Centre of Mass, Momentum and Collision Concepts Page - 1

Definition

Centre of Mass

The center of mass of a body or a system of bodies is a mean position of the total weight of the body where the resultant of the forces applied is considered to be acted upon such that forces, momentum and energy are conserved. The body or system of bodies is balanced around the center of mass and the average of the weighted position coordinates defines its coordinates.

Definition

Center of mass of two point masses

Center of mass of two point masses is given by:
r⃗ COM=m1r⃗ 1+m2r⃗ 2m1+m2${\overrightarrow{r}}_{COM}={\displaystyle \frac{m_1{\overrightarrow{r}}_1+m_2{\overrightarrow{r}}_2}{m_1+m_2}}$
Example:
In a carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance 1.2×10−10m$1.2\times {10}^{-10}m$. Then find the radius of carbon atom is (Assume both atoms touch each other).Solution:
If centre of mass is taken as origin, then r1r2=m2m1${\displaystyle \frac{r_1}{r_2}}={\displaystyle \frac{m_2}{m_1}}$
⇒r1r1+r2=m2m1+m2$\Rightarrow {\displaystyle \frac{r_1}{r_1+r_2}}={\displaystyle \frac{m_2}{m_1+m_2}}$
⇒r1=1612+16×1.2×10−10=0.68×10−10m$\Rightarrow r_1={\displaystyle \frac{16}{12+16}}\times 1.2\times {10}^{-10}=0.68\times {10}^{-10}m$  

Formula

Centre of Mass Formula

Formula for centre of mass is:
X=m1x1+m2x2........+mnxnm!+m2+.....+mn=ΣmixiΣmi$\mathrm{X}={\displaystyle \frac{m_1{x}_1+m_2{x}_2........+m_n{x}_n}{m_{!}+m_2+.....+m_n}}={\displaystyle \frac{\mathrm{\Sigma }{m}_i{x}_i}{\mathrm{\Sigma }{m}_i}}$

Definition

Center of mass of more than two point masses

Center of mass for a multiple mass system is given by:
r⃗ COM=∑r⃗ imi∑mi,i∈1,2,3,...,n${\overrightarrow{r}}_{COM}={\displaystyle \frac{\sum {\overrightarrow{r}}_i{m}_i}{\sum m_i}},i\in 1,2,3,...,n$
Example:
Four particles are in x−y$x-y$ plane at
(1)1kg$(1)1kg$ at (0,0)(2)2kg$(0,0)(2)2kg$ at (1,0)(3)3kg$(1,0)(3)3kg$ at (1,2)(4)4kg$(1,2)(4)4kg$ at (2,0)$(2,0)$
Find where the center of mass is located.Solution:
Centre of mass is given by the relation, (∑miXi)/∑mi$(\sum m_i{X}_i)/\sum m_i$ and (∑miYi)/∑mi$(\sum m_i{Y}_i)/\sum m_i$Thus we get ((1(0)+2(1)+4(2)+3(1))(1+2+3+4),1(0)+2(0)+3(2)+4(0)1+2+3+4)$({\displaystyle \frac{(1(0)+2(1)+4(2)+3(1))}{(1+2+3+4)}},{\displaystyle \frac{1(0)+2(0)+3(2)+4(0)}{1+2+3+4}})$
Thus we get the coordinates as (1.3,0.6)$(1.3,0.6)$

Definition

Use principle of symmetry to find centre of mass of system of discrete masses

For a discrete set of particles numbered 1,2,3,4,........,i,.........N$1,2,3,4,........,i,.........N$ each of mass m1,m2,m3,.....mi,.....,mN$m_1,m_2,m_3,.....m_i,.....,m_N$ at positions (x1,y1,z1),(x2,y2,z2),.....(xi,yi,zi),......(xN,yN,zN)$(x_1,y_1,z_1),(x_2,y_2,z_2),.....(x_i,y_i,z_i),......(x_N,y_N,z_N)$. Let the total mass of the system be M.
The x-coordinate of centre of mass of the system is given by:
X=∑mixiNi=1M$X={\displaystyle \frac{\sum _N^{m_i{x}_i}i=1}{M}}$
The y-coordinate of centre of mass of the system is given by:
Y=∑miyiNi=1M$Y={\displaystyle \frac{\sum _N^{m_i{y}_i}i=1}{M}}$
The z-coordinate of centre of mass of the system is given by:
Z=∑miziNi=1M$Z={\displaystyle \frac{\sum _N^{m_i{z}_i}i=1}{M}}$

