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

- ganeshie8

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

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- ganeshie8

|dw:1443928609926:dw|

- imqwerty

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)

- ganeshie8

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`

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## More answers

- imqwerty

yes :)

- UnkleRhaukus

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

- ganeshie8

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..

- ganeshie8

Just noticed, you're representing the empty region by "negative area" and seems its working perfectly ! xD

- imqwerty

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 :)

- ganeshie8

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 \(x = \dfrac{R}{3}\)

- imqwerty

yes correct :)

- ganeshie8

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 \(M=m_1+m_2+\cdots\)

- imqwerty

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
\[R_{CM}=\frac{ 1 }{ M }\sum_{}^{}m_{1}r_{1}\]

- UnkleRhaukus

\[m=\varrho A\]

- ganeshie8

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 ?

- imqwerty

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 :)

- ganeshie8

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 ?

- IrishBoy123

you are using integrals, just easy ones that don't need doing

- ganeshie8

Maybe... but im not using calculus directly, im using intuition... but im looking for a proof too.. :)

- ganeshie8

Suppose I want to find the center of mass of below weird shape.
|dw:1443948450651:dw|

- ganeshie8

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

- IrishBoy123

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 :-)

- IrishBoy123

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....

- IrishBoy123

make that
\[\sum \delta a_i \delta r_i \sigma_i\]

- ParthKohli

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.

- ganeshie8

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

- ganeshie8

I think using plumbline is not allowed because it is more of experimental method...

- ganeshie8

Also its not so exciting because we can always find the center of mass of any object using plumblines

- ParthKohli

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

- ganeshie8

Exactly! expecting this kindof proof... please keep going...

- ParthKohli

\[\vec{\rm CM'} = \frac{M_1 \vec{\rm CM_1} + M_2 \vec{CM_2}}{M_1 + M_2} = \frac{\sum_{i = 1}^{n} m_i \vec{r_i} + \sum_{i=n+1}^{N}m_i \vec{r_i }}{M} = \frac{\sum_{i=1}^N m_i \vec{r_i }}{M}\]

- ParthKohli

\[\Rightarrow \vec{\rm CM'} = \vec{\rm CM} \]

- ganeshie8

wot the frog

- ganeshie8

that looks really nice!

- IrishBoy123

yeah, that's good!

- ganeshie8

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

- ParthKohli

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.

- ganeshie8

Ahh nice! nice! I sew it now!

- ganeshie8

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

- ParthKohli

o.o

- ganeshie8

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

- ganeshie8

when you see the relationship for the first time, this is really a beautiful result!

- ParthKohli

We get that\[\rm \vec{CM} = CM( CM_1, CM_2, \cdots, CM_n) \]

- ParthKohli

I just proved it for \(n=2\) but you can see what's happening here.

- ganeshie8

Yup, by induction it follows for n parts :)

- ParthKohli

let me give you a nice question in another thread maybe

- Jhannybean

*

- Empty

I wonder if this works for finding the electric dipole moment as well?

- anonymous

Negative mass system!!

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