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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 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.
$$\mathrm{Motion}\:\mathrm{in}\:\mathrm{two}\:\mathrm{dimensions},\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{studied}\:\mathrm{by}\:\mathrm{expressing}\:\mathrm{position}, \\ $$$$\mathrm{velocity}\:\mathrm{and}\:\mathrm{acceleration}\:\mathrm{as}\:\mathrm{vectors}\:\mathrm{in} \\ $$$$\mathrm{Cartesian}\:\mathrm{co}-\mathrm{ordinates}\:\overset{\rightarrow} {{A}}\:=\:{A}_{{x}} \overset{\wedge} {{i}}\:+\:{A}_{{y}} \overset{\wedge} {{j}} \\ $$$$\mathrm{where}\:\overset{\wedge} {{i}}\:\mathrm{and}\:\overset{\wedge} {{j}}\:\mathrm{are}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{along}\:{x} \\ $$$$\mathrm{and}\:{y}\:\mathrm{directions},\:\mathrm{respectively}\:\mathrm{and}\:{A}_{{x}} \\ $$$$\mathrm{and}\:{A}_{{y}} \:\mathrm{are}\:\mathrm{corresponding}\:\mathrm{components} \\ $$$$\mathrm{of}\:\overset{\rightarrow} {{A}}\:\left(\mathrm{Figure}\right).\:\mathrm{Motion}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be} \\ $$$$\mathrm{studied}\:\mathrm{by}\:\mathrm{expressing}\:\mathrm{vectors}\:\mathrm{in}\:\mathrm{circular} \\ $$$$\mathrm{polar}\:\mathrm{co}-\mathrm{ordinates}\:\mathrm{as}\:\overset{\rightarrow} {{A}}\:=\:{A}_{{r}} \overset{\wedge} {{r}}\:+\:{A}_{\theta} \overset{\wedge} {\theta} \\ $$$$\mathrm{where}\:\overset{\wedge} {{r}}\:=\:\frac{\overset{\rightarrow} {{r}}}{{r}}\:=\:\mathrm{cos}\:\theta\:\overset{\wedge} {{i}}\:+\:\mathrm{sin}\:\theta\:\overset{\wedge} {{j}}\:\mathrm{and}\:\overset{\wedge} {\theta}\:= \\ $$$$−\mathrm{sin}\:\theta\:\overset{\wedge} {{i}}\:+\:\mathrm{cos}\:\theta\:\overset{\wedge} {{j}}\:\mathrm{are}\:\mathrm{unit}\:\mathrm{vectors}\:\mathrm{along} \\ $$$$\mathrm{direction}\:\mathrm{in}\:\mathrm{which}\:'{r}'\:\mathrm{and}\:'\theta'\:\mathrm{are} \\ $$$$\mathrm{increasing}. \\ $$$$\left({a}\right)\:\mathrm{Express}\:\overset{\wedge} {{i}}\:\mathrm{and}\:\overset{\wedge} {{j}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\overset{\wedge} {{r}}\:\mathrm{and}\:\overset{\wedge} {\theta} \\ $$$$\left({b}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{both}\:\overset{\wedge} {{r}}\:\mathrm{and}\:\overset{\wedge} {\theta}\:\mathrm{are}\:\mathrm{unit} \\ $$$$\mathrm{vectors}\:\mathrm{and}\:\mathrm{are}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{each} \\ $$$$\mathrm{other}. \\ $$$$\left({c}\right)\:\mathrm{Show}\:\mathrm{that}\:\frac{{d}}{{dt}}\left(\overset{\wedge} {{r}}\right)\:=\:\omega\overset{\wedge} {\theta}\:\mathrm{where} \\ $$$$\omega\:=\:\frac{{d}\theta}{{dt}}\:\mathrm{and}\:\frac{{d}}{{dt}}\left(\overset{\wedge} {\theta}\right)\:=\:−\omega\overset{\wedge} {{r}} \\ $$$$\left({d}\right)\:\mathrm{For}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{along}\:\mathrm{a}\:\mathrm{spiral} \\ $$$$\mathrm{given}\:\mathrm{by}\:\overset{\rightarrow} {{r}}\:=\:\alpha\theta\overset{\wedge} {{r}},\:\mathrm{where}\:\alpha\:=\:\mathrm{1}\:\left(\mathrm{unit}\right), \\ $$$$\mathrm{find}\:\mathrm{dimensions}\:\mathrm{of}\:'\alpha'. \\ $$$$\left({e}\right)\:\mathrm{Find}\:\mathrm{velocity}\:\mathrm{and}\:\mathrm{acceleration}\:\mathrm{in} \\ $$$$\mathrm{polar}\:\mathrm{vector}\:\mathrm{representation}\:\mathrm{for}\:\mathrm{particle} \\ $$$$\mathrm{moving}\:\mathrm{along}\:\mathrm{spiral}\:\mathrm{described}\:\mathrm{in}\:\left({d}\right) \\ $$$$\mathrm{above}. \\ $$
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^� } .
