Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates A^→ = A_x i^∧ + A_y j^∧ where i^∧ and j^∧ are unit vector along x and y directions, respectively and A_x and A_y are corresponding components of A^→ (Figure). Motion can also be studied by expressing vectors in circular polar co-ordinates as A^→ = A_r r^∧ + A_θ θ^∧ where r^∧ = (r^→ /r) = cos θ i^∧ + sin θ j^∧ and θ^∧ = −sin θ i^∧ + cos θ j^∧ are unit vectors along direction in which ′r′ and ′θ′ are increasing. (a) Express i^∧ and j^∧ in terms of r^∧ and θ^∧ (b) Show that both r^∧ and θ^∧ are unit vectors and are perpendicular to each other. (c) Show that (d/dt)(r^∧ ) = ωθ^∧ where ω = (dθ/dt) and (d/dt)(θ^∧ ) = −ωr^∧ (d) For a particle moving along a spiral given by r^→ = αθr^∧ , where α = 1 (unit), find dimensions of ′α′. (e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.


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Question Number 18749 by Tinkutara last updated on 29/Jul/17

$Motion in two dimensions, in a plane$ $can be studied by expressing position,$ $velocity and acceleration as vectors in$ $Cartesian co - ordinates \vec{A} = A_{x} \overset{\land}{i} + A_{y} \overset{\land}{j}$ $where \overset{\land}{i} and \overset{\land}{j} are unit vector along x$ $and y directions, respectively and A_{x}$ $and A_{y} are corresponding components$ $of \vec{A} (Figure) . Motion can also be$ $studied by expressing vectors in circular$ $polar co - ordinates as \vec{A} = A_{r} \overset{\land}{r} + A_{θ} \overset{\land}{θ}$ $where \overset{\land}{r} = \frac{\vec{r}}{r} = \cos θ \overset{\land}{i} + \sin θ \overset{\land}{j} and \overset{\land}{θ} =$ $- \sin θ \overset{\land}{i} + \cos θ \overset{\land}{j} are unit vectors along$ $direction in which^{'} r^{'} and^{'} θ^{'} are$ $increasing .$ $(a) Express \overset{\land}{i} and \overset{\land}{j} in terms of \overset{\land}{r} and \overset{\land}{θ}$ $(b) Show that both \overset{\land}{r} and \overset{\land}{θ} are unit$ $vectors and are perpendicular to each$ $other .$ $(c) Show that \frac{d}{d t} (\overset{\land}{r}) = ω \overset{\land}{θ} where$ $ω = \frac{d θ}{d t} and \frac{d}{d t} (\overset{\land}{θ}) = - ω \overset{\land}{r}$ $(d) For a particle moving along a spiral$ $given by \vec{r} = α θ \overset{\land}{r}, where α = 1 (unit),$ $find dimensions of^{'} α^{'} .$ $(e) Find velocity and acceleration in$ $polar vector representation for particle$ $moving along spiral described in (d)$ $above .$
Commented by Tinkutara last updated on 29/Jul/17

