CH_11

# CH_11 - Chapter 11 CHAPTER 11 General Rotation 1(a For the...

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Chapter 11 Page 1 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 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 k , our thumb points along – j . i k = – j . When we rotate j k , our thumb points along i 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 AB . This is expected, because B A A B . 3. The magnitude of the tangential acceleration is a tan r . From the diagram we see that , r and a tan are all perpendicular, and rotating r gives a vector in the direction of a tan . Thus we have a tan r . The magnitude of the radial acceleration is a R 2 r r v . From the diagram we see that , v and a R are all perpendicular, and rotating v gives a vector in the direction of a R . Thus we have a R 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 . 5. For the limiting process we have d( A B ) d t = lim t 0 ( A B ) t t 0 [( A + A ) ( B + B )] – ( A B ) t t 0 [( A B )+( A B A B A B )]– ( A B ) t t 0 A B t + A t B + A B t = A d B d t + d A d t B . The last term is dropped because it is the product of two differentials. 6. ( a ) When we use the component forms for the vectors, we have A B = ( A x i A y j A z k ) ( B x i B y j B z k A x B x ( i i ) + A x B y ( i j A x B z ( i k ) + x y z i j k axis r v a tan a R

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Chapter 11 Page 2 A y B x ( j i ) + A y B y ( j j A y B z ( j k A z B x ( k i A z B y ( k j A z B z ( k k ) = 0 + A x B y ( k A x B z (– j A y B x (– k ) + 0 + A y B z ( i A z B x ( j A z B y (– i ) + 0 = ( 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 . ( b ) When we use the rules for evaluating a determinant, we get i j k A x A y A z B x B y B z = i A y B z A z B y j A x B z A z B x + k A x B y A y B x = i A y B z A z B y + j A z B x A x B z + k A x B y A y B x = A B . 7. ( a ) For the vectors A = 7.0 i – 3.5 j and B = – 8.5 i + 7.0 j + 2.0 k , we have A B = i j k 7.0 –3.5 0 – 8.5 7.0 2.0 = i (– 3.5)(2.0)– (0)(7.0) j (7.0)(2.0)–(0)(– 8.5) + k (7.0)(7.0)–(– 3.5)(–8.5) = –7.0 i –14 . 0 j +19.3 k .
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CH_11 - Chapter 11 CHAPTER 11 General Rotation 1(a For the...

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