7.0 Preview:

• In this chapter, we shall study the motion of extended bodies or system of particles. We will first begin by defining the types of motion of a rigid body: 1) Purely translational motion: The motion of a body is said to be purely translational if its motion is such that all the particles of the body are moving together, that is, every particle of the body has the same velocity at any instant of time. 2) Rotational Motion: motion of a body is said to be rotational if all particles of the body move in a circle, the centres of which are situated on a straight line called axis of rotation.
• The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation. The motion of a rigid body which is pivoted or fixed in some way is rotation. The rotation may be about an axis that is fixed (e.g. a ceiling fan) or moving (e.g. an oscillating table fan).
• We will learn how the centre of mass of a system of particles moves as if all the mass of the system was concentrated at that point and all the external forces were applied at that point. Further, we will learn how internal forces in a system do not have any effect on the trajectory traced by an object.
• We will then define law of conservation of momentum and its effect. When the total external force acting on a system of particles is zero, the total linear momentum of the system is constant. This is the law of conservation of the total linear momentum.
• We shall also study vector product in some detail. Vector product: The product of two real vectors in three dimensions which itself is a vector at right angles to both the original vectors. Its magnitude is the product of the magnitudes of the original vectors times the sine of the angle between their directions (=ab sinθ).
• We will then establish relation between Linear Velocity $(v_i)$ and Angular Velocity (ω) (and with $r_i$ being the distance from central axis of an object in rotation) as $v_i= ωr_i$ ; angular acceleration, α as the time rate of change of angular velocity; Torque as the rotational analogue of Force in translation motion.
• Along the lines of conservation of momentum, we shall study Mechanical equilibrium as the state of a rigid body when the total momentum on a body (momentum due to translational and rotational) is zero.
• We shall then establish relations to define the principle of moments with the illustration of a level; Moment of inertia as the tendency of an object to stay in its state of rotational motion. We shall also find moment of inertia of a thin ring and a rigid massless rod.
• We shall study two useful theorems relating to moment of inertia: Perpendicular axis theorem states that the moment of inertia of a planar body (lamina) about an axis perpendicular to the plane of the lamina is equal to the sum of moments of inertia of the lamina about the two axes at right angles to each other in its own plane and intersecting each other at the point where the perpendicular axis passes through it. Next, Parallel axis theorem which states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
• Kinematic relations of motion for a rotating body will be listed analogous to that derived for translation motion like, relationships between velocity and acceleration and their various forms; Similarly, Dynamic relations like work done by an external torque analogous to the work done due to an external force in translation will be dealt with.
• Further, we shall learn about rolling motion as a combination of rotation and translation. Consequently, we will find that the total kinetic energy of a rolling body as the sum of translational and rotational kinetic energies.

7.1 Introduction

So far, we have studied the motion of a point object (as you already know, a particle with negligible size is point object) and extended this to the study of motion of bodies of finite size (assuming that motion of such bodies can be described in terms of the motion of a particle).

Any real body which we encounter in daily life has a finite size. In dealing with the motion of bodies of finite size, often, the idealised model of considering point objects is inadequate. In this chapter, we shall try to go beyond this inadequacy and attempt to build an understanding of the motion of “extended bodies”. An extended body, in the first place, is a system of particles or otherwise called rigid body. Ideally, a rigid body is a body with a perfectly definite and unchanging shape. Although no body is truly rigid (wheels, tops, steel beams, molecules and planets undergo warping, bending or vibrate) and do deform under the influence of forces, the deformations are negligible.

7.1.1 What kind of motion can a rigid body have?

• Let us begin with a rectangular block sliding down an inclined plane without any sidewise movement. The block is a rigid body. Refer figure, 7.1 below.
• Its motion down the plane is purely translational motion. The motion is such that all the particles of the body are moving together, that is, every particle of the body has the same velocity at any instant of time.

Fig 7.1 Translational (sliding) motion of a block down an inclined plane. (Any point like $P_1$ or $P_2$ of the block moves with the same velocity at any instant of time.)

• Consider now, the rolling motion of a solid metallic or wooden cylinder down the same inclined plane (Fig. 7.2). The rigid body in this problem, would be a cylinder; shifts from the top to the bottom of the inclined plane, and thus, has translational motion.
• This brings us to the definition of rotational motion; motion of a body is said to be rotational if all particles of the body move in a circle, the centres of which are situated on a straight line called axis of rotation.
• But in this case, as Fig. 7.2 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 with ‘something else’.

In order to understand what this ‘something else’ is, let us take a rigid body constrained from having translational motion.

In order to understand how a constrained body is, let us consider the body to be fixed along a straight line (such as, the blades of a ceiling fan). The only possible motion of such a rigid body is rotation.

The line along which the body is fixed is termed as its axis of rotation. For the same example of fan, the axis of rotation would be the line passing through the centre point of the small circular disk at the centre of a fan.

Other examples of rotation about an axis could be observed in a potter’s wheel, a giant wheel, a merry-go-round and so on (Fig 7.3(a) and (b)).

Let us try to account for what constitutes rotation. Again, only for our understanding, let us picture a fan as the rotational body, we will 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. The centre of the body (or bodies, as is the case for blades) converges at the axis of rotation.

Figure: 7.4: A rigid body rotation about the z-axis (Each point of the body such as $P_1$ or $P_2$ describes a circle with its centre $(C_1$ or $C_2)$ on the axis. The radius of the circle $(r_1$or $r_2)$ is the perpendicular distance of the point $(P_1$ or $P_2)$ from the axis. A point on the axis like $P_3$ remains stationary).Fig.7.4 shows the rotational motion of a rigid body about a fixed axis (the z-axis of the frame of reference).

• Refer figure, 7.4 above. Let $P_1$ be a particle of the rigid body, arbitrarily chosen and at a distance r1 from fixed axis. The particle $P_1$ describes a circle of radius $r_1$ with its centre $C_1$ on the fixed axis. The circle lies in a plane perpendicular to the axis.
• The figure also shows another particle $P_2$ of the rigid body, $P_2$ is at a distance $r_2$ from the fixed axis. The particle $P_2$ moves in a circle of radius $r_2$ and with centre $C_2$ on the axis. This circle, too, lies in a plane perpendicular to the axis.
• Note that the circles described by $P_1$ and $P_2$ may lie in different planes; both these planes, however, are perpendicular to the fixed axis.
• For any particle on the axis, like $P_3$, r = 0. Any such particle remains stationary while the body rotates. This is because, the axis is fixed.

• In some examples of rotation, however, the axis may not be fixed. A prominent example of this kind of rotation is a top spinning in place as shown in Fig. 7.5(a) above. (We assume that the top does not slip from place to place and so, does not have translational motion although the top may slip down straight, along a slope).
• Thus, the axis of such a spinning top moves around the vertical through its point of contact, sweeping out a cone as shown in Fig. 7.5(a) above. (This movement of the axis of the top around the vertical is termed precession.)
• Note that the point of contact of the top with ground is fixed. The axis of rotation of the top at any instant passes through the point of contact (here, the point is O).
• Another simple example of this kind of rotation is the oscillating table fan or a pedestal fan. You may have observed that the axis of rotation of such a fan has an oscillating (sidewise) movement in a horizontal plane; about the fixed vertical and the horizontal plane passes through the point at which the axis is pivoted (the point about which rotation takes place) on the vertical (here, point O in Fig. 7.5(b) above).
• While the fan rotates and its axis moves sidewise, this point remains fixed. Thus, in most cases of rotation, such as the rotation of a top or a pedestal fan, one point and not one line, of the rigid body is fixed.
• In this case, the (horizontal) axis is not fixed, though it always passes through the fixed point.
• We are only going to deal with the case of rotation in which the axis is fixed.
• The rolling motion of a cylinder down an inclined plane is a combination of rotation about a fixed axis (the line passing through the centre of the cylinder when placed vertically up) and translation (the movement of the cylinder along a straight path along the slope or inclined path).
• Thus, the ‘something else’ in the case of rolling motion which we referred to earlier is rotational motion.
• Both, Fig. 7.6(a) and (b) figures show motion of the same body along identical translational trajectory. In one case, Fig. 7.6(a), the motion is a pure translation; in the other case [Fig. 7.6(b)] it is a combination of translation and rotation. (You may try to reproduce in real the two types of motion shown using a rigid object like a heavy book to understand better.)

Thus, we can conclude that:

• The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation.
• The motion of a rigid body which is pivoted or fixed in some way is rotation. The rotation may be about an axis that is fixed (e.g. a ceiling fan) or moving (e.g. an oscillating table fan).

7.2 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 two-particle system.
• The two particles are considered to be joined by 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
  $$ X = \frac {m_1 x_1+m_2x_2}{m_1+m_2} ………….(7.1) $$

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
$$ X = \frac {mx_1+mx_2}{2m} = \frac {x_1+x_2}{2} $$

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
$$ X = \frac {m_1x_1+m_2x_2+……..+m_nx_n}{m_1+m_2+……+m_n}=\frac {\Sigma m_ix_i}{\Sigma m_i} ………………..(7.2)$$

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 Σ (the Greek letter sigma) denotes summation, in this case over n particles. The sum $ \Sigma m_t = M $ 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 $(x_1,y_1)$, $(x_2,y_2)$ 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
$$ X =\frac {m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} ……………..(7.3a) $$
$$ Y = \frac {m_1y_1 + m_2y_2 + m_3y_3}{m_1+m_2+m_3} ……………..(7.3 b) $$

For the particles of equal mass m = $m_1 = m_2= m_3$,

$$ X = \frac {m(x_1+x_2+x_3)}{3m} = \frac {x_1+x_2+x_3}{3} $$

$$ Y = \frac {m(y_1+y_2+y_3)}{3m}= \frac {x_1+x_2+x_3}{3} $$

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

$$ X = \frac {\Sigma m_ix_i}{M} ……….(7.4a) $$

$$ Y = \frac {\Sigma m_iy_i}{M} ………..(7.4 b) $$

And $ Z= \frac { \Sigma m_i z_i}{M} ……..(7.4c) $

Here M = Σ $m_i$ 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 $(x_i, y_i, z_i)$.

