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-vecto

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 \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}

\dots \dots . \dots . \dots \dots \dots \dots . \dots .

(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 .

\dots . \dots . \dots . \dots . . \dots . \dots . . \dots .

(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} .

\dots . \dots . . \dots . \dots . \dots . . \dots . \dots .

(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} .

\dots . \dots . \dots . \dots . \dots . . \dots . . \dots .

(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!