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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 16

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online