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:

$$ r_i= x_i \hat i + y_i \hat j + z_i \hat k $$
$$ and \space R = X \hat i + Y \hat j + Z \hat k $$
$$ Then \space R = \frac { \Sigma m_i r_i}{M} ……..(7.4d) $$

The 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 $Σm_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; $Δm_1$, $Δm_2$… $Δm_n$ ; the $i^{th}$ element $Δm_i$ is taken to be located about the point $(x_i, y_i, z_i)$. The coordinates of the centre of mass are then approximately given by $ X = \frac { \Sigma ( \Delta m_i)x_i}{\Sigma \Delta m_i},Y=\frac {\Sigma (\Delta m_i)y_i}{\Sigma \Delta m_i},Z=\frac { \Sigma (\Delta m_i)z_i}{\Sigma \Delta m_i}$

As we make n bigger and bigger and each $Δm_i$ smaller and smaller, these expressions become exact. In that case, we denote the sums over i by integrals. Thus,
$$ \Sigma \Delta m_i \rightarrow \int dm = M , $$
$$ \Sigma (\Delta m_i )x_i \rightarrow \int x \space dm, $$
$$ \Sigma ( \Delta m _i)y_i \rightarrow \int y \space dm , $$
$$ and \space ( \Delta m_i)z_i \rightarrow \int z \space dm $$

Here M is the total mass of the body. The coordinates of the centre of mass now are
$$ X = \frac {1}{M} \int x \space dm,Y= \frac {1}{M} \int y \space dm \space and \space Z = \frac {1}{M} \int z \space dm …….(7.5a)$$

The vector expression equivalent to these three scalar expressions is
$$ R =\frac {1}{M} \int r \space dm ……..(7.5b)$$

If we choose, the centre of mass as the origin of our coordinate system, R(x,y,z)=0
$$ i.e.., \int r \space dm = 0 $$
$$ or \space \int x \space dm = \int y \space dm = \int z \space dm = 0 ……..(7.6) $$

Example 7.1: 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.

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,
$$ X = \frac { m_1x_1 + m_2x_2 + m_3x_3}{m_1+m_2+m_3} $$
$$ = \frac { [100(0)+150(0.5)+200(0.25)] \space gm}{(100+150+200) \space g} $$
$$ = \frac {75+50}{450} \space m = \frac {125}{450} m = \frac {5}{18 } m $$
$$ Y = \frac { [100(0) + 150(0)+200(0.25 \sqrt {3})] \space gm}{450 \space g} $$
$$ = \frac {50 \sqrt {3}}{450} m =\frac { \sqrt 3}{9} m = \frac {1}{3\sqrt3} m $$
The centre of mass C is shown in the figure. Note that it is not the geometric centre of the triangle OAB. Why?

Example 7.2: Find the centre of mass of a triangular lamina.

Solution: The lamina (ΔLMN) 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.

Example 7.3: 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.

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 1m. The mass of each square is 1kg, 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 ,
$$ X = \frac { [1 (1/2) +1(3/2)+1(1/2)]kg \space m}{(1+1+1)kg} = \frac {5}{6} m $$
$$ Y = \frac { [1 (1/2) +1(1/2)+1(3/2)]kg \space m}{(1+1+1)kg} = \frac {5}{6} m $$
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?

Question for sections, 7.1.1 and 7.2:

1. Define:

a) Purely translational

b) Rotational motion

2. Explain rotation about a fixed axis with the necessary figure.

3. Give the equation for centre of mass of a system located at a distance X from O (equation 7.1). Also, arrive at the equation for coordinates (X and Y) of centre of mass. (That is, 7.4a and 7.4b)

7.3 Motion of Centre of Mass

• Now that we know what centre of mass of a body is, we can discuss its physical importance for a system of particles. We may rewrite Equation-7.4d as:
  $ MR= \Sigma m_ir_i =m_1r_1+m_2r_2+……….+m_nr_n …………(7.7) $

Differentiating the two sides of the equation with respect to time we get $ M \frac {dR}{dt}=m_1 \frac {dr_1}{dt}+m_2\frac {dr_2}{dt}+…..+m_n \frac {dr_n}{dt} $

Or

$$MV=m_1v_1+m_2v_2+…….+m_nv_n …………. (7.8) $$

where $v_1 =(dr_1/dt )$ is the velocity of the first particle $ v_2 =(dr_2/ dt)$ is the velocity of thesecond 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 \frac {dV}{dt}=m_1\frac {dv_1}{dt}+m_2\frac {dv_2}{dt}+….+ \frac {dv_n}{dt} $

Or

$$ MA=m_1a_1+m_2 a_2+….+m_na_n …………(7.9) $$

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 $F_1=m_1a_1$. The force acting on the second particle is given by $F_2=m_2a_2$ and so on. Eq. (7.9) may be written as
$ MA=F_1 +F_2 + …..+F_n …………… (7.10)$

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

$ MA = F_{ext} ……(7.11) $

• Where, $F_{ext}$ represents the sum of all external forces acting on the particles of the system.
• Equation 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 that point and all the external forces were applied at that point.
• The internal forces are disregarded while we determine the motion of the centre of mass.
• Thus, we need to know only the external forces.
• To obtain Eq. (7.11) we did not have to specify the nature of the system of particles. “nature” here refers to systems such as: collection of particles in which there may be all kinds of internal motions, or a rigid body which has either pure translational motion, or a combination of translational and rotational motion. No matter how the system is and the motion of its individual particles are, the centre of mass moves according to Eq. (7.11).
• By treating an object as system of particles, we can obtain the translational component of motion of the system, that is, motion of 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 acting on the system to be acting at the centre of mass.
• Thus, by this procedure of treating a body as a system of particles, we can find 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; as you will learn from the example below:

• A projectile, following the usual parabolic trajectory, explodes into fragments midway in air (as would a rocket or a fire-cracker in air).
• The explosion is caused due to internal forces. However, the forces do not contribute 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. Thus, gravitational force does not affect the motion of the projectile.
• But, the centre of mass under the influence of the external force continues. Even if the explosion had not occurred, the same parabolic trajectory would have been followed by the projectile. Thus, the motion of centre of mass is governed by an external force and any internal forces (such as, that force which causes explosion) has no effect on the centre of mass of a rigid body.
• This shows that internal forces do not affect the motion of centre of mass of the system; only the external force does.

7.4 Linear Momentum of a System of Particles

Let us recall that the linear momentum of a particle is defined as:

$$ p =mv ……..(7.12)$$
Let us also recall that Newton’s second law written in symbolic form for a single particle is
$$ F =\frac {dp}{dt} …….(7.13) $$
where F is the force on the particle. Let us consider a system of n particles with masses
$m_1, m_2,…m_n$ respectively and velocities $v_1,v_2 ,……. v_n$ respectively. The particles may be interacting and have external forces acting on them. The linear momentum of the first particle is $m_1 v_1$ , of the second particle is$m_2 v_2$ and so on.
For the system of n particles, the linear momentum of the system is defined to be the
vector sum of all individual particles of the system,
$$ P =p_1+p_2+….+p_n$$
$$ = m_1v_1+m_2v_2+…….+m_nv_n ……(7.14)$$
Comparing this with Eq. (7.8)
P= MV (7.15)
Thus, the total momentum of a system of particles is equal to the product of the
total mass of the system and the velocity of its centre of mass. Differentiating Eq. (7.15)
with respect to time,
$$ \frac {dP}{dt}=0 \space or \space P =Constant ……(7.18a)$$

• From equation, 7.18(a) we can say that when the total external force acting on a system of particles is zero, the total linear momentum of the system is constant. This is the law of conservation of the total linear momentum of a system of particles.
• From equation, 7.15, we can say that when the total external force on the system is zero, the velocity of the centre of mass remains constant.
• We assume throughout the discussion on systems of particles in this chapter that the total mass of the system remains constant.
• Due to the internal forces (or) the forces exerted by the particles on one another, the individual particles may have complicated trajectories.
• If the total external force acting on the system is zero, the centre of mass moves with a constant velocity, (that is, it moves uniformly in a straight line like a free particle), irrespective of the trajectories the individual particles of the system follow.
• The vector equation (7.18a) is equivalent to three scalar equations;

$$P_x = c_1, P_y = c_2 \space and \space P_z = c_3 ……………….. (7.18 b)
$$

Here, $P_x, P_y$ and $P_z$ are the components of the total linear momentum vector P along the x, y and z axes respectively; $c_1$, $c_2$ and $c_3$ are constants.

• In astronomy, binary (double) stars is a common occurrence. When there are no external forces, the centre of mass of a double star moves like a free particle, as shown in Fig.7.14 (a).
• The trajectories that the two stars (of equal mass) trace are also shown in the figure.
• If we choose the centre of mass as the frame of reference, then we find that the two stars are moving in a circle, about the centre of mass (which is at rest).
• At any point, the position of the stars will be diametrically opposite to each other [Fig. 7.14(b)].
• Thus, with respect to our frame of reference, the trajectories of the stars are a combination of
  (i) uniform motion in a straight line of the centre of mass and (ii) circular orbits of the stars about the centre of mass.
• As can be seen from the two examples, separating the motion of different parts of a system into motion of the centre of mass and motion about the centre of mass is a very useful technique that helps in understanding the motion of the system.

Questions from the sections: 7.3 to 7.6:

1. Arrive at the expression for forces acting on a system of particles in the form: F= MA (equation, 7.10).
2. Explain how internal forces do not affect the movement of centre of mass of a system with the illustration of an object in projectile motion.
3. Arrive at the equation for total momentum. (In the form of equation 7.15)
4. Explain the motion of centre of mass and motion about the centre of mass with the illustration of radioactive decay process and binary stars.

7.5 Vector Product of Two Vectors

Concept box: Vector product: The product of two real vectors in three dimensions which itself is a vector at right angles to both the original vectors. Its magnitude is the product of the magnitudes of the original vectors times the sine of the angle between their directions (=ab sinθ).

You are already familiar with scalar or dot product of two vectors in the previous chapter. In this section, we shall learn about another product of two vectors. This product is a vector.

Mathematical definition of Vector Product: A vector product of two vectors a and b is a vector c such that
(i) Magnitude of c = ab sinθ;
Where, “a” and “b” are magnitudes of vectors, a and b and “θ” is the angle between the two vectors.
(ii) c is perpendicular to the plane containing a and b.
(iii) it follows right handed screw rule, that is, 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.
This right-handed screw rule is illustrated in Fig. 7.15a.

Alternately, the right handed screw rule can be stated as: 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, as shown in Fig. 7.15b.

• 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.
• It should be remembered that there are two angles between any two vectors a and b as shown in Fig. 7.15 (a) or (b). They are “θ” and “3600– θ”.
• While applying either of the above rules, the rotation angle, θ should be considered (whose value is less than 1800) between a and b.
• Because of the cross used to denote the vector product, it is also referred to as cross product.

