SYSTEM OF PARTICLES AND ROTATIONAL MOTION
1. Introduction
Any real body which we encounter in daily life has a
finite size. In dealing with the motion of extended bodies (bodies of finite size) often the idealised model of a particle is inadequate. In this chapter we shall try to go beyond this
inadequacy. We shall attempt to build an understanding of the motion of extended bodies. An extended body, in the
first place, is a system of particles. We shall begin with the consideration of motion of the system as a whole. The centre of mass of a system of particles will be a key concept here. We shall discuss the motion of the centre of mass of a system
of particles and usefulness of this concept in understanding the motion of extended bodies.
A large class of problems with extended bodies can be
solved by considering them to be rigid bodies. Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of particles of such a body do not change. It is evident from
this definition of a rigid body that no real body is truly rigid, since real bodies deform under the influence of forces. But in
many situations the deformations are negligible. In a number of situations involving bodies such as wheels, tops, steel
beams, molecules and planets on the other hand, we can ignore that they warp (twist out of shape), bend or vibrate and treat them as rigid.
1.2. What kind of motion can a rigid body have?
Let us try to explore this question by taking some examples
of the motion of rigid bodies. Let us begin with a rectangular block sliding down an inclined plane without any
sidewise movement. The block is taken as a rigid
body. Its motion down the plane is such that all
the particles of the body are moving together, i.e. they have the same velocity at any instant of time. The rigid body here is in pure translational motion.
In pure translational motion at any
instant of time, all particles of the body have the same velocity.
Consider now the rolling motion of a solid metallic or wooden cylinder down the same inclined plane. The rigid body in this problem, namely the cylinder, shifts from the top to the bottom of the inclined plane, and thus,
seems to have translational motion. But as shows, all its particles are not moving with
the same velocity at any instant.
The body,
therefore, is not in pure translational motion. Its motion is translational plus ‘something else.’ In order to understand what this ‘something else’ is, let us take a rigid body so constrained that it cannot have translational motion. The most common way to constrain a rigid body so that it does not have translational motion is to fix it along a straight line. The only possible motion of such a rigid body is rotation. The line or fixed axis about which the body is rotating is its axis of rotation. If you look around, you will come across many examples of rotation about an axis, a ceiling fan, a potter’s wheel, a giant wheel in a fair, a merry-go-round and so on.
Let us try to understand what rotation is,
what characterises rotation. You may notice that in rotation of a rigid body about a fixed axis,every particle of the body moves in a circle,
which lies in a plane perpendicular to the axis and has its centre on the axis.
The rotational motion of a rigid body about a fixed
axis (the z-axis of the frame of reference). Let P1 be a particle of the rigid body, arbitrarily chosen and at a distance r1 from fixed axis. The particle P1 describes a circle of radius r1 with its centre C1 on the fixed axis. The circle lies in a plane perpendicular to the axis. The figure also shows
another particle P2 of the rigid body, P2 is at a distance r2 from the fixed axis. The particle P2 moves in a circle of radius r2 and with centre C2 on the axis. This circle, too, lies in a plane perpendicular to the axis. Note that the circles described by P1 and P2 may lie in different planes; both these planes, however, are perpendicular to the fixed axis. For any particle on the axis like P3, r = 0. Any such particle remains stationary while the body rotates. This is expected since the axis of rotation is fixed.
2. Center 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 x1 and x2 respectively from some origin O. Let m1
and m2 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 mx mx X
m m+ = + .In Eq. (7.1), X can be regarded as the mass- weighted mean of x1 and x2. If the two particles have the same
Thus, for two particles of equal mass the
centre of mass lies exactly midway between
them.
If we have n particles of masses m1, m2, ...mn 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.
X mx mx mx
mm m
m x
m
m x
where x1, x2,...xn are the distances of the
particles from the origin; X is also measured from
the same origin. The symbol (the Greek letter
sigma) denotes summation, in this case over n particles. The sum m M i =
is the total mass of the system. 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 (x1,y1), (x2,y2) and (x3,y3) respectively.
Let the masses of the three particles be m1, m2 and m3 respectively. The centre of mass C of the system of the three particles is defined and located by the coordinates (X, Y) given by
mx mx mx X
mmmm
+ + = + +
11 2 2 3 3
123
my my my Y
mmm
+ + = + +
For the particles of equal mass m = m1 = m2
= m3,
Thus, for three particles of equal mass, the centre of mass coincides with the centroid of the triangle formed by the particles.
Results 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
m xi i X
M =
m yi i Y
M =
and
m zi i Z
M =
Here M = mi is the total mass of the
system. The index i runs from 1 to n; mi
is the
mass of the i th particle and the position of the i th particle is given by (xi , yi , zi ).
Eqs. can be
combined into one equation using the notation of position vectors. Let i r be the position vector
of the i
th particle and R be the position vector of the centre of mass:
r i jk ii i i =++ xyz
and XY Z R i jk =++
Then
mi i
M = r
R . he sum on the right hand side is a vector sum.
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 mi i r = 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. 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 issmall, we can treat the body as a continuous distribution of mass. We subdivide the body into
n small elements of mass; ∆m1, ∆m2... ∆mn; the
i
th element ∆mi is taken to be located about the point (xi , yi , zi ).
3. MOTION OF CENTRE OF MASS
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 F1 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 their contribution is zero. Only the external forces contribute to the equation. We can then rewrite as
MA F= ext .
where F ext represents the sum of all external
forces acting on the particles of the system.
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.
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 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.
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 of the 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.
This is the procedure that we followed earlier
in analysing forces on bodies and solving problems without explicitly outlining and
justifying the procedure. 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.
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.
4. VECTOR PRODUCT OF TWO VECTORS
We are already familiar with vectors and their
use in physics. In (Work, Energy,
Power) we defined the scalar product of two
vectors. An important physical quantity, work,
is defined as a scalar product of two vector quantities, force and displacement.
We shall now define another product of two
vectors. This product is a vector. Two important quantities in the study of rotational motion, namely, moment of a force and angular momentum, are defined as vector products.
4.2 Definition of Vector Product
A vector product of two vectors a and b is a
vector c such that
(i) magnitude of c = c = ab sinθ where a and b
are magnitudes of a and b and θ is the
angle between the two vectors.
(ii) c is perpendicular to the plane containing
a and b.
(iii) if we take a right handed screw with its head
lying in the plane of a and b and the screw perpendicular to this plane, and if we turn the head in the direction from a to b, then the tip of the screw advances in the direction of c.
Alternately, if one curls up the fingers of
right hand around a line perpendicular to the
plane of the vectors a and b and if the fingers
are curled up in the direction from a to b, then
the stretched thumb points in the direction of c,
A simpler version of the right hand rule is
the following : Open up your right hand palm
and curl the fingers pointing from a to b. Your
stretched thumb points in the direction of c.
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