SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

for a × b can be put in a determinant form which is easy to remember. Example . Find the scalar and vector products of two vectors. a = and b = Answer ( ) ( ) = − = − Note = − ⊳ .

ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY In this section we shall study what is angular velocity 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 these two quantities involves a vector product which we learnt about in the last section.

Let us go back to Fig. . . As said above, in rotational motion of a rigid body about a fixed axis, every particle of the body moves in a circle, Fig.

. 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.) which lies in a plane perpendicular to the axis and has its centre on the axis. In Fig.

. we redraw Fig. . , 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 , the perpendicular distance of the point P from the axis. We also show the linear velocity vector v of the particle at P. It is along the tangent at P to the circle.

Let P ′ be the position of the particle after an interval of time ∆ t (Fig. . ). The angle PCP ′ describes the angular displacement ∆ θ of the particle in time ∆ t .

The average angular velocity of the particle over the interval ∆ t is ∆ θ / ∆ t. As ∆ t tends to zero (i.e. takes smaller and smaller values), the ratio ∆ θ / ∆ t approaches a limit which is the instantaneous angular velocity d θ /d t of the particle at the position P. 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 υ , where r is the radius of the circle. We observe that at any given instant the relation v 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 ( . ) The index i runs from to n , where n is the total number of particles of the body.

For particles on the axis, , and hence v = ω r = . Thus, particles on the axis are stationary. This verifies that the axis is fixed . Note that we use the same angular velocity ω for all the particles.

We therefore, refer to ωωωωω as the angular velocity of the whole body . We have characterised pure translation of a body by all parts of the body having the same velocity at any instant of time. Similarly, we may characterise pure rotation by all parts of the body having the same angular velocity at any instant of time . Note that this characterisation of the rotation of a rigid body about a fixed axis is just another way of saying as in Sec.

. 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. In our discussion so far the angular velocity appears to be a scalar. In fact, it is a vector.

We shall not justify this fact, but we shall accept it. 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. .17a).

The magnitude of this vector is dt referred as above. Fig. . (a) If the head of a right handed screw rotates with the body, the screw advances in the direction of the angular velocity ωωωωω .

If the sense (clockwise or anticlockwise) of rotation of the body changes, so does the direction of ωωωωω . Fig. . (b) The angular velocity vector ωωωωω is directed along the fixed axis as shown.

The linear velocity of the particle at P is v = ωωωωω × r . It is perpendicular to both ω ω and r and is directed along the tangent to the circle described by the particle. We shall now look at what the vector product ωωωωω × r corresponds to. Refer to Fig.

. (b) which is a part of Fig. . reproduced to show the path of the particle P.

The figure shows the vector ωωωωω 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 ωωωωω × r = ωωωωω × OP = ωωωωω × (OC + CP) But ωωωωω × OC = as ω ω is along OC Hence ωωωωω × r = ωωωωω × CP The vector ωωωωω × CP is perpendicular to ωωωωω , i.e. to the z -axis and also to CP , 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 ωωωωω × CP is ω (CP) since ωωωωω and CP are perpendicular to each other. We shall denote CP by ⊥ r and not by r , as we did earlier. Thus, ωωωωω × r is a vector of magnitude ω r ⊥ 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, v = ω ω × r ( . ) In fact, the relation, Eq. ( .

), holds good even for rotation of a rigid body with one point fixed, such as the rotation of the top [Fig. . (a)]. In this case r represents the position vector of the particle with respect to the fixed point taken as the origin.

We note that for rotation about a fixed axis, the direction of the vector ωωωωω does not change with time. Its magnitude may, however, change from instant to instant. For the more general rotation, both the magnitude and the direction of ω ω may change from instant to instant. .

. Angular acceleration You may have noticed that we are developing the study of rotational motion along the lines of the study of translational motion with which we are already familiar. Analogous to the kinetic variables of linear displacement (s) and velocity ( v ) in translational motion, we have angular displacement ( θθθθθ ) and angular velocity ( ωωωωω ) in rotational motion. It is then natural to define in rotational motion 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 ααααα as the time rate of change of angular velocity. Thus, d t ( . ) If the axis of rotation is fixed, the direction of ω ω and hence, that of ααααα is fixed. In this case the vector equation reduces to a scalar equation d t α = ( .

) . TORQUE AND ANGULAR MOMENTUM In this section, we shall acquaint ourselves with two physical quantities (torque and angular momentum) which are defined as vector products of two vectors. These as we shall see, are especially important in the discussion of motion of systems of particles, particularly rigid bodies. .

. Moment of force (Torque) We have learnt that the motion of a rigid body, in general, is a combination of rotation and translation. If the 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, i.e.

to produce linear acceleration. We may then ask, what is the analogue of force in the case of rotational motion? To look into the question in a concrete situation 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.

What makes the door rotate? It is clear that unless a force is applied the door does not rotate. 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.

It is not the force alone, but how and where the force is applied is important in rotational motion. The rotational analogue of force in linear motion is moment of force . It is also referred to as torque or couple . (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. We shall also relate it to a change in the state of rotational motion, i.e. is angular acceleration of a rigid body. Fig.

. τ = τ = τ = τ = τ = r × F , τ τ τ τ τ is perpendicular to the plane containing r and F , and its direction is given by the right handed screw rule. 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. .

), the moment of the force acting on the particle with respect to the origin O is defined as the vector product τττττ = r × F ( . ) The moment of force (or torque) is a vector quantity. The symbol τ τ τ τ τ stands for the Greek letter tau . The magnitude of τ τ τ τ τ is τ = r F sin θ ( .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. Moment of force has dimensions M L T - . Its dimensions are the same as those of work or energy. It is, however, a very different physical quantity than work.

Moment of a force is a vector, while work is a scalar. The SI unit of moment of force is newton metre (N m). The magnitude of the moment of force may be written ( sin ) r F ⊥ ( .24b) or sin r F rF ⊥ ( .24c) where r ⊥ = r sin θ is the perpendicular distance of the line of action of F from the origin and sin ) ⊥ = is the component of F in the direction perpendicular to r . Note that τ = if r = , F = or θ = or .

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. 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.

. . Angular momentum of a particle Just as the moment of a force is the rotational analogue of force in linear motion, the quantity angular momentum is the rotational analogue of linear momentum. We shall first define angular momentum for the special case of a single particle and look at its usefulness in the context of single particle motion.

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 ( .25a) The magnitude of the angular momentum vector is sin l r p ( .26a) where p is the magnitude of p and θ is the angle between r and p . We may write l r p ⊥ or r p ⊥ ( .26b) where r ⊥ (= r sin θ ) is the perpendicular distance of the directional line of p from the origin and sin ) ⊥ = is the component of p in a direction perpendicular to r . We expect the angular momentum to be zero ( l = ), if the linear momentum vanishes ( p = ), if the particle is at the origin ( r = ), or if the directional line of p passes through the origin θ = or .

The physical quantities, moment of a force and angular momentum, have an important relation between them. It is the rotational analogue of the relation between force and linear momentum. For deriving the relation in the context of a single particle, we differentiate l = r × p with respect to time, d ( l Applying the product rule for differentiation to the right hand side, Now, the velocity of the particle is v = d r / dt and p = m v Because of this d , t × as the vector product of two parallel vectors vanishes. Further, since d p