Properties of vector or cross product:

• Note that, scalar product of two vectors is commutative as we studied in the previous chapter, a.b = b.a; the vector product, however, is not commutative, i.e. a × b ≠ b × a.
• The magnitude of both “a × b” is equal to that of “b × a” and are both equal to: “ab sinθ”. The product obtained is perpendicular to a and b.
• But the rotation of the right-handed screw in case of “a × b” is from a to b, whereas in case of “b × a”, it is from b to a. This means the two vector products, c obtained in each case are in opposite directions (or);
  a × b = −b × a
• Another interesting property of a vector product is its behaviour under reflection. Under reflection (i.e. on taking the mirror image) we have x→ −x, y→−y and z→ −z. Thus, all the components of a vector change sign and thus a→ −a, b→ −b.
• What happens to the product of a × b under reflection?
  a × b→(−a)×(−b) = a × b
  Thus, a × b does not change sign under reflection.
• Vector product is distributive with respect to vector addition; as is a scalar product.
  Thus,
  a.(b+c)=a.b+a.c
  a×(b+c)=a×b+a×c
• We may write c = a × b in the component form. For this, we first need to obtain some elementary cross products:
  (i) a × a = 0 (0 is a null vector, i.e. a vector with zero magnitude)
  This follows since magnitude of a × a will be given by “asinθ x asinθ= asin 0 x asin 0 = $a^2$ sin 0°= 0”. “θ” is zero since there is no angle formed when the two vectors coincide with each other.

From this follow the results
$$ \hat i \times \hat i =0, \hat j \times \hat j=0,\hat k \times \hat k =0 $$
$$ (ii) \hat i \times \hat j = \hat k $$

Note that the magnitude of $ \hat i \times \hat j $ is sin$90^0$ or 1, since $ \hat i $ and $ \hat j $ both have unit magnitude and the angle between them is $90^0$.Thus, $ \hat i \times \hat j $ is a unit vector. A unit vector perpendicular to the plane of $ \hat i $ and $ \hat j $ and related to them by the right hand screw rule is $ \hat k $. Hence, the above result. You may verify similarly,
$$ \hat j \times \hat k = \hat i \space and \space \hat k \times \hat i = \hat j $$

From the rule for commutation of the cross product, it follows:
$$ \hat j \times \hat i = – \hat k, \hat k \times \hat j = -\hat i , \hat i \times \hat k = – \hat j $$

Note if $ \hat i,\hat j ,\hat k $ occur cyclically in the abovevvector product relation, the vector product is positive. If $ \hat i,\hat j ,\hat k $ do not occur in cyclic order,the vector product is negative.

Now,
$$ a \times b = (a_x \hat i + a_y \hat j+a_z \hat k)\times (b_x \hat i +b_y \hat j + b_z \hat k )$$
$$ =a_x b_y \hat k-a_xb_z \hat j-a_yb_x \hat k-a_yb_z \hat i +a_zb_x \hat j-a_zb_y \hat i $$
$$=(a_yb_z-a_zb_x)\hat i+(a_zb_x –a_xb_z)\hat j+(a_xb_y-a_yb_x)\hat k $$

We have used the elementary cross products in obtaining the above relation. The expression for a × b can be put in a determinant form which is easy to remember.
$$ a \times b =\begin{vmatrix} \hat i & \hat j & \hat k \ a_x & a_y & a_z \ b_x & b_y & b_z \end{vmatrix} $$

Example 7.4: Find the scalar and vector products of two vectors. $ a = (3 \hat i – 4 \hat j + 5 \hat k )$ and $ b = (– 2 \hat i + \hat j – 3 \hat k )$

Solution:

$$ a.b = (3 \hat i – 4 \hat j + 5 \hat k ).( – 2 \hat i + \hat j – 3 \hat k ) $$

$$ = -6-4-15 $$

$$ = -25 $$

$$ a ×b = [
M=
\begin{bmatrix}
\hat i & \hat j & \hat k \
3 &-4 & 5 \
-2 & 1 & -3
\end{bmatrix}
]

Note $ b×a=−7\hat i+ \hat j+5 \hat k $

7.6 Angular Velocity and its Relation with Linear Velocity

• In this section, we shall study what angular velocity is and its role in rotational motion. We have seen that every particle of a rotating body moves in a circle. The linear velocity of the particle is related to the angular velocity.
• The relation between Linear Velocity and Angular Velocity involves a vector product which we learnt about in the last section.
• Let us go back to Fig. 7.4.; In rotational motion of a rigid body about a fixed axis, every particle of the body (picture blades of a fan and assume a particle or point on it) moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis.

Fig. 7.16 Rotation about a fixed axis. (A particle (P) of the rigid body rotating about the fixed z-axis moves in a circle with centre (C) on the axis.)

• In Fig. 7.16, we redraw Fig. 7.4, showing a typical particle (at a point P) of the rigid body rotating about a fixed axis (taken as the z-axis). The particle describes a circle with a centre C on the axis. The radius of the circle is r (which is also the perpendicular distance of the point P from the axis).
• We also show the linear velocity vector v of the particle at P. As you must have learnt from the preceding chapters, the linear velocity is along the tangent at P to the circle.
• Let P′ be the position of the particle after an interval of time Δt (refer Fig. 7.16). Also, the particle sweeps an angle, PCP′ with an angular displacement, Δθ in time Δt.
• The average angular velocity of the particle (over the interval Δt) is: Δθ/Δt.
• As “Δt” tends to zero (that is, as Δt takes smaller and smaller values), the ratio “Δθ/Δt” approaches a limiting value which is the instantaneous angular velocity; given by, dθ/dt of the particle at the position P. (it would be useful to recollect differentiation at this point, differentiating displacement with respect to time would give: velocity/time which is equal to acceleration.)
• We denote the instantaneous angular velocity by “ω” (the Greek letter omega).
• We know from our study of circular motion that the magnitude of linear velocity “v” of a particle moving in a circle is related to the angular velocity of the particle, ω by the simple relation v =ωr, where r is the radius of the circle.
• We observe that at any given instant, the relation v=ωr applies to all particles of the rigid body.
• Thus, for a particle at a perpendicular distance $r_i$ from the fixed axis, the linear velocity at a given instant $v_i$ is given by:
  $$v_i= ωr_i ……. (7.19)$$
• Where the value of “i” varies from 1 to n; “n” being the total number of particles of the body.
• For particles which make up the axis, r = 0 (since distances are measured from the axis itself), and hence v = ω r = 0. We can conclude that particles on the axis are stationary. This also verifies that the axis is fixed.
• Since we use the same angular velocity ω for all the particles, ω refers to the angular velocity of the whole body as well.
• Like in pure translation motion (in which all particles of the have the same velocity at any instant of time), a body in pure rotational motion has the same angular velocity at any instant of time.
• Note that this characterisation of the rotation for a rigid body about a fixed axis is just another way of saying that each particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has the centre on the axis, as already discussed.

Note box:   1. Angular velocity is a vector. 2. For rotation about a fixed axis, the angular velocity vector lies along the axis of rotation and points out in the direction in which a right handed screw would advance, if the head of the screw is rotated with the body. (See Fig. 7.17a). The magnitude of this vector is: ω=dθ/dt.

We shall now look at what the vector product ω × r corresponds to.

• Refer to Fig. 7.17(b) above, which is a part of Fig. 7.16 reproduced to show the path of the particle P.
• The figure shows the vector, “ω” is directed along the fixed z-axis and also the position vector r = OP of the particle at P of the rigid body with respect to the origin O.
• Note that the origin is chosen to be on the axis of rotation.

Now, ω × r = ω × OP = ω × (OC+CP)
But, ω × OC=0 ……as ω is along OC
Hence, ω × r = ω × CP

• The vector ω × CP is perpendicular to ω, hence, to CP, the z-axis and also to the radius of the circle described by the particle at P. It is therefore, along the tangent to the circle at P.
• Also, the magnitude of ω × CP is ω(CP) since ω and CP are perpendicular to each other. We shall denote CP by r⊥ and not by r, as we did earlier.
• Thus, ω × r is a vector of magnitude ωr⊥ and is along the tangent to the circle described by the particle at P.
• The linear velocity vector v at P has the same magnitude and direction.
  Thus, from last two statements, v=ω × r….………… (7.20)
• In fact, the relation, Eq. (7.20), holds good even for rotation of a rigid body with one point fixed, such as the rotation of the top [refer Fig. 7.6(a)]. In this case, “r” represents the position vector of the particle with respect to the fixed point taken as the origin.

Note box: 1. For rotation about a fixed axis, the direction of the vector, “ω” does not change with time. Its magnitude may, however, change from instant to instant. 3. For the more general rotation, like we saw in a spinning top, both the magnitude and the direction of ω may change from instant to instant.

Questions from sections 7.5 and 7.6:

1. What is a vector product? Also give its mathematical definition.
2. List the properties of a vector product.
3. State the right hand screw rule. With the help of figure, 7.17 explain how angular velocity is measured and also, how the direction is determined.

7.6.1 Angular acceleration

• As you may have noticed, we are developing the study of rotational motion along the lines of the study of translational motion which we are already familiar.
• Analogous to the kinetic variables of linear displacement and velocity (v) in translational motion, we have angular displacement and angular velocity (ω) in rotational motion.
• Similarly, we will study the concept of angular acceleration in analogy with linear acceleration defined as the time rate of change of velocity in translational motion. We define angular acceleration, α as the time rate of change of angular velocity; Thus, $$ α = \frac {d ω}{dt} ……….(7.21) $$

If the axis of rotation is fixed, the direction of ω and hence, that of α is fixed. In this case, the vector equation reduces to a scalar equation,
$$ α = \frac {d ω}{dt} ….(7.22)$$

Q and A:

When the axis is fixed, why do you think the direction of angular velocity and angular fixed (or the same throughout)?
The direction of the angular velocity is always at 900 to the axis of rotation of the body. If the axis is now fixed, the direction of angular velocity is along the axis of rotation. Hence, the angular acceleration is also fixed (in terms of direction). Acceleration is only a function of velocity and time (and time is a scalar), angular acceleration is also along the direction of angular velocity. Hence, for a fixed axis, the direction of angular velocity and the direction of angular acceleration are constant.

7.7.1 Moment of force (Torque)

• If a body is fixed at a point or along a line, it has only rotational motion.
• We know that force is needed to change the translational state of a body in order to produce linear acceleration. Similarly, let us try to understand in rotational motion, the force that produces angular acceleration.

