Question

Obtain expressions in component form for the position vectors having the following polar coordinates.
(a) 12.6 m, 140° counterclockwise from the +x axis
R with arrow =

(b) 4.00 cm, 50.0° counterclockwise from the +x axis
R with arrow =

(c) 20.0 in., 220° counterclockwise from the +x axis
R with arrow =

Answer 1

Answer on Question #81061 - Physics - Mechanics, Relativity

Obtain expressions in component form for the position vectors having the following polar coordinates.

(a) $12.6\mathrm{m}$ , $140{}^{\circ}$ counterclockwise from the $+x$ axis

(b) $4.00\mathrm{cm}$ , $50.0{}^{\circ}$ counterclockwise from the $+x$ axis

(c) 20.0 in., $220{}^{\circ}$ counterclockwise from the $+x$ axis

Solution.

We denote the length of the vector $\vec{\mathbf{R}}$ by $\mathbf{R}$ and the angle between the positive direction of $x$ -axis and the vector $\vec{\mathbf{R}}$ by $\theta$ . Let $\vec{i}$ and $\vec{j}$ be the unit vectors directed along the $x$ -axis and $y$ -axis respectively.

The projections of the vector $\vec{\mathbf{R}}$ on the axis:

$$R _ {x} = R \cos \theta$$ $$R _ {y} = R \sin \theta$$

Vector in the component notation:

$$\vec {\mathrm {R}} = \left(\mathrm {R} _ {\mathrm {x}}, \mathrm {R} _ {\mathrm {y}}\right) = \vec {\mathrm {i}} \mathrm {R} _ {\mathrm {x}} + \vec {\mathrm {j}} \mathrm {R} _ {\mathrm {y}}$$ $$\vec {R} = (R \cos \theta , R \sin \theta) = \vec {i} R \cos \theta + \vec {j} R \sin \theta$$

(a).

$$R = 1 2. 6 \mathrm {m}; \theta = 1 4 0 {}^ {\circ}$$$$R \cos \theta = 1 2. 6 \times \cos 1 4 0 {}^ {\circ} \approx - 9. 7 \mathrm {m}$$ $$R \sin \theta = 1 2. 6 \times \sin 1 4 0 {}^ {\circ} \approx 8. 1 \mathrm {m}$$ $$\vec {\mathrm {R}} = (- 9. 7 \mathrm {m}, 8. 1 \mathrm {m}) = - \vec {\mathrm {i}} \times (9. 7 \mathrm {m}) + \vec {\mathrm {j}} \times (8. 1 \mathrm {m})$$

Answer: $\vec{\mathrm{R}} = (-9.7\mathrm{m}, 8.1\mathrm{m}) = -\vec{\mathrm{i}} \times (9.7\mathrm{m}) + \vec{\mathrm{j}} \times (8.1\mathrm{m})$

(b).

$$R = 4. 0 0 \mathrm {c m}; \theta = 5 0 {}^ {\circ}$$$$R \cos \theta = 4. 0 0 \times \cos 5 0 {}^ {\circ} \approx - 2. 5 7 \mathrm {c m}$$ $$R \sin \theta = 4. 0 0 \times \sin 5 0 {}^ {\circ} \approx 3. 0 6 \mathrm {c m}$$ $$\vec {R} = (- 2. 5 7 \mathrm {c m}, 3. 0 6 \mathrm {c m}) = - \vec {\mathrm {i}} \times (2. 5 7 \mathrm {c m}) + \vec {\mathrm {j}} \times (3. 0 6 \mathrm {c m})$$

Answer: $\vec{\mathsf{R}} = (-2.57\mathrm{cm},3.06\mathrm{cm}) = -\vec{\mathsf{i}}\times (2.57\mathrm{cm}) + \vec{\mathsf{j}}\times (3.06\mathrm{cm})$

(c).

$$R = 2 0. 0 \mathrm {i n}; \theta = 2 2 0 {}^ {\circ}$$$$R \cos \theta = 2 0. 0 \times \cos 2 2 0 {}^ {\circ} \approx - 1 5. 3 \mathrm {i n}$$ $$R \sin \theta = 2 0. 0 \times \sin 2 2 0 {}^ {\circ} \approx - 1 2. 9 \mathrm {i n}$$ $$\vec {R} = (- 1 5. 3 \mathrm {i n}, - 1 2. 9 \mathrm {i n}) = - \vec {\mathrm {i}} \times (1 5. 3 \mathrm {i n}) - \vec {\mathrm {j}} \times (1 2. 9 \mathrm {i n})$$

Answer: $\vec{\mathsf{R}} = (-15.3\mathrm{in}, - 12.9\mathrm{in}) = -\vec{\mathsf{i}}\times (15.3\mathrm{in}) - \vec{\mathsf{j}}\times (12.9\mathrm{in})$

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