142 PHYSICS Fig 7.1 Translational (sliding) motion of a block down an inclined plane. (Any point like P1 or P2 of the block moves with the same velocity at any instant of time.) movement. The block is a rigid body. Its motion down the plane is such that all the particles of the body are moving together, i.e. they have the same velocity at any instant of time. The rigid body here is in pure translational motion (Fig. 7.1). In pure translational motion at any instant of time all particles of the body have the same velocity. Consider now the rolling motion of a solid metallic or wooden cylinder down the same inclined plane (Fig. 7.2). The rigid body in this problem, namely the cylinder, shifts from the top to the bottom of the inclined plane, and thus, has translational motion. But 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 translation. Its motion is translation plus ‘something else.’ Fig. 7.2 Rolling motion of a cylinder It is not pure translational motion. Points P, P,P and P 12 34 have different velocities (shown by arrows) at any instant of time. In fact, the velocity of the point of contact P3 is zero at any instant, if the cylinder rolls without slipping. In order to understand what this ‘something else’ is, let us take a rigid body so constrained that it cannot have translational motion. The most common way to constrain a rigid body so that it does not have translational motion is to fix it along a straight line. The only possible motion of such a rigid body is rotation. The line along which the body is fixed is termed as its axis of rotation. If you look around, you will come across many examples of rotation about an axis, a ceiling fan, a potter’s wheel, a giant wheel in a fair, a merry-go-round and so on (Fig 7.3(a) and (b)). (a) (b) Fig. 7.3 Rotation about a fixed axis (a)A ceiling fan (b)A potter’s wheel. Let us try to understand what rotation is, what characterises rotation. You may notice that in rotation of a rigid body about a fixed SYSTEMS OF PARTICLES AND ROTATIONAL MOTION Fig. 7.4 A rigid body rotation about the z-axis (Each point of the body such as P1 or P2 describes a circle with its centre (C1 or C2) on the axis. The radius of the circle (r1or r2 ) is the perpendicular distance of the point (P1 or P2 ) from the axis. A point on the axis like P3 remains stationary). axis, every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis. Fig. 7.4 shows the rotational motion of a rigid body about a fixed axis (the z-axis of the frame of reference). Let P1 be a particle of the rigid body, arbitrarily chosen and at a distance r1 from fixed axis. The particle P describes a circle of radius 1rwith its centre C on the fixed axis. The circle 1 1lies in a plane perpendicular to the axis. The figure also shows another particle P2 of the rigid body, P2 is at a distance r2 from the fixed axis. The particle P moves in a circle of radius r and 2 2 with centre C2 on the axis. This circle, too, lies in a plane perpendicular to the axis. Note that the circles described by P1 and P2 may lie in different planes; both these planes, however, are perpendicular to the fixed axis. For any particle on the axis like P3, r = 0. Any such particle remains stationary while the body rotates. This is expected since the axis is fixed. Fig. 7.5 (a) A spinning top (The point of contact of the top with the ground, its tip O, is fixed.) Fig. 7.5 (b) An oscillating table fan. The pivot of the fan, point O, 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 [Fig. 7.5(a)]. (We assume that the top does not slip from place to place and so does not have translational motion.) We know from experience that the axis of such a spinning top moves around the vertical through its point of contact with the ground, sweeping out a cone as shown in Fig. 7.5(a). (This movement of the axis of the top around the vertical is termed precession.) Note, 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. 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 146 PHYSICS small, we can treat the body as a continuous distribution of mass. We subdivide the body into n small elements of mass; Δm1, Δm2... Δm n; the ith element Δmi is taken to be located about the point (xi, yi, zi). The coordinates of the centre of mass are then approximately given by (Δm )x (Δm )y (Δm )z∑ ii ∑ ii ∑ iiX = ,Y = ,Z = ∑Δmi ∑Δmi ∑Δmi As we make n bigger and bigger and each Δmi smaller and smaller, these expressions become exact. In that case, we denote the sums over i by integrals. Thus, Δm → dm = M,∑ i∫ (Δm )x → x dm,∑ ii ∫ ∑(Δm )y → y dm,ii ∫and ∑(Δmi )zi →∫z dm Here M is the total mass of the body. The coordinates of the centre of mass now are 11 1 X= x dmY , =∫y dmand Z= z dm (7.5a)∫∫MM M The vector expression equivalent to these three scalar expressions is 1 R =∫ rdm (7.5b)M If we choose, the centre of mass as the origin of our coordinate system, R = 0 i.e., ∫ rdm = 0 or x dm = y dm = z dm = 0 (7.6)∫∫ ∫ Often we have to calculate the centre of mass of homogeneous bodies of regular shapes like rings, discs, spheres, rods etc. (By a homogeneous body we mean a body with uniformly distributed mass.) By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres. Let us consider a thin rod, whose width and breath (in case the cross section of the rod is rectangular) or radius (in case the cross section of the rod is cylindrical) is much smaller than its length. Taking the origin to be at the geometric centre of the rod and x-axis to be along the length of the rod, we can say that on account of reflection symmetry, for every element dm of the rod