Example box Let us take the example of opening or closing of a door. A door is a rigid body which can rotate about a fixed vertical axis passing through the hinges. A force is needed to cause this rotation. But any force does not do the job. A force applied to the hinge line cannot produce any rotation at all; whereas, a force of given magnitude applied at right angles to the door at its outer edge is most effective in producing rotation. Thus, it is not just the force alone, but how and where the force is applied is important in rotational motion.

The rotational analogue of force (in translation motion) is moment of force. It is also referred to as torque. (We shall use the words moment of force and torque interchangeably.)

We shall first define the moment of force for the special case of a single particle. Later on, we shall extend the concept to systems of particles including rigid bodies.

• If a force acts on a single particle at a point P, whose position with respect to the origin, O is given by the position vector, r (Fig. 7.18), the moment of the force (or torque) acting on the particle with respect to the origin, O is defined as the vector product of:
  τ = r × F …… (7.23)
  Alternatively, Torque about a reference point is defined as the cross product of position vector of point of application of the force and the force.

• The moment of force (or torque) is a vector quantity. The symbol τ stands for the Greek letter, tau. The magnitude of τ is
  τ = rF sinθ ………… (7.24a)
  Where, “r” is the magnitude of the position vector r, i.e. the length OP; “F” is the magnitude of force F and θ is the angle between r and F as shown.

Note box: 1. Moment of force has dimensions $ML^2T^{-2}$. Its dimensions are the same as those of work or energy. However, moment of a force is a vector, while work done is a scalar. 2. The SI unit of moment of force is Newton-metre (Nm).

• The magnitude of the moment of force may be written as,
  τ=(rsinθ)F=r⊥F …………… (7.24b)
  (or) τ=rF sinθ=rF⊥ ……….. (7.24c)

Where, r⊥ = r sinθ is the perpendicular distance of the line of action of F from the origin and F⊥(=Fsinθ) is the component of F in the direction perpendicular to r.

Note box: τ = 0 if r = 0 (or) F = 0 (or) θ = (00 or 1800)

• Thus, the moment of a force vanishes if either the magnitude of the force is zero, or if the line of action of the force passes through the origin.
• One may note that since r × F is a vector product, properties of a vector product of two vectors apply to it.
• Refer figure 7.18 again, if the direction of F is reversed, the direction of the moment of force is reversed. If directions of both r and F are reversed, the direction of the moment of force remains the same. Insert the value and direction (by using signs) in the formula for finding moment of force to know how they remain same or differ in other cases.

Questions from sections: 7.6 to 7.7.1:

1. Define mathematically, angular acceleration.
2. Define mathematically, torque.
3. Give the SI unit of moment of force (torque) and also its dimensions.

7.7.2 Angular momentum of a particle

• We shall now study angular momentum for rotation as we studied linear momentum for translation. We shall first define angular momentum for the special case of a single particle. We shall then extend the definition of angular momentum to systems of particles including rigid bodies.
• Like moment of a force, angular momentum is also a vector product. It could also be referred to as moment of linear momentum. From this term one could guess how angular momentum is defined.
• Consider a particle of mass “m” and linear momentum “p” at a position “r” relative to the origin, “O”. The angular momentum, “l” of the particle with respect to the origin “O” is defined to be

l = r × p ……………… (7.25a)

The magnitude of the angular momentum vector is;
l = r p sinθ ………….. (7.26a)

Where, p is the magnitude of p and θ is the angle between r and p. We may write

l=rp⊥ (or) =r⊥p ………. (7.26b)

Where, r⊥ (= r sinθ) is the perpendicular distance of the directional line of p from the origin and p⊥ (=p sinθ) is the component of p in a direction perpendicular to r.

• We expect the angular momentum to be zero (l = 0), if the linear momentum vanishes (p = 0), if the particle is at the origin (r = 0), or if the directional line of p passes through the origin θ = 00 or 1800.
• Just as force and linear momentum are related, moment of a force and angular momentum have an important relation between them as well. For deriving the relation in the context of a single particle, we differentiate l = r × p with respect to time,

$$ \frac {dl}{dt} = \frac {d}{dt}(r \times p)$$
Applying the product rule for differentiation to the right hand side,
$$ \frac {d}{dt} (r \times p)= \frac {dr}{dt} \times p+r \times \frac {dp}{dt}$$
Now, the velocity of the particle is v = dr/dt and p = m v
Because of this $ \frac {dr}{dt} \times p = v \times m \space v = 0 $
as the vector product of two parallel vectors vanishes. Further, since dp / dt = F,
$$ r \times \frac {dp}{dt} =r \times F =τ $$
Hence $ \frac {d}{dt}(r \times p )= τ $
Or $ \frac {dl}{dt}=τ …….(7.27)$

• Thus, the time rate of change of the angular momentum of a particle is equal to the torque acting on it. This is the rotational analogue of the equation F = dp/dt, which is in accordance with Newton’s second law for the translational motion of a single particle.

Torque and angular momentum for a system of particles

• To get the total angular momentum of a system of particles about a given point we need to add vectorially (in other words, by method of vector addition), the angular momenta of individual particles. Thus, for a system of n particles,

$$ \vec L = \vec l_1+ \vec l_2+……+ \vec l_n= \Sigma_{t=1}^{n} \vec l_i $$

• The angular momentum of the $i^{th}$ particle is given by;
  $$ \vec l_i = \vec r_i \times \vec p_i $$
  Where, $r_i$ is the position vector of the $i^{th}$ particle with respect to a given origin and p = ($m_iv_i$) is the linear momentum of the particle. (The particle has mass $m_i$ and velocity $v_i$). We may write the total angular momentum of a system of particles as;

$$ \vec L = \Sigma \vec l_i = \Sigma \vec r_i \times \vec p_i ……(7.25b)$$

• The above equation is thus clearly the definition of angular momentum (Eq. 7.25a) for a single particle – extended to a system of particles.
• Using Eqs. (7.23) and (7.25b), we get;

$$ \frac {d \vec L}{dt}= \frac {d}{dt} (\Sigma \vec l_i)=\Sigma_i \frac {d \vec l_i}{dt}=\Sigma_i \vec τi …..(7.28a)$$ Where $\vec τ_i$ is the torque acting on the $i^{th}$ particle: $$ \vec τ_i = \vec r_i \times \vec F_i $$ The force $\vec F_i$ on the $i^{th}$ particle is the vector sum of external forces $\vec F_i^{ext}$ acting on the particle and the internal forces $\vec F_i^{int}$ exerted on it by the other particles of the system .We may therefore separate the contribution of the external and the internal forces to the total torque. $$ \vec τ = \Sigma_i \vec τ_i = \Sigma_i \vec r_i \times \vec F_i $$ $$ \vec τ = \vec τ{ext} + \vec τ{int} $$ Where $ \vec τ{ext} = \Sigma_i \vec r_i \times \vec F_i^{ext} $
And $ \vec τ_{int} = \Sigma \vec r_i \times \vec F_f^{int} $

• We shall now assume Newton’s third law that, the forces between any two particles of the system is equal and opposite and also assume that these forces are directed along the line joining the system of particles.
• In this case, the torque due to total internal forces is zero since the torque resulting from each action-reaction pair of forces is zero. Therefore, $ϯ{int}$ = 0 and thus, $ϯ = ϯ{ext}$ (that is, total torque is equal to the torque caused due to the external force).

Since $ \vec τ = \Sigma \vec r_i $, it follows from Eq.(7.28 a)
That $ \frac {d \vec L}{dt} = \vec τ_{ext} ……(7.28b)$

• Thus, the time rate of change of total angular momentum acting on the system taken about the same point as obtained in equation, 7.28b is the extension of single particle case as obtained earlier in the form of equation, 7.23.
• Also, the equation is a rotational analogue of :
  $$dP/dt=F_{ext} …………. (7.17) $$

Note box: 1. A system constituted of one single particle does not experience any force or torque. 2. As does equation 7.17 hold good for translation, equation 7.28b also holds good for any system of particles: either rigid, or even if the internal particles of the system have all kinds of internal motion.

Conservation of angular momentum:

If $ \vec τ_{ext}=0$.Eq.(7.28b) reduces to
$$ \frac {d \vec L}{dt}=0 $$
Or $ \vec L $ = constant (7.29a)
Thus,if the total external torque on a system of particles is zero,then the angular momentum of the system is conserved i.e, remains constant.Eq (7.29a) is equivalent to three scalar equations,
$$ L_x=K_1,L_y=K_2 \space and \space L_z = K_3 \hspace{10mm} (7.29b)$$

Here, $K_1$, $K_2$, $K_3$ are constants. $L_x$, $L_y$, $L_z$ are the components of total angular momentum vector, L along the x, y and z axis respectively.

• Thus, by saying, “total angular momentum of the system is conserved”, we must understand that it means: the angular momentum along these three components is conserved.
• Equation, 7.29a is the rotational analogue of 7.18a (which is law of conservation of linear momentum) for a system of particles. We shall study the applications of this later in this chapter.

Example 7.5: Find the torque of a force $ 7\hat i+ 3 \hat j-5 \hat k $ about the origin .The force acts on a particle whose position vector is $ \hat I – \hat j + \hat k $.

Solution: Here $ r = \hat I – \hat j + \hat k $ and F =$ 7\hat i+ 3 \hat j-5 \hat k $
$ \tau $ =[
M=
\begin{bmatrix}
\hat i & \hat j & \hat k \
1 & -1 & 1 \
7 & 3 & -5
\end{bmatrix}
]
$ =(5-3) \hat i-(-5-7) \hat j + (3-(-7)) \hat k $
$ or \space \tau = 2 \hat i + 12 \hat j +10 \hat k $

Example 7.6: Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

Solution: Let the particle with velocity v be at point P at some instant t. We want to calculate the angular momentum of the particle about an arbitrary point O.

The angular momentum is l = r × mv. Its magnitude is mvr sinθ, where θ is the angle between r and v as shown in Fig. 7.19. Although the particle changes position with time, the line of direction of v remains the same and hence OM = r sin θ. is a constant. Further, the direction of l is perpendicular to the plane of r and v. It is into the page of the figure. This direction does not change with time. Thus, l remains the same in magnitude and direction and is therefore conserved. Is there any external torque on the particle?

