Question

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

Answer 1

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)

Answer 2

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`

Answer 3

yes :)

Answer 4

If instead, the origin is taken as the left most point on the circles

\[\begin{align}
CM_x &= \frac{r_0A_0+r_1A_1}{A_0+A_1}\\
&= \frac{R(-\pi R^2)+2R(\pi(2R)^2)}{-\pi R^2+\pi(2R)^2}\\
&= \frac{\pi R^3(-1+8)}{\pi R^2(-1+4)}\\
&= \tfrac73R
\end{align}\]

Answer 5

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..

Answer 6

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

Answer 7

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 :)

Answer 8

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

Answer 9

yes correct :)

Answer 10

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

Answer 11

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

Answer 12

\[m=\varrho A\]

Answer 13

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 ?

Answer 14

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 :)

Answer 15

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 ?

Answer 16

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

Answer 17

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

Answer 18

Suppose I want to find the center of mass of below weird shape.