$$\left(\mathrm{a}\right)\mathrm{Let}\:\:\:\:\:\hat {\mathrm{i}}=\lambda\hat {\mathrm{r}}+\mu\hat {\theta} \\ $$$$\:\:\hat {\mathrm{i}}=\lambda\left(\mathrm{cos}\:\theta\hat {\mathrm{i}}+\mathrm{sin}\:\theta\hat {\mathrm{j}}\right)+\mu\left(−\mathrm{sin}\:\theta\hat {\mathrm{i}}+\mathrm{cos}\:\theta\hat {\mathrm{j}}\right) \\ $$$$\Rightarrow\:\:\:\:\:\left[\:\lambda\mathrm{cos}\:\theta−\mu\mathrm{sin}\:\theta=\mathrm{1}\:\:\right]×\mathrm{cos}\:\theta \\ $$$$\mathrm{and}\:\:\left[\:\lambda\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta=\mathrm{0}\:\:\right]×\mathrm{sin}\:\theta \\ $$$$\mathrm{Adding},\:\mathrm{we}\:\mathrm{obtain} \\ $$$$\:\:\:\:\lambda=\mathrm{cos}\:\theta\:\:;\:\:\mu=−\mathrm{sin}\:\theta \\ $$$$\mathrm{So},\:\:\:\hat {\mathrm{i}}=\left(\mathrm{cos}\:\theta\right)\hat {\mathrm{r}}−\left(\mathrm{sin}\:\theta\right)\hat {\theta} \\ $$$$\:\:\:\:\mathrm{Let}\:\:\:\:\:\hat {\mathrm{j}}=\rho\hat {\mathrm{r}}+\epsilon\hat {\theta} \\ $$$$\mathrm{or}\:\hat {\mathrm{j}}=\rho\left(\mathrm{cos}\:\theta\hat {\mathrm{i}}+\mathrm{sin}\:\theta\hat {\mathrm{j}}\right)+\epsilon\left(−\mathrm{sin}\:\theta\hat {\mathrm{i}}+\mathrm{cos}\:\theta\hat {\mathrm{j}}\right) \\ $$$$\Rightarrow\:\:\left[\:\rho\mathrm{cos}\:\theta−\epsilon\mathrm{sin}\:\theta=\mathrm{0}\:\right]×\mathrm{cos}\:\theta\:,\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\left[\:\rho\mathrm{sin}\:\theta+\epsilon\mathrm{cos}\:\theta=\mathrm{1}\:\right]×\mathrm{sin}\:\theta \\ $$$$\mathrm{Adding}\:\mathrm{we}\:\mathrm{get}, \\ $$$$\:\:\:\:\:\:\:\:\rho=\mathrm{sin}\:\theta\:;\:\:\epsilon=\mathrm{cos}\:\theta \\ $$$$\mathrm{Hence}\:\:\hat {\mathrm{j}}=\left(\mathrm{sin}\:\theta\right)\hat {\mathrm{r}}+\left(\mathrm{cos}\:\theta\right)\hat {\theta} \\ $$$$…\:\:\:\:….\:\:\:\:….\:\:\:…\:\:\:…\:\:\:…\:\:\:….\:\:\:…. \\ $$$$\left(\mathrm{b}\right)\:\mid\hat {\mathrm{r}}\mid=\sqrt{\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{sin}\:^{\mathrm{2}} \theta}\:=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{hence}\:\mathrm{a}\:\mathrm{unit}\:\mathrm{vector} \\ $$$$\:\:\:\:\:\:\:\:\mid\hat {\theta}\mid=\sqrt{\left(−\mathrm{sin}\:\theta\right)^{\mathrm{2}} +\left(\mathrm{cos}\:\theta\right)^{\mathrm{2}} }\:=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:\mathrm{unit}\:\mathrm{vector}. \\ $$$$\:\hat {\mathrm{r}}.\hat {\theta}=\left(\mathrm{cos}\:\theta\hat {\mathrm{i}}+\mathrm{sin}\:\theta\hat {\mathrm{j}}\right).\left(−\mathrm{sin}\:\theta\hat {\mathrm{i}}+\mathrm{cos}\:\theta\hat {\mathrm{j}}\right) \\ $$$$\:\:\:\:\:\:\:=−\mathrm{cos}\:\theta\mathrm{sin}\:\theta+\mathrm{sin}\:\theta\mathrm{cos}\:\theta\:=\mathrm{0} \\ $$$$\:\Rightarrow\:\:\:\:\hat {\mathrm{r}}\:\mathrm{and}\:\hat {\theta}\:\mathrm{are}\:\bot\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$$$….\:\:\:….\:\:\:\:….\:\:\:…..\:\:\:\:….\:\:\:\:…..