7.8 Equilibrium of a Rigid Body

• We are now going to concentrate on the motion of rigid bodies rather than on the motion of general systems of particles. We shall recapitulate what effect the external forces have on a rigid body.
• A rigid body is said to be in mechanical equilibrium, if both its linear momentum and angular momentum are not changing with time, or equivalently, the body has neither linear acceleration nor angular acceleration. This means;
  (1) The total force, that is, the vector sum of the forces, on the rigid body is zero;
  $$ \vec F_1 + \vec F_2 +…….+ \vec F_n = \Sigma_{i=1}^{n} \vec F_i = 0 ……..(7.30a)$$
  If the total force on the body is zero, then, total linear momentum of the body does not change with time. Equation 7.30a gives the condition for the translational equilibrium of the body.
  (2) The total torque, i.e. the vector sum of the torques on the rigid body is zero,
  $$ \vec τ1 + \vec τ_2+……..+ \vec τ_n = \Sigma{i=1}^{n} \vec τ_i = 0 ……..(7.30b) $$
  If the total torque on the rigid body is zero, the total angular momentum of the body does not change with time.
• Equation 7.30b gives the condition for the rotational equilibrium of the body.

Q and A:

Does the rotational equilibrium condition [Eq. 7.30(b)] remain valid even if the origin with respect to which the torque is determined is shifted?
One can show that if the translational equilibrium condition [Eq. 7.30(a)] holds for a rigid body, then such a shift of origin does not matter. In other words, the rotational equilibrium condition is independent of the location of the origin about which the torques are taken as long as the translational equilibrium condition is satisfied. We shall learn about it in detail in the example problems that will follow.
$$ \Sigma_{i=1}^{n} F_{ix} = 0 , \Sigma_{t=1}^{n} F_{iy} = 0 \space and \space \Sigma_{t=1}^{n} F_{iz}=0 …..(7.31a) $$
Where $F_{ix}$,$F_{iy}$ and $F_{iz}$ are respectively the x,y and z components of the forces $\vec F_i$.Similaly.Eq.(7.30b) is equivalent to three scalar equations
$$ \Sigma_{i=1}^{n} τ{ix}=0,\Sigma{i=1}^{n} τ{iy} =0 \space and \space \Sigma{i=1}^{n} τ{iz} =0 ……(7.31b) $$ Where $ τ{ix}, τ{iy}$ and $ τ{iz}$ are respectively the x,y and z components of the torque $ \vec τ_i$

• Equations, (7.31a) and (7.31b) gives six independent conditions that have to be satisfied for mechanical equilibrium of a rigid body.
• For forces acting in the same plane (coplanar), we need only three conditions to be satisfied for mechanical equilibrium.
• Two of these conditions correspond to translational equilibrium; the sum of components of forces along any two perpendicular axes in the plane must be zero.
• The third condition corresponds to rotational equilibrium. The sum of the components of the torques along any axis perpendicular to the plane of the forces must be zero.
• Thus, for equilibrium of a particle, the vector sum of all the forces on it must be zero. Since all these forces act on the single particle, they must be concurrent.
• A body may be in partial equilibrium, that is, it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium.
• For instance, consider a light (i.e. of negligible mass) rod (AB), at the two ends (A and B) of which two parallel forces both equal in magnitude are applied perpendicular to the rod as shown in Fig. 7.20(a).

• Let C be the midpoint of AB, CA = CB = a. the moment of the forces at A and B will both be equal in magnitude (aF), but opposite in sense as shown. The net moment on the rod will be zero. The system will be in rotational equilibrium, but it will not be in translational equilibrium; ΣF≠0.

• The force at B in Fig. 7.20(a) is reversed in Fig. 7.20(b). Thus, we have the same rod with two equal and opposite forces applied perpendicular to the rod, one at end A and the other at end B.
• Here, the moments of both the forces are equal, but they are not opposite. However, they act in the same manner and cause anticlockwise rotation of the rod.
• The total force on the body is zero; so the body is in translational equilibrium (that is, the body is either not moving or is moving at a constant velocity); but it is not in rotational equilibrium although the rod is not fixed in any way and hence undergoes pure rotation (i.e. rotation without translation).

Note Box: Line of action: Line through the point at which the force is applied.

• A pair of equal and opposite forces with different lines of action is known as a couple. A couple produces rotation without translation.

Questions from sections, 7.7.2 to 7.8:

1. Give the expression for angular momentum.
2. When is angular momentum zero?
3. Arrive at the equation 7.27 which is expression for torque on a single particle. Also derive from it, the same for a system of particles in the form of equation, 7.28a.
4. When is a body said to be in mechanical equilibrium? Explain with the two necessary equations.
5. When do you say that a body is in partial equilibrium?
6. Define couple; explain with an example, its effect.

7.8.1 Principle of moments

• An ideal lever is essentially a light (that is, of negligible mass) rod pivoted at a point along its length. This point is called the fulcrum.
• A seesaw on the children’s playground is a typical example of a lever. Two forces $F_1$ and $F_2$, parallel to each other and usually perpendicular to the lever, as shown in figure, 7.23, act on the lever at distances d1 and d2 respectively from the fulcrum.

Note box: 1. Upward forces such as reaction force (here, R) will be denoted by a positive sign; likewise, downward with a negative. 2. The anti-clockwise moments are taken as positive and clockwise as negative.

• The lever is a system in mechanical equilibrium.
• Let R be the reaction of the support at the fulcrum; R is directed opposite to the forces $F_1$ and $F_2$. For translational equilibrium,
  $$R – F_1 – F_2 = 0 ……… (i) $$
• For considering rotational equilibrium we take the moments about the fulcrum; the sum of moments must be zero,
  $$d_1F_1 – d_2F_2 = 0 …….. (ii)$$
• R acts at the fulcrum itself; thus has zero moment about the fulcrum (since distances are taken from the point of fulcrum itself).
• In the case of the lever, force, $F_1$ is usually some weight to be lifted. It is called the load and its distance from the fulcrum, $d_1$ is called the load arm. Force, F2 is the effort applied to lift the load. Distance, $d_2$ is where the force or the “effort” is applied from the fulcrum and is called, the effort arm.

Eq. (ii) can be written as $d_1F_1 = d_2 F_2$ ………… (7.32a)
(or) load arm × load = effort arm × effort

The above equation expresses the principle of moments for a lever.

By re-arranging the above expression, we get the ratio: $F_1/F_2$ which is called, the Mechanical Advantage (M.A.). Thus, we obtain the expression:

$$  MA= (F_1/F_2) = (d_2/d_1) …………..       (7.32b)$$

• We can infer from the equation that, if the effort arm, $d_2$ is larger than the load arm, the mechanical advantage is greater than one.
• The Mechanical advantage being greater than one means that a small effort can be used to lift a large load.
• There are several examples of a lever around you besides the see-saw. The beam of a balance is a lever.

Note box: The principle of moment holds good even when the parallel forces $F_1$ and $F_2$ are not perpendicular to the lever, but act at some angle.

7.8.2 Centre of gravity

Therefore, the total gravitational torque about the CG is zero, that is,
$$ τ_g = \Sigma τ_i = \Sigma r_i \times m_i g = 0 ……(7.33) $$

• We will now define the Centre of Gravity for a body as that point where the total gravitational torque on the body is zero. We notice that in Eq. (7.33), g is the same for all particles, and hence it comes out of the summation. This gives (since g is nonzero),
  $$ \Sigma m_ir_i= 0 $$
• The position vectors ($r_i$) is taken with respect to the CG. In accordance with Equation-7.4a (go back and refer section- 7.2, equation- 7.4a): if $X=\Sigma(m_ix_i)/M=0$,the origin must be the centre of mass of the body (since, the only possible variable in the equation that can take a zero-value is $x_i$and CG would be the origin itself because distances along x-axis are taken from the origin). Thus, the centre of gravity of the body coincides with the centre of mass.
• This is true because the body being small, g (or CG) does not vary from one point of the body to the other. If the body is so extended (or is big enough) that g varies from part to part of the body, then, the centre of gravity and centre of mass will not coincide. Basically, the two are different concepts. The centre of mass has nothing to do with gravity. The CG depends only on the distribution of mass of the body.
• In Section 7.2, we found out the position of the centre of mass of several regular, homogeneous objects. Obviously the method used there gives us also the centre of gravity of these bodies, if they are small enough.

Example 7.8: A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg weight is suspended at 30 cm from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.)

Solution:

Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP = 30 cm, PG = 5 cm, $AK_1= BK_2$ = 10 cm and $ K_1G = K_2G $ = 25 cm. Also, W= weight of the rod =4.00 kg and $W_1$= suspended weight = 6.00 kg; $R_1$ and $R_2$ are the normal reactions of the support at the knife edges.

For translational equilibrium of the rod,
$$ R_1+R_2 –W_1 –W = 0 \hspace{20mm} (i) $$

Note $W_1$ and W act vertically down and $R_1$ and $R_2$ act vertically up.

For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of $R_2$ and $W_1$ are anticlockwise (+ve), whereas the moment of $R_1$ is clockwise (-ve).
For rotational equilibrium,
$$ –R_1 (K_1G) + W_1 (PG) + R_2 (K_2G) = 0 \hspace{10mm} (ii) $$
It is given that W = 4.00g N and $W_1$ = 6.00g N, where g = acceleration due to gravity. We take g = 9.8 m/s$^2$.
With numerical values inserted, from (i)
$$ R_1 + R_2 – 4.00g – 6.00g = 0 $$
$$ or \space R_1 + R_2 = 10.00g \space N \hspace{10mm} (iii) $$
$$= 98.00 \space N $$
From (ii) $– 0.25 \space R_1 + 0.05 \space W_1 + 0.25 \space R_2 = 0 $
$$ or \space R_2 – R_1 = 1.2g \space N = 11.76 \space N \space (iv) $$
From (iii) and (iv), $R_1 $= 54.88 N,$R_2$ = 43.12 N
Thus the reactions of the support are about
55 N at $K_1$ and 43 N at $K_2$.

Example 7.9: A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig.7.27. Find the reaction forces of the wall and the floor.

