Ch11Word - Chapter 11 CHAPTER 11 General Rotation 1 z(a For...

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Page 1 Chapter  11 CHAPTER 11 – General Rotation 1. ( a ) For the magnitudes  of the vector products we have i   ×   i  =  i   i  sin 0 °  = 0; j   ×   j  =  j   j  sin 0 °  = 0; k   ×   k  =  k   k  sin 0 °  = 0. ( b ) For the magnitudes  of the vector products we have i   ×   j  =  i   j  sin 90 °  = (1)(1)(1) = 1; i   ×   k  =  i   k  sin 90 °  = (1)(1)(1) = 1; j   ×   k    =  j   k  sin 90 °  = (1)(1)(1) = 1. From the right hand  rule, if we rotate our fingers from  i  into  j , our thumb  points in the direction of  k Thus  i   ×   j   =  k . Similarly, when  we rotate  i  into  k , our thumb  points along –  j .  Thus  i   ×   k   = –  j . When we rotate  j  into  k , our thumb  points along  i .  Thus  j   ×   k   =  i . 2. ( a ) We have  A  = –  A i  and   B  =  B k .  For the direction of  A   ×   B  we have –  i   ×   k   = – (–  j ) =   j ,        the positive  y -axis . ( b ) For the direction of  B   ×   A  we have k   ×  (–  i )  = – ( k   ×   i ) = – ( j ) = –   j ,        the negative  y -axis . ( c ) For the magnitude  of  A   ×   B  we have A   ×   B  =  A   B  sin 90 °  =        AB . For the magnitude  of  B   ×   A  we have B   ×   A  =  B   A  sin 90 °  =        AB . This is expected, because  B   ×   A  = –  A   ×   B . 3. The magnitude  of the tangential acceleration  is  a tan  =  α r .   From the diagram  we see that  α r  and   a tan  are all perpendicular,  and  rotating  α  into  r  gives a vector in the direction of  a tan   .   Thus we have  a tan  =  α   ×   r . The magnitude  of the radial acceleration is  a R  =  ϖ 2 r  =  ϖ r ϖ  =  ϖ v .   From the diagram  we see that  ϖ v  and   a R  are all perpendicular,  and  rotating  ϖ  into  v  gives a vector in the direction of  a R .   Thus we have  a R  =  ϖ   ×   v . 4. When we use the component  forms for the vectors, we have A   ×  ( B  +  C )  = [ A y ( B z  +  C z ) –  A z ( B y  +  C y )] i  + [ A z ( B x  +  C x ) –  A x ( B z  +  C z )] j   + [ A x ( B y  +  C y ) –  A y ( B x  +  C x )] k   = ( A y B z  –  A z B y ) i  + ( A z B x  –  A x B z ) j  + ( A x B y  –  A y B x ) k  +  ( A y C z  –  A z C y ) i  + ( A z C x  –  A x C z ) j  + ( A x C y  –  A y C x ) k A   ×   B  +  A   ×   C .
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