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

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

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

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