CHAPTER - I V THE KINEMATICS OF RIGID BODY Unit 1: Rigid Body: Introduction: In this chapter we define a rigid body and describe how the number of degrees of freedom of a rigid body with N particles is determined. There are two types of motion involved in the case of rigid body viz.; the translation and the rotation. Various sets of variables have been used to describe the orientation of rigid body. We will discuss in this chapter how the Eulerian angles and the complex Cayley-Klein parameters can be used for the description of rigid body with one point fixed. Geometrically, matrix represents rotation; we will find the matrix of transformation in terms of Eulerian angles and Cayley-Klein parameters and establish the relation between them. This unit is devoted to the study of orthogonal transformations and its properties. Rigid Body : A rigid body is regarded as a system of many (at least three) non-collinear particles whose positions relative to one another remain fixed. i.e., distance between any two of them remains constant through out the motion. The internal forces holding the particles at fixed distances from one another are known as forces of constraint. These forces of constraint obey the Newton’s third law of motion. Classical Mechanics Page No. 225 Worked Examples • Example 1 : Explain how the generalized co-ordinates of a rigid body with N particles reduce to six for its description. Solution : • Generalized co-ordinates of rigid body : A system of N particles free from constraints can have 3N degrees of freedom and hence 3N generalized co-ordinates. But the constraints involved in rigid body with N particles are holonomic and scleronomic and are given by raijij= ij , ≠ = 1,2,...., N . (1) th th where rij denotes the distance between the i and j particles, and aij are constants. Equation (1) is symmetric in i and j and i≠ j as the distance of the ith from itself is zero, therefore, the possible number of constraints is N( N −1) NC = . (2) 2 2 N( N −1) We notice that for N > 7, > 3N . Therefore the actual number of 2 degrees of freedom cannot be obtained simply by subtracting the number of constraints from 3N. This is simply because all constraints in equation (1) are not independent. To show how the generalized co-ordinates of a rigid C body with N particles reduce to six for its P description, let a rigid body be regarded as a system A of at least three non-collinear particles whose positions relative to one another remain fixed. Thus a B system of 3 particles free from constraints has 9 degrees of freedom but there involves 3 constraints. Hence the number of generalized co-ordinates reduces to six. Thus the total number of degrees of freedom for three non-collinear particles A, B, Classical Mechanics Page No. 226 and C of a rigid body is equal to six. This is because each particle has 3-degrees of freedom and less three equations of constraints. The position of each further particle say P requires three more co-ordinates for its description, but there will be three equations of constraints for this particle, because the distance of P from A, B, C is fixed. Thus three co-ordinates for P and less three equations of constraints for P gives zero degrees of freedom. Thus any other particle apart from A, B, C taken to specify the configuration of the rigid body will not add any degrees of freedom. Once the positions of three of the particles of the rigid body are determined the constraints fix the positions of all remaining particles. Thus the configuration of the rigid body would be completely specified by only three particles i.e., by six degrees of freedom, no matter how many particles it may contain. Example 2 : Describe the motion of the rigid body. Solution: A rigid body can have two types of motion (i) a translational motion and (ii) a rotational motion. Thus a rigid body in motion can be completely specified if its position and orientation are known. However, if one of the points of a rigid body is fixed, the translation motion of the body is absent and the body rotates about any line through the fixed point. Again, if we fix up a second point, then the motion of the body is restricted to rotate about the line joining the two fixed points. Further, if we also fix the third point of the body non-collinear with other two, the position of the body is fixed and there is no motion of any kind. The co-ordinates of the third