  1. ganeshie8
    • one year ago
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    |dw:1443928609926:dw|

  2. imqwerty
    • one year ago
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    the area of bigger circle=4pi(R)^2 the area of smaller circle=pi(R)^2 the cordinates of centre of mass of bigger circle(x1,y1) =0,0 taking its centre to be origin the cordinates of COM of small cricle(x2,y2)=(-R,0) \[X_{CM}=\frac{ A_{1}\times x_{1} -A_{2} \times x_{2} }{A_{1}-A_{2} }\] \[X_{CM}=\frac{ 4\pi R^2 \times 0 -\pi R^2 \times -R }{ 4\pi R^2-\pi R^2 }\]\[X_{CM}=\frac{ \pi R^3 }{ 3\pi R^2 }\]\[X_{CM}=\frac{ R }{ 3 }\] since both y1 and y2 =0 Ycm=0 so cordinates of centre of mass=(R/3,0)

  3. ganeshie8
    • one year ago
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    Awesome! so does that trick always work ? is below statement true in general ? `center of mass of a composite shape = center of mass of center of masses of simpler parts`

  4. imqwerty
    • one year ago
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    yes :)

  5. UnkleRhaukus
    • one year ago
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    If instead, the origin is taken as the left most point on the circles \[\begin{align} CM_x &= \frac{r_0A_0+r_1A_1}{A_0+A_1}\\ &= \frac{R(-\pi R^2)+2R(\pi(2R)^2)}{-\pi R^2+\pi(2R)^2}\\ &= \frac{\pi R^3(-1+8)}{\pi R^2(-1+4)}\\ &= \tfrac73R \end{align}\]

  6. ganeshie8
    • one year ago
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    Ahh that looks neat! If I am interpreting it correctly, it basically uses the same trick : `center of mass of a composite shape = center of mass of center of masses of simpler parts` If I have some weird shape, I am allowed to break it into multiple simpler pieces, work center of mass of each shape and finally work the center of mass of the center of masses... wonder if it is easy to prove above method..

  7. ganeshie8
    • one year ago
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    Just noticed, you're representing the empty region by "negative area" and seems its working perfectly ! xD

  8. imqwerty
    • one year ago
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    we can also think like this- suppose that circle is made of cardboard we put a circle of radius R at the place where there is a gap now we have a complete circle but we put another circle of radius R on top of that place so that the system be the same :)

  9. ganeshie8
    • one year ago
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    Right! 1) center of mass of smaller circular disc = \((-R,0)\) 2) Assume center of mass of circular disc with hole = \((x, 0)\) 3) center of mass full bigger circular disc = \((0,0) = \left( \dfrac{-R(\pi R^2)+x(4\pi R^2-\pi R^2)}{4\pi R^2}, 0\right)\) 4) solving does give me \(x = \dfrac{R}{3}\)

  10. imqwerty
    • one year ago
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    yes correct :)

  11. ganeshie8
    • one year ago
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    do you have a proof for above method ? could we attempt to prove it using the definition of center of mass : \[\text{center of mass}{} = \dfrac{1}{M}\sum\limits_{i} m_i\vec{r_i}\] where \(M=m_1+m_2+\cdots\)

  12. imqwerty
    • one year ago
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    ok lets say that a force F is acting on a system with masses m1, m2 , m3....mn inside it :) \[F=m_{1}a_{1}+m_{2}a_{2}......m_{n}a_{n}\] we can write this as- \[F=\frac{ d^2 }{dt }[m_{1}r_{1}+m_{2}r_{2}....m_{n}r_{n}]\]\[F=(m_{1}+m_{2}..m_{n})\frac{ d^2 }{ dt }\frac{ [m_{1}r_{1}+m_{2}r_{2}...m_{n}r_{n}] }{ [m_{1}+m_{2}..+m_{n}]}\] now F=m(a) we can see the total mass on LHS of eq and the rest part under the differentiation is the acceleration of system the acceleration of COM =acceleration of system so \[F=M \frac{ d^2 }{ dt }R_{CM}\] hence we can conclude that \[R_{CM}=\frac{ 1 }{ M }\sum_{}^{}m_{1}r_{1}\]

  13. UnkleRhaukus
    • one year ago
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    \[m=\varrho A\]

  14. ganeshie8
    • one year ago
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    Nice! I think that shows we can treat center of mass as a point where all the external forces on a system act. How does that prove that we can break a continuous solid shape into multiple pieces to find the center of mass ?

  15. imqwerty
    • one year ago
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    yes, we can say that but we should say that the centre of mass is the point whose acceleration is equal to the acceleration of system :) i donno the proper reason y we say this but our proff. said this :)

  16. ganeshie8
    • one year ago
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    Nice! I think that shows we can treat center of mass as a point where all the external forces on a system act; Also the mass of center of mass equals the sum of masses of all the masses in the system. How does that prove that we can break a continuous solid shape into multiple pieces to find the center of mass ?

  17. IrishBoy123
    • one year ago
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    you are using integrals, just easy ones that don't need doing

  18. ganeshie8
    • one year ago
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    Maybe... but im not using calculus directly, im using intuition... but im looking for a proof too.. :)

  19. ganeshie8
    • one year ago
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    Suppose I want to find the center of mass of below weird shape. |dw:1443948450651:dw|

  20. ganeshie8
    • one year ago
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    How to prove that below method works ? 1) break that shape into 2 pieces : circle, segment; 2) find the center of mass of each piece; 3) then finally, find the center of mass of center of masses of individual pieces to get the center of mass of the composite shape

  21. IrishBoy123
    • one year ago
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    yes, sir, that works but imagine there was a density function \(\sigma (x,y)\) that you had to factor in. everything we know from the vanilla wiki page on centroids would be redundant. we'd have to re-cast \(\sum m_i \delta r_i \) as \(\sum m_i \delta r_i \sigma_i\). all bets would be off. perhaps i am splitting hairs but does the fact that \(\sigma = 1\) mean that calculus no longer applies, or does it mean that we just use simpler integration? dunno.... if you really want to do it without calculus, you need to use a plumbline :-)

  22. IrishBoy123
    • one year ago
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    and the proof would follow i think from the \(\sum m_i r_i \) idea which i suppose maths bods could do just using summation notation....

