Chapter-04 - 111 Chapter 4 1. The initial position vector r...

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Unformatted text preview: 111 Chapter 4 1. The initial position vector r r o satisfies r r r r r r- = o ∆ , which results in o ? ? ? ? (3.0j 4.0k)m (2.0i 3.0j 6.0k)m ( 2.0 m) i (6.0 m) j ( 10 m) k r r r =- ∆ =--- + = - + + - r r r . 2. (a) The position vector, according to Eq. 4-1, is ? = ( 5.0 m) i + (8.0 m)j r- r . (b) The magnitude is 2 2 2 2 2 2 | | + + ( 5.0 m) (8.0 m) (0 m) 9.4 m. r x y z = =- + + = r (c) Many calculators have polar ↔ rectangular conversion capabilities which make this computation more efficient than what is shown below. Noting that the vector lies in the xy plane and using Eq. 3-6, we obtain: 1 8.0 m tan 58 or 122 5.0 m θ-   = = - ° °  -   where the latter possibility (122° measured counterclockwise from the + x direction) is chosen since the signs of the components imply the vector is in the second quadrant. (d) The sketch is shown on the right. The vector is 122° counterclockwise from the + x direction. (e) The displacement is r r r ′ ∆ =- r r r where r r is given in part (a) and ˆ (3.0 m)i. r ′ = r Therefore, ? (8.0 m)i (8.0 m)j r ∆ =- r . (f) The magnitude of the displacement is 2 2 | | (8.0 m) ( 8.0 m) 11 m. r ∆ = + - = r (g) The angle for the displacement, using Eq. 3-6, is 1 8.0 m tan = 45 or 135 8.0 m-  - ° °  -   where we choose the former possibility (- 45°, or 45° measured clockwise from + x ) since the signs of the components imply the vector is in the fourth quadrant. A sketch of r ∆ r is shown on the right. CHAPTER 4 112 3. (a) The magnitude of r r is 2 2 2 | | (5.0 m) ( 3.0 m) (2.0 m) 6.2 m. r = + - + = r (b) A sketch is shown. The coordinate values are in meters. 4. We choose a coordinate system with origin at the clock center and + x rightward (towards the “3:00” position) and + y upward (towards “12:00”). (a) In unit-vector notation, we have 1 2 ? (10 cm)i and ( 10 cm)j. r r = = - r r Thus, Eq. 4-2 gives 2 1 ? ( 10 cm)i ( 10 cm)j. r r r ∆ =- = - + - r r r and the magnitude is given by 2 2 | | ( 10 cm) ( 10 cm) 14 cm. r ∆ =- + - = r (b) Using Eq. 3-6, the angle is 1 10 cm tan 45 or 135 . 10 cm θ--   = = °- °  -   We choose 135- ° since the desired angle is in the third quadrant. In terms of the magnitude-angle notation, one may write 2 1 ? ( 10 cm)i ( 10 cm)j (14 cm 135 ). r r r ∆ =- = - + - → ∠ - ° r r r (c) In this case, we have 1 2 ? ? ( 10 cm)j and (10 cm)j, and (20 cm)j. r r r = - = ∆ = r r r Thus, | | 20 cm. r ∆ = r (d) Using Eq. 3-6, the angle is given by 1 20 cm tan 90 . 0 cm θ-   = = °     (e) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero. (f) The corresponding angle for a full-hour sweep is also zero....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.

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Chapter-04 - 111 Chapter 4 1. The initial position vector r...

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