question_answer

       Discuss whether or not ; angular displacement is a vector quantity?         

Answer:

                           Angular displacement is a vector quantity provided angle \[\theta \] is small because the commutative law of vector addition for large angle is not valid, where as for small angles, the law is valid i.e., \[{{\theta }_{1}}+{{\theta }_{2}}\ne {{\theta }_{2}}+{{\theta }_{1}}\], but \[\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}}=\Delta {{\theta }_{2}}+\Delta {{\theta }_{1}}\].The same is explained below.     Consider a book with front page F lying in the plane of paper. It can be rotated about two mutually      perpendicular axes 1 and 2 as shown in Fig. 2(d).25(a).                                 (a)                    (b)                      (c)                        (d)                (e) (i) Let the book be rotated through an angle \[{{\theta }_{1}}\left( ={{90}^{o}} \right)\]in the clockwise direction about axis 1. It takes the position as shown in Fig. 2(d).25(b). On further rotating the book through an angle \[{{\theta }_{2}}\left( ={{90}^{o}} \right)\]in anticlockwise direction about axis 2, the book occupies the position as shown in Fig. 2(d).25(c).                      (ii) From the initial position of book, [Fig 2(d).25(a)], if the book is rotated through\[{{\theta }_{1}}\left( ={{90}^{o}} \right)\] in the  anticlockwise direction first about the axis 2, it occupies the position as shown in Fig. 2(d).25(d). On further rotating the book through\[{{\theta }_{2}}\left( ={{90}^{o}} \right)\] in clockwise direction about axis 1, the book occupies the position shown in Fig. 2(d).25(e).   From above we note that the final positions of the book shown in Fig. 2(d).25(c) and Mg.  are not the same. Hence          \[{{\theta }_{1}}+{{\theta }_{2}}\ne {{\theta }_{2}}+{{\theta }_{1}}\] This shows that \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] are not vectors as they do not obey the commutative law of vector addition. If the book is rotated through a smaller angle \[\Delta {{\theta }_{1}}\] and \[\Delta {{\theta }_{2}}\] (say 2° or 3°), the final positions of the book in the two cases discussed above would almost be the same. As \[\Delta {{\theta }_{2}}\to 2\], the final positions of the book will become indistinguishable, hence \[\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}}=\Delta {{\theta }_{2}}+\Delta {{\theta }_{1}}\]