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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_{y}\hat{j} where \hat{i} and \hat{j} are unit vector long x and y directions,

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_{y}\hat{j} where \hat{i} and \hat{j} are unit vector long x and y directions, respectively and A_{x} and A_{y} are corresponding components of A (fig.4.9). 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{r}{r}=\cos \theta \hat{i}+\sin \theta \hat{j} and \hat{\theta }=-\sin \theta \hat{i}+\cos \theta \hat{j} are unit vectors along direction in which 'r' and '\theta ' 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})=\omega \hat{\theta } where  \omega =\frac{d\theta }{dt} and \frac{d}{dt}(\theta )=-\omega \hat{r}

(d) For particle moving along a spiral given by r=a\theta \hat{r}, where a=1 (unit), find dimensions of 'a'.

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

 

Answers (1)

a) \hat{r}=\cos\; \theta \hat{i}+\sin\; \theta \hat{J}\; \; \; \; \; \; -------(1)

\hat{\theta }=-\sin\; \theta \hat{i}+\cos \theta \hat{J}\; \; \; \; ----------(2)

When we multiply 1 by \sin \theta and 2 by \cos \theta, we get :

\hat{r}\sin \theta +\hat{\theta }\cos \theta =(\sin^{2}\theta +\cos^{2}\theta )\hat{J}

\hat{r}\sin \theta +\hat{\theta }\cos \theta =\hat{J}\; \; \; \; \; \; \; \; ---------(3)

When we multiply 1 by \cos \theta and 2 by \sin \theta we get :

\hat{r}\cos \theta =\cos^{2}\theta \hat{i}+\sin \theta \cos \theta \hat{J}\; \; \; \; \; -----(4)

\hat{\theta }\sin \theta =-\sin^{2}\theta \hat{i}+\sin \theta \cos \theta \hat{J}\; \; \; \; \; -----(5)

Subtracting equation 5 from 4 and comparing the coefficients we get,

\hat{r}\cos\theta -\hat{\theta }\sin \theta =\hat{i}

\hat{r}\sin\theta -\hat{\theta }\cos \theta =\hat{J}

b) from equation 1 and 2, through the dot product method we get,

\hat{r}.\hat{\theta }=\left ( \cos\; \theta \hat{i}+\sin \theta \hat{J} \right ).\left ( -\sin\; \theta \hat{i}+\cos\; \theta \hat{J} \right )

\left | \hat{r} \right |\left | \hat{\theta } \right |\cos\; \theta =0

Since LHS elements cannot be zero,

\cos \theta =0

and \theta =90

c) \hat{r}=\cos\; \theta \hat{i}+\sin \theta \hat{J}

\frac{d\; \hat{r}}{dt}=\frac{d(\cos\; \theta \hat{i}+\sin \theta \hat{J})}{dt}

=\left ( -\sin \theta \hat{i} +\cos \theta \hat{J}\right )\frac{d\; \theta }{dt}

Since, \omega =\frac{d\theta }{dt}

\frac{d\hat{r}}{dt}=\omega \hat{\theta }

d) \hat{r}=\left | \hat{a} \right |\left | \hat{\theta } \right |\hat{r}

now looking at the dimensions of the quantities on the LHS and the RHS,

[a]=\frac{[\hat{r}]}{[\hat{\theta }][\hat{r}]}

=\frac{[M^{0}L^{1}T^{0}]}{[M^{0}L^{0}T^{0}][M^{0}L^{0}T^{0}]}

=[M^{0}L^{1}T^{0}]

e) a=1

\hat{r}=\hat{\theta }[\cos\; \theta \hat{i}+\sin\; \theta \hat{J}]

V=\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=\omega \hat{r}+\theta .\hat{\theta }\omega

\vec{a}=\frac{dv}{dt}

=\frac{d(\omega \hat{r}+\theta .\hat{\theta }\omega )}{dt}

=d^{2}\frac{\theta }{dt^{2}}\hat{r}+d\frac{\theta }{dt}d\; \hat{r}\frac{\hat{r}}{dt}+d^{2}\frac{\theta }{dt^{2}}(\theta .\hat{\theta })+d\frac{\theta }{dt}(\theta .\hat{\theta })

=d^{2}\frac{\theta }{dt^{2}}\; \hat{r}+\omega \left ( \frac{-\sin\; \theta \hat{i}\; d}{dt}+\frac{\cos\; \theta \hat{J}\; d\; \theta }{dt} \right )+\frac{d^{2}\theta }{dt^{2}}(\theta .\hat{\theta })+\frac{\omega \; d\; \theta }{dt}(\theta .\hat{\theta })

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

\vec{a}=\left ( \frac{d^{2}\theta }{dt^{2}-\omega ^{2}} \right )\hat{r}+\left ( \frac{2\omega ^{2+}d^{2}\theta }{dt^{2}\theta } \right )\hat{\theta }

 

Posted by

infoexpert23

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