  23. IrishBoy123
    • one year ago
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    make that \[\sum \delta a_i \delta r_i \sigma_i\]

  24. ParthKohli
    • one year ago
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    This shape is simply a superposition of: - a disk of mass \(m\) and radius \(2R\) with mass-density \(m/\pi(2R)^2\) - a disk of mass \(-m/4\) and radius \(R\) with the same mass-density. Thus, subtracting the two centers of mass can be viewed as adding the center of mass of a negative mass.

  25. ganeshie8
    • one year ago
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    Hmm I don't mind using integrals @IrishBoy123 , its just they are not covered at 11th grade level yet. The textbook still uses summation notation and deliberately avoids using the integrals. Do you have a proof for below statement ? `center of mass of a composite shape = center of mass of center of masses of simpler parts` and yes, since it is a 2D object, we can assume that the density is uniform i guess

  26. ganeshie8
    • one year ago
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    I think using plumbline is not allowed because it is more of experimental method...

  27. ganeshie8
    • one year ago
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    Also its not so exciting because we can always find the center of mass of any object using plumblines

  28. ParthKohli
    • one year ago
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    \[\vec{\rm CM } = \frac{\sum_{i=1}^{N} m_i \vec{r_i} }{M}\]If we divide a body into two parts\[\vec{\rm CM}_1 = \frac{\sum_{i=1}^n m_i \vec{r_i}}{M_1 }\]\[\vec{\rm CM}_2 = \frac{\sum_{i=n+1}^{N} m_i \vec{r_i}}{M_2 }\]

  29. ganeshie8
    • one year ago
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    Exactly! expecting this kindof proof... please keep going...

  30. ParthKohli
    • one year ago
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    \[\vec{\rm CM'} = \frac{M_1 \vec{\rm CM_1} + M_2 \vec{CM_2}}{M_1 + M_2} = \frac{\sum_{i = 1}^{n} m_i \vec{r_i} + \sum_{i=n+1}^{N}m_i \vec{r_i }}{M} = \frac{\sum_{i=1}^N m_i \vec{r_i }}{M}\]

  31. ParthKohli
    • one year ago
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    \[\Rightarrow \vec{\rm CM'} = \vec{\rm CM} \]

  32. ganeshie8
    • one year ago
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    wot the frog

  33. ganeshie8
    • one year ago
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    that looks really nice!

  34. IrishBoy123
    • one year ago
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    yeah, that's good!

  35. ganeshie8
    • one year ago
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    Oh one sec.. what do we get after below step ? \[\vec{\rm CM} = \frac{\sum_{i=1}^N m_i \vec{r_i }}{M}\\~\\= \frac{\sum_{i = 1}^{n} m_i \vec{r_i} + \sum_{i=n+1}^{N}m_i \vec{r_i }}{M} \\~\\= \frac{M_1 \vec{\rm CM_1} + M_2 \vec{CM_2}}{M_1 + M_2} \\~\\= ?\]

  36. ParthKohli
    • one year ago
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    That shows us that we can treat the simpler parts of the composite body as point-masses situated at the center of masses of the simpler parts, then find the center of mass of those point masses in order to find the center of mass of the composite body.

  37. ganeshie8
    • one year ago
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    Ahh nice! nice! I sew it now!

  38. ganeshie8
    • one year ago
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    \[\vec{\rm CM} = \frac{\sum_{i=1}^N m_i \vec{r_i }}{M}\\~\\= \frac{\sum_{i = 1}^{n} m_i \vec{r_i} + \sum_{i=n+1}^{N}m_i \vec{r_i }}{M} \\~\\= \frac{M_1 \vec{\rm CM_1} + M_2 \vec{CM_2}}{M_1 + M_2} \\~\\= CM(CM_1,CM_2)\]

  39. ParthKohli
    • one year ago
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    o.o

  40. ganeshie8
    • one year ago
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    \[\vec{\rm CM} = \frac{\sum_{i=1}^N m_i \vec{r_i }}{M}\\~\\= \frac{\sum_{i = 1}^{n} m_i \vec{r_i} + \sum_{i=n+1}^{N}m_i \vec{r_i }}{M} \\~\\= \frac{M_1 \vec{\rm CM_1} + M_2 \vec{CM_2}}{M_1 + M_2} \\~\\= \text{center of mass of CM1 and CM2}\]

  41. ganeshie8
    • one year ago
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    when you see the relationship for the first time, this is really a beautiful result!

  42. ParthKohli
    • one year ago
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    We get that\[\rm \vec{CM} = CM( CM_1, CM_2, \cdots, CM_n) \]

  43. ParthKohli
    • one year ago
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    I just proved it for \(n=2\) but you can see what's happening here.

  44. ganeshie8
    • one year ago
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    Yup, by induction it follows for n parts :)

  45. ParthKohli
    • one year ago
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    let me give you a nice question in another thread maybe

  46. Jhannybean
    • one year ago
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    *

  47. Empty
    • one year ago
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    I wonder if this works for finding the electric dipole moment as well?

  48. anonymous
    • one year ago
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    Negative mass system!!

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