	Answered by ajfour last updated on 29/Jul/17
	$(a)Let i^� =λr^� +μθ^� i^� =λ(cos θi^� +sin θj^� )+μ(−sin θi^� +cos θj^� ) ⇒ [ λcos θ−μsin θ=1 ]×cos θ and [ λsin θ+μcos θ=0 ]×sin θ Adding, we obtain λ=cos θ ; μ=−sin θ So, i^� =(cos θ)r^� −(sin θ)θ^� Let j^� =ρr^� +εθ^� or j^� =ρ(cos θi^� +sin θj^� )+ε(−sin θi^� +cos θj^� ) ⇒ [ ρcos θ−εsin θ=0 ]×cos θ , and [ ρsin θ+εcos θ=1 ]×sin θ Adding we get, ρ=sin θ ; ε=cos θ Hence j^� =(sin θ)r^� +(cos θ)θ^� ... .... .... ... ... ... .... .... (b) ∣r^� ∣=(√(cos^2 θ+sin^2 θ)) =1 hence a unit vector ∣θ^� ∣=(√((−sin θ)^2 +(cos θ)^2 )) =1 a unit vector. r^� .θ^� =(cos θi^� +sin θj^� ).(−sin θi^� +cos θj^� ) =−cos θsin θ+sin θcos θ =0 ⇒ r^� and θ^� are ⊥ to each other. .... .... .... ..... .... ..... .... (c) (dr^� /dt)=(d/dt)(cos θi^� +sin θj^� ) =(dθ/dt)(−sin θi^� +cos θj^� ) ⇒ (dr^� /dt)=ωθ^� . (dθ^� /dt)=(d/dt)(−sin θi^� +cos θj^� ) =(dθ/dt)(−cos θi^� −sin θj^� ) ⇒ (dθ^� /dt)=−ωr^� . .... ..... .... .... ..... .... .... (d) r^→ =αθr^� θ is dimensionless, so is unit vector r^� ; so [α]=[∣r^� ∣] =M^0 L^1 T^0 . .... .... .... .... ..... ..... .... (e) v^→ =(dr^→ /dt)=α(d/dt)(θr^� ) =α((dθ/dt)r^� +θ(dr^� /dt)) =αω(r^� +θθ^� ) . a^→ =(dv^→ /dt)=α(d/dt)(ωr^� +ωθθ^� ) =α((dω/dt)r^� +ω(dr^� /dt))+α(θ(dω/dt)θ^� +ω(dθ/dt)θ^� +ωθ(dθ^� /dt)) =α((dω/dt)r^� +ω(ωθ^� ))+α[θ(dω/dt)θ^� +ω(dθ/dt)θ^� +ωθ(−ωr^� )] a^→ =α{((dω/dt)−ω^2 θ)r^� +(2ω^2 +θ(dω/dt))θ^� } .$
	$(a) Let \hat{i} = λ \hat{r} + μ \hat{θ}$ $\hat{i} = λ (\cos θ \hat{i} + \sin θ \hat{j}) + μ (- \sin θ \hat{i} + \cos θ \hat{j})$ $\Rightarrow [λ \cos θ - μ \sin θ = 1] \times \cos θ$ $and [λ \sin θ + μ \cos θ = 0] \times \sin θ$ $Adding, we obtain$ $λ = \cos θ; μ = - \sin θ$ $So, \hat{i} = (\cos θ) \hat{r} - (\sin θ) \hat{θ}$ $Let \hat{j} = ρ \hat{r} + ϵ \hat{θ}$ $or \hat{j} = ρ (\cos θ \hat{i} + \sin θ \hat{j}) + ϵ (- \sin θ \hat{i} + \cos θ \hat{j})$ $\Rightarrow [ρ \cos θ - ϵ \sin θ = 0] \times \cos θ, and$ $[ρ \sin θ + ϵ \cos θ = 1] \times \sin θ$ $Adding we get,$ $ρ = \sin θ; ϵ = \cos θ$ $Hence \hat{j} = (\sin θ) \hat{r} + (\cos θ) \hat{θ}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $(b) ∣ \hat{r} ∣= \sqrt{\cos^{2} θ + \sin^{2} θ} = 1$ $hence a unit vector$ $∣ \hat{θ} ∣= \sqrt{{(- \sin θ)}^{2} + {(\cos θ)}^{2}} = 1$ $a unit vector .$ $\hat{r} . \hat{θ} = (\cos θ \hat{i} + \sin θ \hat{j}) . (- \sin θ \hat{i} + \cos θ \hat{j})$ $= - \cos θ \sin θ + \sin θ \cos θ = 0$ $\Rightarrow \hat{r} and \hat{θ} are ⊥ to each other .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $(c) \frac{d \hat{r}}{dt} = \frac{d}{dt} (\cos θ \hat{i} + \sin θ \hat{j})$ $= \frac{d θ}{dt} (- \sin θ \hat{i} + \cos θ \hat{j})$ $\Rightarrow \frac{d \hat{r}}{dt} = ω \hat{θ} .$ $\frac{d \hat{θ}}{dt} = \frac{d}{dt} (- \sin θ \hat{i} + \cos θ \hat{j})$ $= \frac{d θ}{dt} (- \cos θ \hat{i} - \sin θ \hat{j})$ $\Rightarrow \frac{d \hat{θ}}{dt} = - ω \hat{r} .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $(d) \vec{r} = α θ \hat{r}$ $θ is dimensionless, so is unit$ $vector \hat{r}; so [α] = [∣ \hat{r} ∣] = M^{0} L^{1} T^{0} .$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $(e) \vec{v} = \frac{d \vec{r}}{dt} = α \frac{d}{dt} (θ \hat{r})$ $= α (\frac{d θ}{dt} \hat{r} + θ \frac{d \hat{r}}{dt})$ $= α ω (\hat{r} + θ \hat{θ}) .$ $\vec{a} = \frac{d \vec{v}}{dt} = α \frac{d}{dt} (ω \hat{r} + ω θ \hat{θ})$ $= α (\frac{d ω}{dt} \hat{r} + ω \frac{d \hat{r}}{dt}) + α (θ \frac{d ω}{dt} \hat{θ} + ω \frac{d θ}{dt} \hat{θ} + ω θ \frac{d \hat{θ}}{dt})$ $= α (\frac{d ω}{dt} \hat{r} + ω (ω \hat{θ})) + α [θ \frac{d ω}{dt} \hat{θ} + ω \frac{d θ}{dt} \hat{θ} + ω θ (- ω \hat{r})]$ $\vec{a} = α {(\frac{d ω}{dt} - ω^{2} θ) \hat{r} + (2 ω^{2} + θ \frac{d ω}{dt}) \hat{θ}} .$
	Commented by Tinkutara last updated on 29/Jul/17

	$Thank you very much ajfour Sir!$
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