# Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates

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Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates A = $A_x\hat i+A_x \hat j$ where $\hat i$and $\hat j$ are unit vector along x and y direction, respectively and Aand Ay are corresponding components of A (Fig.). Motion can also be studied by expressing vectors in circular polar co-ordinates as A = $A_r\hat r+A_\theta \hat \theta$ where$\hat r$ = $\frac{\vec r}{\begin{vmatrix} r \end{vmatrix}}$ = cos θ $\hat i$+ sin θ $\hat j$ and $\hat \theta$= −sin θ$\hat i$+ cos θ$\hat j$ are unit vectors along direction in which ‘r’ and ‘θ’ are increasing.

(a) Express $\hat i$ and$\,\hat j$ in terms of$\, \hat r$ and$\,\hat \theta$

(b) Show that both$\,\hat r \,$and$\,\hat \theta \,$are unit vectors and are perpendicular to each other.

(c) Show that $\frac{d}{dt}$$\hat r$) = ω$\hat r$

(d) For a particle moving along a spiral given by r = aθ$\hat r$ where a = 1 (unit), find dimensions of ‘a’.

(e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.

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(a) unit vector, $\hat r$= cos θ$\hat i$+ sin θ$\hat j$   (1)

$\hat \theta$ = −sin θ$\hat i$+ cos θ$\hat j$   (2)

Multiplying eq. (1) by sin θ and eq. (2) by cos θ,

Adding ⇒ $\hat r$sin θ + $\hat \theta$ cos θ =$\hat j$

Now, multiplying eq. (1) by cos θ and eq. (2) by xsin θ

n($\hat r$cos θ $\hat \theta$ sin θ ) = $\hat i$

(b) $\hat r.\hat \theta= (cos\,\theta\hat i+sin\,\theta\hat j).(-sin\,\theta\hat i+cos\,\theta\hat j)$

= $-cos\,\theta.sin\,\theta+sin\,\theta.cos\,\theta=0$

θ = 90º

(c) $\frac{dr}{dt}=\frac{d}{dt}(cos\,\theta\hat i+sin\,\theta\hat j)$

= $\omega(-sin\,\theta\hat i+cos\,\theta\hat j)$ ∵ ( $\frac{d\theta}{dt}=\omega$)

(d) As, r = aθ$\hat r$

⇒ [a] = L = [M0L1T0]

(e) $v=\frac{dr}{dt}=\frac{d\theta}{dt} \hat r+\theta\frac{d\hat r}{dt}$

= $\frac{d\theta}{dt}\hat r+\theta[(-sin\,\theta\hat i+cos\,\theta\hat j)\frac{d\theta}{dt}]$

v =$\frac{d\theta}{dt}\hat r+\theta\hat \theta \omega=\omega\hat r+\omega\theta\hat \theta$

a = $\frac{dv}{dt}=\frac{d}{dt}[\omega\hat r+\omega\theta\,\hat\theta]=\frac{d}{dt}[\frac{d\theta}{dt} \hat r+\frac{d\theta}{dt}(\theta\,\hat \theta)]$

= $\frac{d^r \theta}{dt^2}\hat r+\omega^2\hat\theta+\frac{d^2 \theta}{dt^2}\times \theta\,\hat \theta+\omega^2\hat\theta+w^2\theta(-\hat r)$

= $(\frac{d^2\theta}{dt^2}-\omega^2\theta)\,\hat r$ +$(2\omega^2+\frac{d^2\theta}{dt^2}\hat \theta)$