`●` We shall first see what the centre of mass of a system of particles is and then discuss its significance. For simplicity we shall start with a two particle system.
`●` We shall take the line joining the two particles to be the x- axis.
`●` Let the distances of the two particles be `x_1` and `x_2` respectively from some origin `O`. Let `m_1` and `m_2` be respectively the masses of the two particles. The centre of mass of the system is that point `C` which is at a distance `X` from `O`, where `X` is given by
` color{blue}{X = (m_1x _1 + m_2 x _2)/( m_1 + m_2)....................(7.1)}`
`=>` In Eq. (7.1), `X` can be regarded as the mass weighted mean of `x_1` and `x_2`. If the two particles have the same mass `m_1 = m_2 = m,` then
` X = ( mx _1 + m x _2)/( 2m) = ( x_1 + x_2)/2`
`color{green}☞` Thus, for two particles of equal mass the centre of mass lies exactly midway between them.
`●` If we have n particles of masses `m_1 , m_2 , ......... m_n` respectively, along a straight line taken as the `x- `axis, then by definition the position of the centre of the mass of the system of particles is given by
` color{blue}{X = ( m_1 x_1 + m_2 x_2 + ...... + m_n x_n)/( m_1 + m_2 + ..... m_n) = ( sum m_t x_t)/(sum m_t)...............(7.2)}`
`=>` where `x_1, x_2 ,... x_n` are the distances of the particles from the origin; `X` is also measured from the same origin. The symbol `sum` (the Greek letter sigma) denotes summation, in this case over `n` particles. The sum
` sum m_t = M` is the total mass of the system
`●` Now Suppose that we have three particles, not lying in a straight line. We may define `x` and `y` axes in the plane in which the particles lie and represent the positions of the three particles by coordinates `(x_1 , y_1 ), (x_2,y_2)` and `(x_3 , y_3)` respectively.
`●` Let the masses of the three particles be `m_1, m_2` and `m_3` respectively. The centre of mass `C` of the system of the three particles is defined and located by the coordinates `(X, Y)` given by
`color{blue} {X = ( m_1 x_1 + m_2 x_2 + m_3 x_3)/(m_1 + m_2 + m_3)........................(7.3 a)}`
`color{blue}{Y = ( m_1 y_1 + m_2 y_2 + m_3 y_3)/(m_1 + m_2 + m_3)....................(7.3 b)}`
For the particles of equal mass `m = m_1 = m_2 = m_3`,
` X = ( m ( x_1 + x_2 + x_3))/(3m) = ( x_1 + x_2 + x_3 )/3`
` Y = ( m ( y_1 + y_2 + y_3))/(3m) = ( y_1 + y_2 + y_3 )/3`
`●` Thus, for three particles of equal mass, the centre of mass coincides with the centroid of the triangle formed by the particles.
`●` Results of Eqs. (7.3a) and (7.3b) are generalised easily to a system of `n` particles, not necessarily lying in a plane, but distributed in space. The centre of mass of such a system is at `(X, Y, Z )`, where
`color{blue} {x = (sum m_i x_i)/M........................(7.4 a)}`
`color{blue} {y = (sum m_i x_i)/M........................(7.4 b)}`
and ` color{blue}{Z = (sum m_i z_i)/M.................(7.4 c)}`
`●` Here `M = sum m_i ` is the total mass of the system. The index `i` runs from `1` to `n ; m_i` is the mass of the `i^(th)` particle and the position of the `i^(th)` particle is given by `(x_1 , y_1 , z_1 )`.
`=>` Eqs. (7.4a), (7.4b) and (7.4c) can be combined into one equation using the notation of position vectors. Let `r_i` be the position vector of the `i^(th)` particle and `R` be the position vector of the centre of mass:
`r_i = x_i hat i + y_i hat j + z_i hat k`
and ` R= X hat i +Y hat j+ Z hat k`
Then ` color{blue}{R = ( sum m_i r_i)/M....................(7.4 d)}`
The sum on the right hand side is a vector sum.
`color{green}☞` Note the economy of expressions we achieve by use of vectors. If the origin of the frame of reference (the coordinate system) is chosen to be the centre of mass then ` sum m_i r_i =0` for the given system of particles.
`●` A rigid body, such as a metre stick or a flywheel, is a system of closely packed particles; Eqs. (7.4a), (7.4b), (7.4c) and (7.4d) are therefore, applicable to a rigid body. The number of particles (atoms or molecules) in such a body is so large that it is impossible to carry out the summations over individual particles in these equations.
