Ch11Word - Chapter 11 Page 1 CHAPTER 11 – General Rotation 1 a For the magnitudes of the vector products we have ⏐ i × i ⏐ = ⏐ i ⏐ ⏐ i

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Unformatted text preview: 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 ⏐ 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 = ω 2 r = ω r ω = ω v . From the diagram we see that ω , v and a R are all perpendicular, and rotating ω into v gives a vector in the direction of a R . Thus we have a R = ω × 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 → ∆ ( A × B ) ∆ t = lim ∆ t → [( A + ∆ A ) × ( B + ∆ B )] – ( A × B ) ∆ t = lim ∆ t → [( A × B ) + ( A × ∆ B ) + ( ∆ A × B ) + ( ∆ A × ∆ B )] – ( A × B ) ∆ t = lim ∆ t → 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. x y z i j k axis ω r v a tan α a R Chapter 11 Page 2 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 ) + 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 ) = + A x B y ( k ) + A x B z (– j ) +...
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch11Word - Chapter 11 Page 1 CHAPTER 11 – General Rotation 1 a For the magnitudes of the vector products we have ⏐ i × i ⏐ = ⏐ i ⏐ ⏐ i

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