Solution: The ladder AB is 3 m long, its foot A is at distance AC = 1 m from the wall. From Pythagoras theorem, BC = 2 2 m. The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces $F_1$ and $F_2$ of the wall and the floor respectively. Force $F_1$ is perpendicular to the wall, since the wall is frictionless. Force $F_2$ is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.
For translational equilibrium, taking the forces in the vertical direction,
$$ N – W = 0 \hspace{10mm} (i) $$
Taking the forces in the horizontal direction,
$$F – F_1 = 0 \hspace{10mm} (ii) $$
For rotational equilibrium, taking the moments of the forces about A,
$$ 2 \sqrt {2}F_1 − (1/2) W = 0 \hspace{10mm} (iii) $$
Now W = 20 g = 20 × 9.8 N = 196.0 N
From (i) N = 196.0
From (iii) $ F_1=W/ 4 \sqrt {2}=196.0/4 \sqrt {2} = 34.6 \space N $$
From (ii) $ F=F_1 =34.6 \space N $
$$ F_2 = \sqrt { F^2 +N^2 }=199.0 \space N $$
The force $F_2$ makes an angle α with the horizontal,
$ tan \space α=N/ F=4 \sqrt 2 $ , $ α=tan^{−1}(4 \sqrt 2)≈80^o $

Questions from sections: 7.8.1 and 7.8.2:

1. Explain how Mechanical equilibrium is maintained in a lever system with necessary mathematical conditions.
2. Explain: load, load arm and effort arm with figure 7.23.
3. What is mechanical advantage? Give the equations associated with it (equations, 7.32a and 7.32b).
4. When is mechanical equilibrium greater than 1? What does this denote?
5. Define Centre of Gravity for a system of particles with the necessary expression.

7.9 Moment of Inertia

Definition box: Moment of inertia is the inherent property of an object by virtue of which it tends to remain in its state of rotational motion.

• We have already mentioned that we are developing the study of rotational motion parallel to the study of translational motion since we are familiar with it.
• We have yet to answer one major question in this connection. What is the analogue of mass in rotational motion? We shall attempt to answer this question in the present section.
• For this, we shall consider rotation about a fixed axis. Let us try to get an expression for the kinetic energy of a rotating body.
• We know that, for a body rotating about a fixed axis, each particle of the body moves in a circle with linear velocity given by equation, (7.19). (Refer to Fig. 7.16).
  For a particle at a distance from the axis, the linear velocity is $v_i= r_iω$. The kinetic energy of motion of this particle is:

$$ k_i = \frac {1}{2}m_iv_i^2=\frac {1}{2}m_ir^2_i \omega^2 $$
where $m_i$ is the mass of the particle. The total kinetic energy K of the body is then given by the sum of the kinetic energies of individual particles,
$$ K= \Sigma_{i=1}^{n}k_i=\frac {1}{2} \Sigma_{i=1}^{n}(m_ir^2_i\omega^2) $$
Here n is the number of particles in the body.Note ω is the same for all particles. Hence, taking ω out of the sum,
$$ K = \frac {1}{2} ω^2(\Sigma_{i=1}^{n} m_ir^2_i) $$
We define a new parameter characterising the rigid body, called the moment of inertia I , given by
$$ I = \Sigma_{i=1}^{n} m_ir^2_i ……(7.34) $$
With this definition,
$$ K = \frac {1}{2}I ω^2 …….(7.35)$$

Note box: The moment of inertia, “I” is independent of the magnitude of the angular velocity. It is a characteristic of the rigid body and the axis about which it rotates.

• Compare equation, (7.35) for the kinetic energy of a rotating body with the expression for the kinetic energy of a body in linear (translational) motion which is given by;
  $$ K = \frac {1}{2} mv^2 …(7.35)$$
  Here, “m” is the mass of the body and “v” is its velocity.
• Since we have already determined that angular velocity is the rotational analogue of linear velocity and by comparing the two equations of kinetic energy for a body in rotational motion and for a body in translation motion, we can conclude that the rotational analogue of mass is moment of inertia.
• We now apply the definition in equation- 7.34 to calculate the moment of inertia in two simple cases:

(a) Consider a thin ring of radius, R and mass, M, rotating in its own plane around its centre with angular velocity, ω. Each mass element of the ring is at a distance R from the axis, and moves with a speed, Rω. The kinetic energy is therefore,

$$ K = \frac {1}{2} Mv^2 = \frac {1}{2}MR^2 ω^2 $$
Comparing with Eq. (7.35), we get $I = MR^2$ for the ring.

(b) Next, take a rigid massless rod of length, l with a pair of small masses, rotating about an axis through the centre of mass perpendicular to the rod (Fig. 7.28). Each mass M/2 is at a distance l/2 from the axis. The moment of inertia of the masses is therefore given by;
$$ I = (M/2) (l/2)^2 + (M/2) (l/2)^2 $$

Thus, for the pair of masses, rotating about the axis through the centre of mass perpendicular to the rod;
$$ I = Ml^2 / 4$$

• Table 7.1 gives the moment of inertia of various familiar regular shaped solids about specific axis.
• As the mass of a body resists a change in its state of linear motion, mass is regarded a measure of its inertia in linear motion. Similarly, as the moment of inertia about a given axis of rotation resists a change in its rotational motion, it can be regarded as a measure of rotational inertia of the body.
• Rotational inertia is actually a measure of the way in which different parts of the body are distributed at different distances from the axis. Unlike the mass of a body, the moment of inertia is not a fixed quantity but depends on the orientation and position of the axis of rotation with respect to the body as a whole.
• As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we can define a new parameter, the radius of gyration. It is related to the moment of inertia and the total mass of the body.
• Notice from the Table 7.1 that in all cases, we can write $I = Mk^2$, where k has the dimension of length. For a rod, about the perpendicular axis at its midpoint, $k^2=L^2/12$, i.e.$ k^2 $=L/ . Similarly, k = R/2 for the circular disc about its diameter. The length k is a geometric property of the body and axis of rotation. It is called the radius of gyration. The radius of gyration of a body about an axis may be defined as the distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis.
• Thus, the moment of inertia of a rigid body depends on the mass of the body, its shape and size; distribution of mass about the axis of rotation, and the position and orientation of the axis of rotation.
• From the definition, Eq. (7.34), we can infer that the dimensions of moments of inertia we ML2 and its SI units are kg $m^2$.
• The property of this extremely important quantity, I as a measure of rotational inertia of the body has been put to a great practical use. The machines, such as steam engine and the automobile engine, etc., that produce rotational motion have a disc with a large moment of inertia, called a flywheel. Because of its large moment of inertia, the flywheel resists the sudden increase or decrease of the speed of the vehicle. It allows a gradual change in the speed and prevents jerky motions, thereby ensuring a smooth ride for the passengers on the vehicle.

Questions for the section, 7.9:

1. Define moment of inertia.
2. Arrive at the expression relating kinetic energy and moment of inertia of a rigid body in rotational motion in the form: $K= (1/2).I.ω^2$
3. With equations, explain the moment of Inertia of: (a) thin ring (b) rigid massless rod

7.10 Theorems of Perpendicular and Parallel Axes

These are the two useful theorems relating to moment of inertia. We shall first discuss the theorem of perpendicular axes and its application in working out the moments of inertia of some regular-shaped bodies.

Theorem of perpendicular axes:

• This theorem is applicable to bodies which are planar. In practice this means the theorem applies to flat bodies whose thickness is very small compared to their other dimensions (e.g. length, breadth or radius).
  Fig. 7.29 below illustrates the theorem and we shall use it as we define the theorem.
• Perpendicular axis theorem states that the moment of inertia of a planar body (lamina) about an axis perpendicular to the plane of the lamina is equal to the sum of moments of inertia of the lamina about the two axes at right angles to each other in its own plane and intersecting each other at the point where the perpendicular axis passes through it.

Figure 7.29: Theorem of perpendicular axes applicable to a planar body; x and y axes are two perpendicular axes in the plane and the z-axis is perpendicular to the plane.

• The figure shows a planar body. An axis perpendicular to the body through a point O is taken as the z-axis. Two mutually perpendicular axes lying in the plane of the body and concurrent with z-axis, i.e. passing through O, are taken as the x and y-axes. Then, the theorem can mathematically be expressed as:

$$ I_z=I_x + I_y ………… (7.36)$$

Let us look at the usefulness of the theorem through an example.

Example 7.10: What is the moment of inertia of a disc about one of its diameters?

Solution: We assume the moment of inertia of the disc about an axis perpendicular to it and through its centre to be known; it is MR2/2, where M is the mass of the disc and R is its radius (Table 7.1) The disc can be considered to be a planar body. Hence the theorem of perpendicular axes is applicable to it. As shown in Fig. 7.30, we take three concurrent axes through the centre of the disc, O as the x,y,z axes; x and y-axes lie in the plane of the disc and z is perpendicular to it. By the theorem of perpendicular axes,
$$ I_z =I_x +I_y $$
Now, x and y axes are along two diameters of the disc, and by symmetry the moment of inertia of the disc is the same about any diameter. Hence
$$ I_x = I_y $$
$$ and \space I_z = 2I_x $$
$$ But \space I_z = MR^2/2 $$
So finally, $ I_x = I_z/2 = MR^2/4 $
Thus the moment of inertia of a disc about any of its diameter is MR$^2$/4.

7.10.1 Theorem of parallel axes

• This theorem is applicable to a body of any shape. It allows us to find the moment of inertia of a body about any axis, given the moment of inertia of the body about a parallel axis through the centre of mass of the body.
• Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

• As shown in Fig. 7.31 above, z and z′ are two parallel axes separated by a distance, “a”. The z-axis passes through the centre of mass, “O” of the rigid body. Then according to the theorem of parallel axes,

$$ I_{z′}= I_z + Ma^2 ……………………… (7.37)$$
Where, $I_z$ and $I_{z’}$, are the moments of inertia of the body about the z and z′ axes respectively, M is the total mass of the body and “a” is the perpendicular distance between the two parallel axes.

Example 7.11: What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end?
Solution: For the rod of mass M and length l,$ I = Ml^2/12$. Using the parallel axes theorem,
$I′ = I + Ma^2$ with a = l/2 we get,
$$ I’ =M \frac {l^2}{12}+M \bigg( \frac {l}{2} \bigg)^2 = \frac {Ml^2}{3} $$
We can check this independently since I is half the moment of inertia of a rod of mass 2M and length 2l about its midpoint,
$$ I’ = 2M. \frac {4l^2}{12}× \frac {1}{2} = \frac {Ml^2}{3} $$$

Example 7.12: What is the moment of inertia of a ring about a tangent to the circle of the ring?

Solution: The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring. The distance between these two parallel axes is R, the radius of the ring. Using the parallel axes theorem,

$$ I_{tangent} = I_{dia}+MR^2 = \frac {MR^2}{2}+MR^2 = \frac {3}{2}MR^2 $$

7.11 Kinematics of Rotational Motion about a Fixed Axis

We have already indicated the analogy between rotational motion and translational motion. For example, the angular velocity, ω plays the same role in rotation as the linear velocity, v in translation. In this section, we shall take this analogy further.