at x, there is an element of the same mass dm located at –x (Fig. 7.8). The net contribution of every such pair to the integral and hence the integral ∫x dm itself is zero. From Eq. (7.6), the point for which the integral itself is zero, is the centre of mass. Thus, the centre of mass of a homogenous thin rod coincides with its geometric centre. This can be understood on the basis of reflection symmetry. The same symmetry argument will apply to homogeneous rings, discs, spheres, or even thick rods of circular or rectangular cross section. For all such bodies you will realise that for every element dm at a point (x, y, z ) one can always take an element of the same mass at the point (–x, –y, –z). (In other words, the origin is a point of reflection symmetry for these bodies.) As a result, the integrals in Eq. (7.5 a) all are zero. This means that for all the above bodies, their centre of mass coincides with their geometric centre. • 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. Answer Fig. 7.8 Determining the CM of a thin rod. Fig. 7.9 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 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 3 ). Let the masses 100 g, 150g and 200g be located at O, A and B be respectively. Then, mx + mx + mx 11 22 33X = m + m + m123 ⎣ ()+ 150(0.5) + 200(0.25) ⎤⎦ gm ⎡100 0 = (100 + 150 + 200) g 75 + 50 125 5 = m = m = m 450 450 18 ⎡100(0) + 150(0) + 200(0.25 3) ⎤ gm ⎣ ⎦Y = 450 g 50 3 3 1 = m = m = m 450 933 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. Answer The lamina (ΔLMN ) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig. 7.10 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. Answer 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 C1, C2 and C3 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. Fig. 7.11 Hence [] 51(1/2) + 1(3/2) + 1(1/2) kg m X = = m(+ 1 +) 61 1kg ⎡ 1(1/ 2) + 1(1/2) +1(3/ 2) ⎤ kg m ⎣[ ]⎦ 5 Y == m(+ 1+) 61 1kg 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 152 PHYSICS From this follow the results ˜˜˜˜ ˜˜i × i = 0, j× j = 0, k × k = 0 (ii) ˜i × ˜j = k˜ Note that the magnitude of ˜i × ˜j is sin900 or 1, since ˜ and ˜j both have unitimagnitude and the angle between them is 900. Thus, i˜× ˜j is a unit vector. A unit vector perpendicular to the plane of ˜ and ˜j andirelated to them by the right hand screw rule is k˜. Hence, the above result. You may verify similarly, ˜˜˜ ˜˜˜j × k = i and k × i = j From the rule for commutation of the cross product, it follows: ˜˜˜˜˜˜˜˜ ˜j × i =−k, k × j =−i, i × k =− j ˜˜ ˜ Note if ,,i jkoccur cyclically in the above vector product relation, the vector product is ˜˜ ˜ positive. If i jk do not occur in cyclic order, ,,the vector product is negative. Now, ˜˜˜ ˜˜˜a × b = (a i + a j + a k) × (b i + b j + b k)xyz xyz ˜ ˜ ˜˜˜˜= ab k − ab j − ab k + ab i + ab j − ab i xy xzyx yzzxzy = (ab − ab )˜i + (ab − ab )˜j + (ab − ab )k˜ yzzx zxxz xyyx 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. ˜˜˜i jk a × b = aaa xyz bbb xyz • Example 7.4 Find the scalar and vector products of two vectors. a = (3ˆi– 4jˆ + 5ˆk) ˆ )and b = (– 2iˆ + ˆ j– 3kAnswer ˜˜˜ ˜˜˜i i 5 i − i 3ab = (3 − 4j + k)( 2 + j − k) =−6 − 4 − 15 =−25 ˜˜˜i jk ˜˜ ˜a × b = 3 −45 = 7i − j − 5k −2 1 −3 ˜˜ ˜Note b × a =−7i + j + 5k • 7.6 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. 7.4. As said above, in rotational motion of a rigid body about a fixed axis, every particle of the body moves in a circle, 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.) which lies in a plane perpendicular to the axis and has its centre 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 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 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 Δt (Fig. 7.16). The angle PCP′ describes the angular displacement Δθ of the particle in time Δt. The average angular velocity of the particle over the interval Δt is Δθ/Δt. As Δt tends to zero (i.e. takes smaller and smaller values), the ratio Δθ/Δt approaches a limit which is the instantaneous angular velocity dθ/dt of the particle at the position P. We denote the instantaneous angular velocityby 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=ωr , where r is the radius of the circle. We observe that at any given instant the relation v =ω r applies to all particles of the rigid body. Thus for a particle at a perpendicular distance r from the fixed axis, the linear velocityiat a given instant v is given byiv =ω r (7.19)ω= dθ dt i i The index i runs from 1 to n, where n is the total number of particles of the body. For particles on the axis, , and hencer = 0 v = ω r = 0. 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 ω 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. 7.1 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. 7.17a). The magnitude of this vector is referred as above. Fig. 7.17 (a) If the head of a right handed screw rotates with the body, the screw advances in the direction of the angular velocity ω. If the sense (clockwise or anticlockwise) of rotation of the body changes, so does the direction of ω. Fig. 7.17 (b) The angular velocity vector ωis directed along the fixed axis as shown. The linear velocity of the particle at P is v =ω × r. It is perpendicular to both 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 ω × r corresponds to. Refer to Fig. 7.17(b) which is a part of Fig. 7.16 reproduced to show the path of the particle P. The figure shows the vector ωdirected along the fixed (z–) axis and also the position vector r = OPof 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.