`●` Since the spacing of the particles is small, we can treat the body as a continuous distribution of mass. We subdivide the body into n small elements of mass ; `Delta m_1 , Delta m_2 ... Delta m_n ;` the `i^(th)` element `Delta m_i` is taken to be located about the point `(x_1, y_1, z_1)`. The coordinates of the centre of mass are then approximately given by
` color{blue}{X = (sum ( Delta m_i) x_i)/(sum Delta m_i) , Y = (sum ( Delta m_i) y_i)/(sum Delta m_i) , Z = ( sum ( Delta m_i) y_i)/(sum Delta m_i)}`
`●` As we make `n` bigger and bigger and each `Delta m_i` smaller and smaller, these expressions become exact. In that case, we denote the sums over `i` by integrals. Thus,
` Delta m_i -> int dm = M,`
` ( Delta m_i ) x_i -> x dm `
` ( Delta m_t ) y_t -> y dm `
and ` ( Delta m_i ) z_i -> y dm `
`●` Here `M` is the total mass of the body. The coordinates of the centre of mass now are
`color{blue}{ X = 1/M int x dm , Y = 1/M int y dm " and " Z = 1/M int zdm......................(7.5 a)}`
`●` The vector expression equivalent to these three scalar expressions is
`color{blue} {R = 1/M rdm....................(7.5 b)}`
`\color{blue}➢` If we choose, the centre of mass as the origin of our coordinate system,
`R(x,y,z) = 0`
i.e., `r dm = 0`
or `color{blue}{x dm = y dm = z dm = 0..............(7.6)}`
`=>` Often we have to calculate the centre of mass of homogeneous bodies of regular shapes like rings, discs, spheres, rods etc. (By a homogeneous body we mean a body with uniformly distributed mass.) By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres.
`●` Let us consider a thin rod, whose width and breath (in case the cross section of the rod is rectangular) or radius (in case the cross section of the rod is cylindrical) is much smaller than its length.
`●` Taking the origin to be at the geometric centre of the rod and x-axis to be along the length of the rod, we can say that on account of reflection symmetry, for every element `dm` of the rod at `x`, there is an element of the same mass `dm` located at `-x` (Fig. 7.8).
The net contribution of every such pair to the integral and hence the integral `x dm` itself is zero. From Eq. (7.6), the point for which the integral itself is zero, is the centre of mass. Thus, the centre of mass of a homogenous thin rod coincides with its geometric centre. This can be understood on the basis of reflection symmetry.
`\color{green} ✍️` The same symmetry argument will apply to homogeneous rings, discs, spheres, or even thick rods of circular or rectangular cross section. For all such bodies you will realise that for every element `dm` at a point ` (x, y, z)` one can always take an element of the same mass at the point `(–x, –y, –z)`. (In other words, the origin is a point of reflection symmetry for these bodies.) As a result, the integrals in Eq. (7.5 a) all are zero. This means that for all the above bodies, their centre of mass coincides with their geometric centre.
`●` We shall first see what the centre of mass of a system of particles is and then discuss its significance. For simplicity we shall start with a two particle system.
`●` We shall take the line joining the two particles to be the x- axis.
`●` Let the distances of the two particles be `x_1` and `x_2` respectively from some origin `O`. Let `m_1` and `m_2` be respectively the masses of the two particles. The centre of mass of the system is that point `C` which is at a distance `X` from `O`, where `X` is given by
` color{blue}{X = (m_1x _1 + m_2 x _2)/( m_1 + m_2)....................(7.1)}`
`=>` In Eq. (7.1), `X` can be regarded as the mass weighted mean of `x_1` and `x_2`. If the two particles have the same mass `m_1 = m_2 = m,` then
` X = ( mx _1 + m x _2)/( 2m) = ( x_1 + x_2)/2`
`color{green}☞` Thus, for two particles of equal mass the centre of mass lies exactly midway between them.
`●` If we have n particles of masses `m_1 , m_2 , ......... m_n` respectively, along a straight line taken as the `x- `axis, then by definition the position of the centre of the mass of the system of particles is given by
` color{blue}{X = ( m_1 x_1 + m_2 x_2 + ...... + m_n x_n)/( m_1 + m_2 + ..... m_n) = ( sum m_t x_t)/(sum m_t)...............(7.2)}`
`=>` where `x_1, x_2 ,... x_n` are the distances of the particles from the origin; `X` is also measured from the same origin. The symbol `sum` (the Greek letter sigma) denotes summation, in this case over `n` particles. The sum
` sum m_t = M` is the total mass of the system
`●` Now Suppose that we have three particles, not lying in a straight line. We may define `x` and `y` axes in the plane in which the particles lie and represent the positions of the three particles by coordinates `(x_1 , y_1 ), (x_2,y_2)` and `(x_3 , y_3)` respectively.
`●` Let the masses of the three particles be `m_1, m_2` and `m_3` respectively. The centre of mass `C` of the system of the three particles is defined and located by the coordinates `(X, Y)` given by
`color{blue} {X = ( m_1 x_1 + m_2 x_2 + m_3 x_3)/(m_1 + m_2 + m_3)........................(7.3 a)}`
`color{blue}{Y = ( m_1 y_1 + m_2 y_2 + m_3 y_3)/(m_1 + m_2 + m_3)....................(7.3 b)}`
For the particles of equal mass `m = m_1 = m_2 = m_3`,
` X = ( m ( x_1 + x_2 + x_3))/(3m) = ( x_1 + x_2 + x_3 )/3`
` Y = ( m ( y_1 + y_2 + y_3))/(3m) = ( y_1 + y_2 + y_3 )/3`
`●` Thus, for three particles of equal mass, the centre of mass coincides with the centroid of the triangle formed by the particles.