• We can recall that, for specifying the angular displacement of the rotating body, we take any particle like “P” (refer, Fig.7.33) on the body.
• The angular displacement, θ that the particle sweeps in the plane is the angular displacement of the whole body. θ is measured in a fixed direction in the plane of motion of P, which we take to be the x′ – axis, chosen parallel to the x-axis.
• Note that, as shown, the axis of rotation is the z – axis and the plane of the motion of the particle is the x – y plane. Fig. 7.33 also shows $θ_0$, the angular displacement at t = 0.
• We also recall that the angular velocity is the time rate of change of angular displacement, ω = dθ/dt.

Thus, the angular acceleration is given by,
α = dω/dt

• The kinematic quantities in rotational motion, angular displacement (θ), angular velocity (ω) and angular acceleration (α) respectively correspond to kinematic quantities in linear motion: displacement (x), velocity (v) and acceleration (a).
• We know the kinematic equations of linear motion with uniform (or constant) acceleration:

$$ v=v_o+at ……..(a)$$
$$ x=x_o +v_ot + \frac {1}{2}at^2 ……..(b)$$
$$v^2=v^2_o+2ax ………………..(c) $$
where $x_0$ = initial displacement and $v_0$= initial velocity. The word ‘initial’ refers to values of the quantities at t = 0
The corresponding kinematic equations for rotational motion with uniform angular acceleration are:
$$ ω =ω_0 + \alpha t ……(7.38) $$
$$ θ =θ_o + ω_0t+ \frac {1}{2} \alpha t^2 ……(7.39)$$
$$ and \space ω^2 = ω_0^2+2 \alpha (θ-θ_o) …….(7.40) $$
where $θ_0$= initial angular displacement of the rotating body, and $ω_0$ = initial angular velocity
of the body.

Example 7.13: Obtain Eq. (7.38) from first principles.

Solution: The angular acceleration is uniform, hence
$$ \frac {dω}{dt} =α = constant \hspace{10mm} (i) $$
Integrating this equation,
$$ ω = \int α \space dt +c $$
=αt+c (as α is constant)
At t = 0, ω = $ω_0$ (given)
From (i) we get at t = 0, ω = c = $ω_0$
Thus, ω = αt + $ω_0$ as required.
With the definition of ω = dθ/dt we may integrate Eq. (7.38) to get Eq. (7.39). This
derivation and the derivation of Eq. (7.40) is left as an exercise.

Example 7.14: The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds.
(i) What is its angular acceleration, assuming the acceleration to be uniform?
(ii) How many revolutions does the engine make during this time?

Solution: (i) We shall use ω = $ω_0$ + αt
$ω_0$ =initial angular speed in rad/s
= 2π × angular speed in rev/s
$$ = \frac { 2π×angular \space speed \space in \space rev/min}{60 \space s/min} $$
$$ = \frac {2π×1200}{60} \space rad/s $$
= 40π rad/s
Similarly ω = final angular speed in rad/s
$$ = \frac {2π×3120}{60} \space rad/s $$
= 2π × 52 rad/s
= 104 π rad/s
∴ Angular acceleration
$$ α = \frac { ω –ω_o}{t} =4 π \space rad/s^2 $$
The angular acceleration of the engine = 4π rad/$s^2$
(ii) The angular displacement in time t is given by
$$ θ=ω_ot+ \frac {1}{2} αt^2 $$
$$ = ( 40 π×16+ \frac {1}{2}×4π×16^2) \space rad $$
=(640π+512π) rad
= 1152π rad
$$ Number \space of \space revolutions = \frac {1152π}{2 π} =576 $$

Questions from sections: 7.10 and 7.11:

1. Explain perpendicular and parallel axis theorem with necessary figures.

7.12 Dynamics of Rotational Motion about a Fixed Axis

• In rotational motion about a fixed axis, only those components of torques which are along the direction of the fixed axis need to be considered in our discussion. Only these components can cause the body to rotate about the axis.
• A component of the torque perpendicular to the axis of rotation will tend to turn the axis from its position. We specifically assume that there will arise, necessary forces of constraint to cancel the effect of the perpendicular components of the (external) torques, so that the fixed position of the axis will be maintained. The perpendicular components of the torque therefore, need not be taken into account.
• This means that for our calculation of torques on a rigid body:
  (1) We need to consider only those forces that lie in planes perpendicular to the axis. Forces which are parallel to the axis will give torques perpendicular to the axis and need not be taken into account.
  (2) We need to consider only those components of the position vectors which are perpendicular to the axis. Components of position vectors along the axis will result in torques perpendicular to the axis and need not be taken into account.

Work done by a Torque

• Figure 7.34 shows a cross-section of a rigid body rotating about a fixed axis, which is taken as the z-axis (perpendicular to the plane of the page; see Fig. 7.33).
• As mentioned earlier, we need to consider only those forces which lie in planes perpendicular to the axis.
• Again, let us refer figure 7.34 above. Let F1 be one such typical force acting as shown on a particle of the body at point $P_1$ with its line of action in a plane perpendicular to the axis. For convenience, we call this to be the x′–y′ plane.
• The particle at $P_1$ describes a circular path of radius $r_1$ with centre C on the axis; $CP_1 = r_1$.
• In time Δt, the point moves to the position, $P_1′$. The displacement of the particle is $ds_1$ and has magnitude: $ds_1 = r_1dθ$. The direction at point, $P_1$ is governed by the tangent at $P_1$ to the circular path as shown.
• Here, dθ is the angular displacement of the particle given by;
  $$ dθ = ∠P_1CP_1′$$
• The work done by the force on the particle is;

$$ dW_1 = F_1. Ds_1= F_1ds_1 cosφ_1= F_1(r_1 dθ)sinα_1 $$

• Where, $φ_1$ is the angle between $F_1$ and the tangent at $P_1$, and $α_1$ is the angle between $F_1$ and the radius vector $OP_1$; $φ_1 + α_1 = 90°$
• The torque due to $F_1$ about the origin is $OP_1 × F_1$. Now $OP_1 = OC + OP_1$. Refer to Fig. 7.17(b), (Since OC is along the axis, the torque resulting from it is excluded from our consideration) the effective torque due to $F_1$ is $τ1$= CP × $F_1$. $τ_1$ is directed along the axis of rotation and has a magnitude, $$ τ 1= r_1F_1 sinα;$$
  Also, by definition of work done, $dW_1 = τ_ 1dθ$
• If there are more than one forces acting on the body, the work done by all of them can be added to give the total work done on the body. Denoting the magnitudes of the torques due to the different forces as $τ_1, τ_2$,…etc,
  $$ dW = (τ_1+τ_2+…)dθ $$
• The forces giving rise to the torques act on different particles, thus, the value of torque differs from one particle to another. But the angular displacement, “dθ” is the same for all particles. Since all the torques considered are parallel to the fixed axis, the magnitude, τ of the total torque is just the algebraic sum of the magnitudes of the torques, i.e., τ =$ τ 1 + τ 2$ + …..
  We therefore have,
  dW =τdθ ……….. (7.41)
  This expression gives the work done by the total (external) torque, τ which acts on the body rotating about a fixed axis. It is analogous to the expression, dW= F ds for linear (translational) motion is obvious.
• Dividing both sides of Eq. (7.41) by dt gives,
  $$ P = \frac {dW}{dt}= τ \frac {dθ}{dt} = τ ω $$
  $$ or \space P = τ ω …..(7.42)$$

Where, P is the instantaneous power.

• Let us compare this expression for power in the case of rotational motion about a fixed axis with the expression for power in the case of linear motion as below;
  P = Fv
• In a perfectly rigid body, there is no internal motion. The work done by external torques is therefore, not lost in causing any internal motion but results in the increase in kinetic energy of the body.
• The rate at which work is done on the body is given by equation-7.42. This is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is thus given by,

τ= Iα …………. (7.43)

$$ \frac {d}{dt} \bigg( \frac {I ω^2}{2} \bigg) = I \frac {(2 ω)}{2} \frac {d ω}{dt} $$
We assume that the moment of inertia does not change with time. This means that the mass
of the body does not change, the body remains rigid and also the axis does not change its
position with respect to the body.
Since α=dω/dt, we get
$$ \frac {d}{dt} \bigg( \frac {I ω^2}{2} \bigg) =I ω α $$
Equating rates of work done and of increase in kinetic energy,
τω = Iω α
τ= Iα (7.43)

• Equation, 7.43 is similar to Newton’s second law for linear motion expressed symbolically as:
  F = ma
• Just as force produces acceleration, torque produces angular acceleration in a body. From the equation, we can infer that, the angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body.
• Equation- 7.43 can thus be called Newton’s second law for rotation about a fixed axis.

Example 7.15: A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 7.35. The flywheel is mounted on a horizontal axle with frictionless bearings.
(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when 2m of the cord is unwound.
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
(d) Compare answers to parts (b) and (c).

Solution:

(a) We use I α = τ the torque τ = F R
= 25 × 0.20 Nm (as R = 0.20m)
= 5.0 Nm
I = M. I. of flywheel about its axis= $ \frac {MR^2}{2} $
$$ = \frac { 20.0 × (0.2)^2}{2} = 0.4 \space kg \space m^2 $$
α = angular acceleration
= 5.0 N m/0.4 kg $m^2$ = 12.35 $s^{–2}$

(b) Work done by the pull unwinding 2m of the cord
= 25 N × 2m = 50 J

(c) Let ω be the final angular velocity. The kinetic energy gained = $ \frac {1 }{2}Iω^2 $
since the wheel starts from rest. Now,
$$ ω^2 =ω^2_o +2αθ, ω_o =0 $$

The angular displacement θ = length of unwound string / radius of wheel = 2m/0.2 m = 10 rad
$$ ω^2=2×12.5×10.0=250(rad/s)^2 $$

∴K E gained = $ \frac {1}{2} ×0.4×250 =50 \space J $

(d) The answers are the same, i.e. the kinetic energy gained by the wheel = work done by the force. There is no loss of energy due to friction.

7.13 Angular Momentum in case of Rotation about a Fixed Axis

• We have studied in section 7.7, the angular momentum of a system of particles. We already know from there, that the time rate of total angular momentum of a system of particles about a point is equal to the total external torque on the system taken about the same point.
• When the total external torque is zero, the total angular momentum of the system is conserved.
• The general expression for the total angular momentum of the system is;
  $$ L = \Sigma_{i=1}^{N} r_i \times p_i ……(7.25 b) $$
  We will now study the angular momentum for a special case of rotation about a fixed axis as below;
• We first consider the angular momentum for a particle of the rotating rigid body.
• We then sum up the contributions of individual particles to get momentum, L of the whole body.
• For any chosen particle on the body, l = r × p. As seen in the last section r = OP = OC + CP [refer Fig. 7.17(b)].