\:\:\:\:…. \\ $$$$\:\left(\mathrm{c}\right)\:\frac{\mathrm{d}\hat {\mathrm{r}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{cos}\:\theta\hat {\mathrm{i}}+\mathrm{sin}\:\theta\hat {\mathrm{j}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{d}\theta}{\mathrm{dt}}\left(−\mathrm{sin}\:\theta\hat {\mathrm{i}}+\mathrm{cos}\:\theta\hat {\mathrm{j}}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:\frac{\mathrm{d}\hat {\mathrm{r}}}{\mathrm{dt}}=\omega\hat {\theta}\:. \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{d}\hat {\theta}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(−\mathrm{sin}\:\theta\hat {\mathrm{i}}+\mathrm{cos}\:\theta\hat {\mathrm{j}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{d}\theta}{\mathrm{dt}}\left(−\mathrm{cos}\:\theta\hat {\mathrm{i}}−\mathrm{sin}\:\theta\hat {\mathrm{j}}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\:\frac{\mathrm{d}\hat {\theta}}{\mathrm{dt}}=−\omega\hat {\mathrm{r}}\:.\:\:\:\: \\ $$$$….\:\:\:…..\:\:\:….\:\:\:\:….\:\:\:\:…..\:\:\:….\:\:\:…. \\ $$$$\left(\mathrm{d}\right)\:\:\overset{\rightarrow} {\mathrm{r}}=\alpha\theta\hat {\mathrm{r}} \\ $$$$\theta\:\mathrm{is}\:\mathrm{dimensionless},\:\mathrm{so}\:\mathrm{is}\:\mathrm{unit}\: \\ $$$$\mathrm{vector}\:\hat {\mathrm{r}}\:;\:\mathrm{so}\:\:\left[\alpha\right]=\left[\mid\hat {\mathrm{r}}\mid\right]\:=\mathrm{M}^{\mathrm{0}} \mathrm{L}^{\mathrm{1}} \mathrm{T}^{\mathrm{0}} \:. \\ $$$$….\:\:\:….\:\:\:\:….\:\:\:\:….\:\:\:\:…..\:\:\:\:…..\:\:\:…. \\ $$$$\left(\mathrm{e}\right)\:\overset{\rightarrow} {\mathrm{v}}=\frac{\mathrm{d}\overset{\rightarrow} {\mathrm{r}}}{\mathrm{dt}}=\alpha\frac{\mathrm{d}}{\mathrm{dt}}\left(\theta\hat {\mathrm{r}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\alpha\left(\frac{\mathrm{d}\theta}{\mathrm{dt}}\hat {\mathrm{r}}+\theta\frac{\mathrm{d}\hat {\mathrm{r}}}{\mathrm{dt}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\alpha\omega\left(\hat {\mathrm{r}}+\theta\hat {\theta}\right)\:. \\ $$$$\:\:\:\overset{\rightarrow} {\mathrm{a}}=\frac{\mathrm{d}\overset{\rightarrow} {\mathrm{v}}}{\mathrm{dt}}=\alpha\frac{\mathrm{d}}{\mathrm{dt}}\left(\omega\hat {\mathrm{r}}+\omega\theta\hat {\theta}\right) \\ $$$$=\alpha\left(\frac{\mathrm{d}\omega}{\mathrm{dt}}\hat {\mathrm{r}}+\omega\frac{\mathrm{d}\hat {\mathrm{r}}}{\mathrm{dt}}\right)+\alpha\left(\theta\frac{\mathrm{d}\omega}{\mathrm{dt}}\hat {\theta}+\omega\frac{\mathrm{d}\theta}{\mathrm{dt}}\hat {\theta}+\omega\theta\frac{\mathrm{d}\hat {\theta}}{\mathrm{dt}}\right) \\ $$$$=\alpha\left(\frac{\mathrm{d}\omega}{\mathrm{dt}}\hat {\mathrm{r}}+\omega\left(\omega\hat {\theta}\right)\right)+\alpha\left[\theta\frac{\mathrm{d}\omega}{\mathrm{dt}}\hat {\theta}+\omega\frac{\mathrm{d}\theta}{\mathrm{dt}}\hat {\theta}+\omega\theta\left(−\omega\hat {\mathrm{r}}\right)\right] \\ $$$$\overset{\rightarrow} {\mathrm{a}}=\alpha\left\{\left(\frac{\mathrm{d}\omega}{\mathrm{dt}}−\omega^{\mathrm{2}} \theta\right)\hat {\mathrm{r}}+\left(\mathrm{2}\omega^{\mathrm{2}} +\theta\frac{\mathrm{d}\omega}{\mathrm{dt}}\right)\hat {\theta}\right\}\:. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 29/Jul/17
Thank you very much ajfour Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{ajfour}\:\mathrm{Sir}! \\ $$

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