`●` Results of Eqs. (7.3a) and (7.3b) are generalised easily to a system of `n` particles, not necessarily lying in a plane, but distributed in space. The centre of mass of such a system is at `(X, Y, Z )`, where
`color{blue} {x = (sum m_i x_i)/M........................(7.4 a)}`
`color{blue} {y = (sum m_i x_i)/M........................(7.4 b)}`
and ` color{blue}{Z = (sum m_i z_i)/M.................(7.4 c)}`
`●` Here `M = sum m_i ` is the total mass of the system. The index `i` runs from `1` to `n ; m_i` is the mass of the `i^(th)` particle and the position of the `i^(th)` particle is given by `(x_1 , y_1 , z_1 )`.
`=>` Eqs. (7.4a), (7.4b) and (7.4c) can be combined into one equation using the notation of position vectors. Let `r_i` be the position vector of the `i^(th)` particle and `R` be the position vector of the centre of mass:
`r_i = x_i hat i + y_i hat j + z_i hat k`
and ` R= X hat i +Y hat j+ Z hat k`
Then ` color{blue}{R = ( sum m_i r_i)/M....................(7.4 d)}`
The sum on the right hand side is a vector sum.
`color{green}☞` Note the economy of expressions we achieve by use of vectors. If the origin of the frame of reference (the coordinate system) is chosen to be the centre of mass then ` sum m_i r_i =0` for the given system of particles.
`●` A rigid body, such as a metre stick or a flywheel, is a system of closely packed particles; Eqs. (7.4a), (7.4b), (7.4c) and (7.4d) are therefore, applicable to a rigid body. The number of particles (atoms or molecules) in such a body is so large that it is impossible to carry out the summations over individual particles in these equations.
`●` Since the spacing of the particles is small, we can treat the body as a continuous distribution of mass. We subdivide the body into n small elements of mass ; `Delta m_1 , Delta m_2 ... Delta m_n ;` the `i^(th)` element `Delta m_i` is taken to be located about the point `(x_1, y_1, z_1)`. The coordinates of the centre of mass are then approximately given by
` color{blue}{X = (sum ( Delta m_i) x_i)/(sum Delta m_i) , Y = (sum ( Delta m_i) y_i)/(sum Delta m_i) , Z = ( sum ( Delta m_i) y_i)/(sum Delta m_i)}`
`●` As we make `n` bigger and bigger and each `Delta m_i` smaller and smaller, these expressions become exact. In that case, we denote the sums over `i` by integrals. Thus,
` Delta m_i -> int dm = M,`
` ( Delta m_i ) x_i -> x dm `
` ( Delta m_t ) y_t -> y dm `
and ` ( Delta m_i ) z_i -> y dm `
`●` Here `M` is the total mass of the body. The coordinates of the centre of mass now are
`color{blue}{ X = 1/M int x dm , Y = 1/M int y dm " and " Z = 1/M int zdm......................(7.5 a)}`
`●` The vector expression equivalent to these three scalar expressions is
`color{blue} {R = 1/M rdm....................(7.5 b)}`
`\color{blue}➢` If we choose, the centre of mass as the origin of our coordinate system,
`R(x,y,z) = 0`
i.e., `r dm = 0`
or `color{blue}{x dm = y dm = z dm = 0..............(7.6)}`
`=>` Often we have to calculate the centre of mass of homogeneous bodies of regular shapes like rings, discs, spheres, rods etc. (By a homogeneous body we mean a body with uniformly distributed mass.) By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres.
`●` Let us consider a thin rod, whose width and breath (in case the cross section of the rod is rectangular) or radius (in case the cross section of the rod is cylindrical) is much smaller than its length.
`●` Taking the origin to be at the geometric centre of the rod and x-axis to be along the length of the rod, we can say that on account of reflection symmetry, for every element `dm` of the rod at `x`, there is an element of the same mass `dm` located at `-x` (Fig. 7.8).
The net contribution of every such pair to the integral and hence the integral `x dm` itself is zero. From Eq. (7.6), the point for which the integral itself is zero, is the centre of mass. Thus, the centre of mass of a homogenous thin rod coincides with its geometric centre. This can be understood on the basis of reflection symmetry.
`\color{green} ✍️` The same symmetry argument will apply to homogeneous rings, discs, spheres, or even thick rods of circular or rectangular cross section. For all such bodies you will realise that for every element `dm` at a point ` (x, y, z)` one can always take an element of the same mass at the point `(–x, –y, –z)`. (In other words, the origin is a point of reflection symmetry for these bodies.) As a result, the integrals in Eq. (7.5 a) all are zero. This means that for all the above bodies, their centre of mass coincides with their geometric centre.