Physics CENTRE OF MASS AND MOTION OF CENTRE OF MASS

### Topic covered

star CENTRE OF MASS
star MOTION OF CENTRE OF MASS

### CENTRE OF MASS

● 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.

Q 3189178917

Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long.
Class 11 Chapter 7 Example 0
Solution:

With the x and y axes chosen as shown in Fig.
7.9, the coordinates of points O, A and B forming
the equilateral triangle are respectively (0,0),
(0.5,0), (0.25,0.25 sqrt 3 ). Let the masses 100 g,
150g  and 200g be located at O, A and B be
respectively. Then,

color{blue}{X = ( m_1 x_1 + m_2 x_2 + m_3 x_3)/( m_1 + m_2 + m_3)}

 = ( 100(0) + 150(0.5) + 200(0.25) gm)/( (100 + 150 + 200) g)

 = ( 75 + 50)/(450) m = (125)/(450) m = 5/(18) m

 Y = ( 100(0) + 150(0) + 200(0.25 sqrt 3) gm )/(450 g)

 = ( 50 sqrt 3) /( 450) m = (sqrt3)/9 m = 1/(3 sqrt3 ) m

The centre of mass C is shown in the figure.
Note that it is not the geometric centre of the
triangle OAB. Why?
Q 3119178919

Find the centre of mass of a triangular lamina.
Class 11 Chapter 7 Example 0
Solution:

The lamina (DLMN) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig. 7.10

By symmetry each strip has its centre of mass at its midpoint. If we join the midpoint of all the strips we get the median LP. The centre of mass of the triangle as a whole therefore, has to lie on the median LP. Similarly, we can argue that it lies on the median MQ and NR. This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.
Q 3119180010

Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 kg.
Class 11 Chapter 7 Example 0
Solution:

Choosing the X and Y axes as shown
in Fig. 7.11 we have the coordinates of the
vertices of the L-shaped lamina as given in the
figure. We can think of the
L-shape to consist of 3 squares each of length
1 m. The mass of each square is 1 kg, since the
lamina is uniform. The centres of mass C_1 , C_2
and C_3 of the squares are, by symmetry, their
geometric centres and have coordinates (1//2,1//2),
(3//2,1//2), (1//2,3//2) respectively. We take the
masses of the squares to be concentrated at
these points. The centre of mass of the whole
L shape (X, Y) is the centre of mass of these
mass points.

Hence

 X = ( [ 1 (1//2) + 1 (3//2) + 1 (1//2) ] k g m)/( (1 + 1 + 1 ) kg) = 5/6 m

 Y = ( [ v 1 (1//2) + 1 (1//2) + 1 (3//2) ] kgm)/((1 + 1 + 1 ) kg) = 5/6 m

The centre of mass of the L-shape lies on
the line OD. We could have guessed this without
calculations. Can you tell why ? Suppose, the
three squares that make up the L shaped lamina
of Fig. 7.11 had different masses. How will you
then determine the centre of mass of the
lamina ?

### MOTION OF CENTRE OF MASS

● Here we discuss physical importance for a system of particles. We may rewrite Eq.(7.4d) as

color{blue}{MR = sum m_ i r_i = m_1 r_1 + m_2 r_2 + ... + m_n r_n.............. (7.7)}

Differentiating the two sides of the equation with respect to

 M = (dR)/(dt) = m_1 (dr_1)/(dt) + m_2 (dr_2)/(dt) + ..... + m_n (dr_n)/(dt)

or

color{blue} { M V = m_1 v_1 +m_2 v_2 + .......... + m_n v_n.............(7.8)}

=> where v_1 ( = dr_1 // dt) is the velocity of the first particle v_2 = dr_2 // dt) is the velocity of the second particle etc. and V = dR//dt is the velocity of the centre of mass. Note that we assumed the masses m_1, m_2, ... etc. do not change in time. We have therefore, treated them as constants in differentiating the equations with respect to time. Differentiating Eq.(7.8) with respect to time, we obtain

 M (dV)/(dt) = m_1 (dv_1)/(dt) + m_2 (dv_2)/(dt) + .... m_n (dv_n)/(dt)

or

 color{blue}{MA = m_1 a_1 + m_2 a_2 + m_n a_n ................. (7.9)}

=> where a_1 = (dv_1 // dt) is the acceleration of the first particle, a_2 = (dv_2 // dt) is the acceleration of the second particle etc. and A = (dV //dt ) is the acceleration of the centre of mass of the system of particles. Now, from Newton’s second law, the force acting on the first particle is given by F_1 =m_1 a_1 . The force acting on the second particle is given by  F_2 = m_2 a_2 and so on. Eq. (7.9) may be written as

color{blue}{ MA = F_1 + F_2 + ...... + F_n....................(7.10)}

color{green}☞ Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.

● Note when we talk of the force F_1 on the first particle, it is not a single force, but the vector sum of all the forces on the first particle; likewise for the second particle etc.
● Among these forces on each particle there will be external forces exerted by bodies outside the system and also internal forces exerted by the particles on one another. We know from Newton’s third law that these internal forces occur in equal and opposite pairs and in the sum of forces of Eq. (7.10), their contribution is zero. Only the external forces contribute to the equation. We can then rewrite Eq. (7.10) as

color{blue}{MA = F_(ext)................ (7.11)}

=> where  F_(ext) represents the sum of all external forces acting on the particles of the system.

Eq. (7.11) states that the centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point.

color{green}☞ Notice, to determine the motion of the centre of mass no knowledge of internal forces of the system of particles is required; for this purpose we need to know only the external forces.

● To obtain Eq. (7.11) we did not need to specify the nature of the system of particles. The system may be a collection of particles in which there may be all kinds of internal motions, or it may be a rigid body which has either pure translational motion or a combination of translational and rotational motion. Whatever is the system and the motion of its individual particles, the centre of mass moves according to Eq. (7.11).

● Instead of treating extended bodies as single particles as we have done in earlier chapters, we can now treat them as systems of particles. We can obtain the translational component of their motion, i.e. the motion centre of mass of the system, by taking the mass of the whole system to be concentrated at the centre of mass and all the external forces on the system to be acting at the centre of mass.

● We now realise that in earlier studies we assumed, without saying so, that rotational motion and/or internal motion of the particles were either absent or negligible. We no longer need to do this. We have not only found the justification of the procedure we followed earlier; but we also have found how to describe and separate the translational motion of
(1) a rigid body which may be rotating as well, or
(2) a system of particles with all kinds of internal motion.

● Figure 7.12 is a good illustration of Eq. (7.11). A projectile, following the usual parabolic trajectory, explodes into fragments midway in air. The forces leading to the explosion are internal forces. They contribute nothing to the motion of the centre of mass.
● The total external force, namely, the force of gravity acting on the body, is the same before and after the explosion. The centre of mass under the influence of the external force continues, therefore, along the same parabolic trajectory as it would have followed if there were no explosion.