point alone will be able to locate the rigid body completely in space. It follows that the position of the rigid body is determined by any three non-collinear points of it that is by six degrees of freedom. Classical Mechanics Page No. 227 Of the six generalized co-ordinates, 3 co-ordinates are used to describe translational motion and other three co-ordinates are used to describe rotational motion. Since a rigid body with one point fixed has no translational motion and hence it has 3- degrees of freedom and three generalized co-ordinates-which are used to describe the rotational motion. • Orthogonal Transformation : Example 3 : Define orthogonal transformation. Show that finite rotation of a rigid body about a fixed point of the body is not commutative. Solution : Consider ( x1, x 2 , x 3 ) and ( x1′, x 2′ , x 3′ ) be two co-ordinate systems. The general linear transformation between these two co-ordinate systems is defined by the following set of equations x1′ = ax 111 + ax 122 + ax 133 , x2′ = ax 211 + ax 222 + ax 233 , . (1) x3′ = ax 311 + ax 322 + ax 333 , where a11, a 12 ,..., a 33 are constants. These three equations can be combined in to a single equation as 3 xi′ =∑ ax ij j , i = 1,2,3. (2) j=1 where A= ( a ij ) is called the matrix of transformation. Let a vector r= xi1 + xj 2 + xk 3 defined in ( x1, x 2 , x 3 ) co-ordinate system be transformed to ( x1′, x 2′ , x 3′ ) co-ordinate systems in the form r= xi1′ + xj 2′ + xk 3′ . Since the magnitude of the vector must be the same in both the co-ordinate system, we must have therefore 3 3 2 2 ∑xi′ = ∑ x i . (3) i=1 i = 1 Classical Mechanics Page No. 228 Using equation (2) in equation (3) we get 33 3 3 2 ∑∑axij j ∑ ax ik k = ∑ x i , ij==11 k = 1 i = 1 333 3 ⇒ 2 ∑∑∑aaij ik xx j k= ∑ x i . jki===111 i = 1 Equating the corresponding coefficients on both the sides of the above equation we get 3 ∑ aij a ik= δ jk , . (4) i=1 where δ jk is the Kronecker delta symbol and is defined by δ =0when j ≠ k , jk . (5) =1when j = k . Thus any transformation (2) satisfying (4) is called as an orthogonal transformation. Ex. 4. Show that two successive finite rotations of a rigid body about a fixed point of the body are not commutative. Solution : Consider two successive linear transformations described by the matrices B and A corresponding to two successive displacements of the rigid body. Let the first transformation from x to x′ be denoted by the matrix B and is defined by 3 xk′ =∑ bx kj j , k = 1,2,3, . (6) j=1 where the matrix of transformation is B= ( b kj ) . Let the succeeding transformation from x′ to x′′ be defined by the matrix A= ( a ik ) and is given by 3 xi′′=∑ ax ikk ′ , i = 1,2,3. (7) k=1 Classical Mechanics Page No. 229 Now the transformation from x to x′′ is obtained by combining the two equations (6) and (7) as 3 xi′′ =∑ abx ik kj j , i = 1,2,3. j, k = 1 This may also be written as 3 xi′′ =∑ cx ij j , i = 1,2,3. (8) j=1 where C= ( c ij ) is the matrix of transformation from x to x′′ and the elements of the matrix of transformation are defined as 3 cij= ∑ a ik b kj . (9) k =1 These elements are obtained by multiplying the two matrices A and B. Thus the two successive linear transformations described by A and B is equivalent to a third linear transformation described by the matrix C, defined by C = AB. (10) Since the matrix multiplication is not commutative in general, hence the finite rotations of a rigid body about a fixed point of the body are not commutative. • Properties of orthogonal transformation matrix : Example 5: Prove that the product of two orthogonal transformations is again orthogonal transformation. Solution : Consider two successive orthogonal linear transformations of a rigid body with one point fixed corresponding to two successive displacement of the rigid body and are described by the matrices B and A respectively. We know that the two successive orthogonal transformations is equivalent to a third linear transformation described by the matrix C, where C = AB, and its elements are defined by Classical Mechanics Page No. 230 cij= ∑ a ik b kj . (1) k where Aa=( ik), Bb = ( kj ) are matrices of orthogonal transformations. This implies that ∑ aij a ik= δ jk , . (2) i ∑bij b ik= δ jk . (3) i Consider now ∑ccij ik= ∑∑ ab im mj ∑ ab il lk i im l = ∑ aim b mj ab il lk , i, m , l = ∑ ∑ aim a il b mj b lk , m, l i = ∑δmlb mj b lk , m, l = ∑blj b lk , l ∑cij c ik= δ jk .