Example

Principle of Negative mass to find centre of mass for a discrete mass system

Example: Look at the drawing given in the figure which has been drawn with ink
of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m$m$. The mass of the ink used to draw the outer circle is 6 m$6m$. The coordinates of the centres of the different parts are: outer circle (0, 0) left inner circle (a, a), right inner circle (a, a),vertical line (0, 0) and horizontal line (0, a). The y-coordinate of the centre of mass of the ink in the above drawing is:

Solution: YCM=6m×0+ma+ma+−ma+m×06m+m+m+m+m$Y_{CM}={\displaystyle \frac{6m\times 0+ma+ma+-ma+m\times 0}{6m+m+m+m+m}}$
YCM=ma10m=a10$Y_{CM}={\displaystyle \frac{ma}{10m}}={\displaystyle \frac{a}{10}}$

Example

Principle of Negative mass to find centre of mass

Example: In the figure shown, find out the distance of centre of mass of a system of a uniform circular plate of radius 3R$3R$ from O. On this plate, a hole
of radius R$R$ is cut whose centre is at 2R$2R$ distance from centre of the
plate.

Solution: From law of conservation of momentum,
x¯=m1x1+(−m2)x2m1+(−m2)${\displaystyle \overline{x}={\displaystyle {\displaystyle \frac{m_1{x}_1+(-m_2)x_2}{m_1+(-m_2)}}}}$

=A1x1+(−A2)x2A1+(−A2)$={\displaystyle {\displaystyle \frac{A_1{x}_1+(-A_2)x_2}{A_1+(-A_2)}}}$

A1=π(3R)2$A_1=\pi {\left(3R\right)}^2$,
A2=πR2$A_2=\pi R^2$

x1=0$x_1=0$,
x2=2R$x_2=2R$
∴x¯=−R/4$\therefore \overline{x}=-R/4$

Definition

Principle of symmetry to find gravitational field

Symmetry in the geometrical distribution of a body can be used to calculate the gravitational field at a point much more easily.
Example:
If Gravitational field due to uniform thin hemispherical shell at point P is I, then find the magnitude of gravitational field at Q. (Mass of hemisphere is M, radius R).
Solution:The gravitational field at P and Q due to the whole sphere is given as GM4R2${\displaystyle \frac{GM}{4R^2}}$.
As the gravitational field due to the hemispherical shell at point P is I, using symmetry the gravitational field due to the imaginary hemispherical shell at point Q would also be I.
Thus the gravitational field due to the given hemisphere at point Q will be
(Gravitational field due to whole sphere at Q)-(Gravitational field due to imaginary hemisphere at Q)
=GM4R2−I$={\displaystyle \frac{GM}{4R^2}}-I$

Definition

Linear, Area & Volume Mass Density

The area density of a two-dimensional object is calculated as the mass per unit area. The SI derived unit is: kilogram per square metre.
The linear density of a two-dimensional object is calculated as the mass per unit length. The SI derived unit is: kilogram per metre.
The volume density of a two-dimensional object is calculated as the mass per unit volume. The SI derived unit is: kilogram per metre cube.

Definition

Centre of Mass of a Half Disc

Center of Mass of a Half a Disk

r⃗ c.m.=(xc.m.,yc.m.)=(0,2rπ)${\overrightarrow{r}}_{c.m.}=(x_{c.m.},y_{c.m.})=(0,{\displaystyle \frac{2r}{\pi }})$

yc.m.=∫dmMyring=∫dmM2rπ$y_{c.m.}=\int {\displaystyle \frac{dm}{M}}{y}_{ring}=\int {\displaystyle \frac{dm}{M}}{\displaystyle \frac{2r}{\pi }}$

Surface Density :  The mass per unit area

σ=dmdA⟶dm=σdA$\sigma ={\displaystyle \frac{dm}{dA}}⟶dm=\sigma dA$

dA=π(r+dr)2−π(r)2$dA=\pi (r+dr{)}^2-\pi (r{)}^2$

dA=πr2+2πrdr+π(dr)2−πr2$dA=\pi r^2+2\pi rdr+\pi (dr{)}^2-\pi r^2$

dA=2πrdr+π(dr)2$dA=2\pi rdr+\pi (dr{)}^2$

dA=2πrdr$dA=2\pi rdr$

dm=σdA=σπrdr=MAtotπrdr=M12πR2πrdr$dm=\sigma dA=\sigma \pi rdr={\displaystyle \frac{M}{A_{tot}}}\pi rdr={\displaystyle \frac{M}{{\displaystyle \frac{1}{2}}\pi R^2}}\pi rdr$

dm=MR22rdr

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