With p = mv
l= (OC x mv) + (CP x mv)
The magnitude of the linear velocity v of the particle at P is given by v = ω$r_⊥$ where $r_⊥$ is the length of CP or the perpendicular distance of P from the axis of rotation. Further, v is tangential at P to the circle which the particle describes. Using the right-hand rule one can check that CP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) is $µk$. Hence $$ CP x m v = r⊥ (mv) µk$$ $$ m r^2⊥ \Omega_k $$ since ( v = $\Omega r_⊥ $
Similarly, we can check that OC × v is perpendicular to the fixed axis. Let us denote the part of l along the fixed axis (i.e. the z-axis) by$ l_z$ , then

$$I_Z = CP m v = m r_⊥^2 \Omega µk $$ And $$ l = l_z + OC $$ We note that $l_z$ is parallel to the fixed axis, but l is not. In general, for a particle, the angular momentum l is not along the axis of rotation, i.e. for a particle, l and ω are not necessarily parallel. Compare this with the corresponding fact in translation. For a particle, p and v are always parallel to each other. For computing the total angular momentum of the whole rigid body, we add up the contribution of each particle of the body. Thus $$ L = \Sigma OC_i m_i v_i …..(7.44 a)$$ where $m_i$ and $v_i$ are respectively the mass and the velocity of the $i {th}$ particle and Ci is the centre of the circle described by the particle;
and $$ L_z = \Sigma l_{iz} = \bigg( \Sigma_i m_i r_i^2 \bigg) \Omega µk $$ or $$L_z = I ω µ_k …(7.44b)$$ The last step follows since the perpendicular distance of the i th particle from the axis is $r_i$ ; and by definition the moment of inertia of the body about the axis of rotation is $ I = \Sigma m_i r^2_i$ Note $$ L = L_z +L⊥ $$
The rigid bodies which we have mainly considered in this chapter are symmetric about the axis of rotation, i.e. the axis of rotation is one of their symmetry axes. For such bodies, for a given $O_Ci$ , for every particle which has a velocity $v_i$ , there is another particle of velocity –$v_i$ located diametrically opposite on the circle with centre Ci described by the particle. Together such pairs will contribute zero to$ L_⊥$ and as a result for symmetric bodies $L_⊥$ is zero, and hence

$$ L = L_z = I \Omega µk ….(7.44d)$$ For bodies, which are not symmetric about the axis of rotation, L is not equal to Lz and hence L does not lie along the axis of rotation. Referring to table 7.1, can you tell in which cases L = Lz will not apply? Let us differentiate Eq. (7.44b). Since $µ_k$ is a fixed (constant) vector, we get $$ \frac {d}{dt} (L_z) = \bigg ( \frac {d}{dt} (I \Omega) \bigg) µ_k $$ Now, Eq. (7.28b) states $$ \frac {dL}{dt} = τ $$ As we have seen in the last section, only those components of the external torques which are along the axis of rotation, need to be taken into account, when we discuss rotation about a fixed axis. This means we can take τ = τ$µ_k$. Since L = $L z + L _⊥$ and the direction of $L_z$ (vector $µ_k$) is fixed, it follows that for rotation about a fixed axis,

$$ \frac {dL_z}{dt} = τ µk … (7.45a)$$ And $$ \frac {dL⊥}{dt} = 0 … (7.45b)$$
Thus, for rotation about a fixed axis, the component of angular momentum perpendicular to the fixed axis is constant. As

$$L_z = I \Omega µ_k $$, we get from Eq. (7.45a),
$$ \frac {d}{dt} (I \Omega) = τ ….(7.45c) $$
If the moment of inertia I does not change with time,
$$ \frac {d}{dt} (I \Omega) = I \frac {d \Omega}{dt = I $$
and we get from Eq. (7.45c),
$$ τ = I …(7.43)$$
We have already derived this equation using the work – kinetic energy route.

7.13.1 Conservation of angular momentum

• Let us revisit the principle of conservation of angular momentum in the context of rotation about a fixed axis. From equation, 7.45c, if the external torque is zero, $$ L_z = Iω = constant …………. (7.46)$$
• For symmetric bodies, from equation-7.44d, $L_z$ may be replaced by L. (L and Lz are respectively the magnitudes of L and $L_z$.)
• Equation-7.46 then, is the required form, for rotation about a fixed axis as is equation-7.29a that expresses the general law of conservation of angular momentum of a system of particles.

A circus acrobat and a diver take advantage of this principle. Also, skaters and classical, Indian or western, dancers performing a pirouette on the toes of one foot display ‘mastery’ over this principle. They control the angular speed while they dance to have varied movements.

7.14 Rolling Motion

• Rolling motion is a combination of rotation and translation. The centre of mass of the body may make a translatory motion; whereas, the body itself may rotate about an axis through its centre of mass. For instance, when a football is kicked by a player, the motion of the ball in general is a combination of rotation and translation. In this case, the total kinetic energy of the body is the sum of kinetic energy due to rotation: $(1/2)(Iω^2)$ and due to translation: $(1/2)(mv^2)$. While applying the law of conservation of energy, both rotational and translational motion of the body are to be considered.
• Rolling: Rolling is a common type of motion that we observe in daily life. For example, the wheels of a vehicle will be rolling on a surface. When an object rolls on a surface, if the point on the object in contact with the surface will be momentarily at rest, then the motion is said to be pure rolling.
• Consider a rigid body of mass, M and radius, R rolling on a horizontal surface with a speed, v without slipping. Then the centre of mass moves with a velocity, $v_{cm} = v$.
• Other particles, in addition to the translational velocity, $v_{cm}$ will also rotate about centre of mass with an angular velocity, ω.
• A particle at a distance, r from the centre of mass will have a velocity, $v_r$= r ω along the tangent to the circle.
• The resultant velocity of the particle is $v_{cm}+v_r$. This resultant velocity will be different at different positions of particles as shown in figure, 7.37.
• A particle at the boundary of the object will have a velocity vR= Rω due to rotation. For a particle in contact with the surface at any instant t, the net velocity,
  $$ v_{cm}-Rω=0 ……………… (7.47)$$
  Thus, equation 7.47 gives the condition for rolling without slipping for a disc.
• Sometimes, the rolling object may slide or slip on the surface, in addition to rotating. This happens when the point of contact with the surface may have a non-zero velocity with respect to the surface. Then the motion is not pure rolling.
• Friction between the object and the surface plays an important role in rolling motion. Friction helps in rolling motion. An object cannot role on a perfectly smooth surface.

The figure below will help you understand how particles exhibit different velocities at different positions. Since the friction is greater at the lowest point where the object is in contact with the ground, the velocity is also least here.

7.14.1 Kinetic Energy of Rolling Motion

Our next task will be to obtain an expression for the kinetic energy of a rolling body.

• The kinetic energy of a rolling body can be separated into kinetic energy of translation and kinetic energy of rotation.
• We know that kinetic energy for a system of particles (K) can be separated into kinetic energy of motion of centre of mass (translation) and kinetic energy of rotational motion about the centre of mass of the system of particles (K’).
• Thus, kinetic energy = translational kinetic energy + rotational kinetic energy
  $$ K = \frac {Mv^2}{2}+ K’ ……(7.48)$$
  $$ K = \frac {Mv^2}{2}+ \frac {1}{2} I \omega^2$$
  Generalising this to a rolling body
  $$ K = \frac {1}{2}mv^2_{cm}+ \frac {1}{2} I \omega^2 ….(7.49a)$$
  where m is the mass of the body ,$v_{cm}$ is the speed of the centre of mass,I is the moment of inertia about the axis of rotation and $\omega$ is its angular speed.
  Substituting $I = mk^2$ where k is the corresponding radius of gyration of the body
  $v_{cm} = R \omega $ ,we get
  $$ K = \frac {1}{2}mv^2_{cm}+ \frac {1}{2} (mk^2) \omega^2 \hspace{10mm} But \space \omega = \frac {v_{cm}}{R}$$
  $$ K = \frac {1}{2}mv^2_{cm}+ \frac {1}{2} (mk^2) \bigg( \frac {v^2_{cm}}{R^2} \bigg) $$
  $$ K = \frac {1}{2}mv^2_{cm} \bigg( 1 + \frac {k^2}{R^2} \bigg) ….(7.49b)$$
• Equation (7.49b) applies to any rolling body: a disc, a cylinder, a ring or a sphere.

Example 7.16: Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

Solution: We assume conservation of energy of the rolling body, i.e. there is no loss of energy due to friction etc. The potential energy lost by the body in rolling down the inclined plane (= mgh) must, therefore, be equal to kinetic energy gained. (See Fig.7.38) Since the bodies start from rest the kinetic energy gained is equal to the final kinetic energy of the bodies. From Eq. (7.49b),
$$ K = \frac {1}{2}mv^2 \bigg( 1+ \frac {k^2}{R^2} \bigg) $$ ,
where v is the final velocity of (the centre of mass of) the body. Equating K and mgh,

$$ mgh = \frac {1}{2} mv^2 \bigg( 1+ \frac {k^2}{R^2} \bigg) $$
$$ or \space v^2 \space = \bigg( \frac {2gh}{1+k^2/R^2} \bigg) $$
Note is independent of the mass of the rolling body;
For a ring, $k^2 = R^2$
$$ υ {ring} = \sqrt { \frac {2gh}{1+1} } $$ $$ = \sqrt {gh} $$ For a solid cylinder $k^2 = R^2$/2 $$ υ{disc} =\sqrt { \frac {2gh}{1+1/2} } $$
$$ = \sqrt { \frac {4gh}{3} } $$
For a solid sphere $k^2 = 2R^2$/5
$$ υ_{sphere}=\sqrt { \frac {2gh}{1+2/5} } $$
$$ = \sqrt { \frac {10gh}{7} } $$

From the results obtained it is clear that among the three bodies the sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane. Suppose the bodies have the same mass. Which body has the greatest rotational kinetic energy while reaching the bottom of the inclined plane?

Question from sections 7.12 and 7.13:

1. Give the expression for total work done on a body due to torques resulting from several external forces.
2. Arrive at the expression, P = τω : The expression for Newton’s second law in rotational motion.
3. Explain how angular momentum varies as “I” varies with the example of a person sitting on a rotating chair.
4. Obtain the condition for rolling without slipping, with figure.
5. Obtain the expression for kinetic energy of a rolling body in the form of equation, 7.49b.