question_answer 68)

                  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 along \[x\] and \[y\] directions, respectively and \[{{A}_{x}}\] and \[{{A}_{y}}\] are corresponding components of A (fig). Motion can also be studied by expressing vectors in circular polar co-ordinates at \[A=\,{{A}_{r}}\hat{r}+{{A}_{\theta }}\hat{\theta },\] 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 \[\frac{d}{dt}\,\,(\hat{\theta })\,\,=-\,\omega r\]               (d) For a 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 representation for particle moving along spiral described in (d) above.

Answer:

                  (a) \[\hat{r}\,\,=\,\,\cos \,\theta \,\hat{i}\,+\,\sin \theta \hat{j}\]                      ?. (i)                 \[\hat{\theta }\,=-\,\sin \,\theta \hat{i}+\,\cos \,\theta \hat{j}\]                           ? (ii) Multiplying eqn. (i) by \[\sin \theta \] and eqn. (ii) by \[\cos \theta \] and adding, we get                             \[\hat{j}\,=\,\hat{r}\,\sin \,\,\theta +\,\hat{\theta }\,\,\cos \,\theta \]              ... (iii) Put this value in eqn. (i), we get                 \[\hat{i}=\hat{r}\,\cos \theta -\,\hat{\theta }\,\sin \,\theta \]               ... (iv)                 (b) \[|\hat{r}|\,\,=\hat{r}.\,\hat{r}\,=1\] and \[|\hat{\theta }|\,=\,\hat{\theta }.\,\hat{\theta }\,\,=1\] Since magnitude of both \[\hat{r}\] and \[\hat{\theta }\] in unity (1), so they are unit vectors, \[\hat{r}\] and \[\hat{\theta }\] will be perpendicular to each other if \[\hat{r}\,.\,\hat{\theta }\,=0\] Now \[\hat{r}.\,\hat{\theta }\]                 \[=\,(\cos \,\theta \hat{i}+\,\sin \,\theta \hat{j})\,.\,(-\,\sin \,\theta \hat{i}\,+\,\cos \theta \hat{j})\]                 \[=-\,\sin \,\theta \,\,\cos \,\,\theta +\,\sin \theta \,\cos \theta =0\]                 \[(\because \,\,\,\hat{i}\,.\,\hat{i}\,=\,\hat{j}\,.\,\hat{j}\,=1\,\,\text{and}\,\,i\,.\,\hat{j}\,=\,j.\,\hat{i}\,=0)\] Therefore, \[\hat{r}\] and \[\hat{\theta }\] are perpendicular to each other.                 (c) \[\frac{d}{dt}\,(\hat{r})\,\,=\,\frac{d}{dt}\,(\cos \,\theta \,\hat{i}\,+\,\sin \,\theta \,\hat{j})\] \[=(-\sin \theta )\,\frac{d\theta }{dt}\,\hat{i}\,+\,(\cos \theta )\,\frac{d\theta }{dt}\,\hat{j}\]                 \[=\,\frac{d\theta }{dt}\,[-\sin \theta \,\hat{i}\,+\,\cos \,\theta \,\hat{j}]\]                 Since \[\frac{d\theta }{dt}\,\,=\,\omega \] and \[-\,\sin \,\theta \hat{i}\,+\,\cos \theta \hat{j}\,=\hat{\theta }\]                 \[\therefore \]\[\frac{d}{dt}\,\,(\hat{r})=\,\omega \hat{\theta }\]                 \[\frac{d}{dt}\,(\hat{\theta })=\,\frac{d}{dt}\,(-\,\sin \,\theta \hat{i}\,+\,\cos \theta \hat{j})\]                 \[=-\frac{d\theta }{dt}\,\,\cos \,\theta \,\hat{i}\,-\,\frac{d\theta }{dt}\,\sin \,\theta \,\,\hat{j}\]                 \[=-\,\frac{d\theta }{dt}\,(\cos \theta \,\hat{i}\,+\,\sin \,\theta \hat{j})\]                 \[=-\omega \,\hat{r}\]              [using eqn. (i)]                 (e) \[\vec{r}\,=a\,\,\theta \,\,\,\hat{r}\,=\,\theta \,\,\,\hat{r}\]            \[(\because \,\,a=1)\]                 \[\therefore \] \[\vec{\upsilon }\,=\,\frac{d\,\,\vec{r}}{dt}=\,\theta \,\frac{d\hat{r}}{dt}+\,\hat{r}\,\frac{d\theta }{dt}\]            ... (i)                 Since \[\frac{d\hat{r}}{dt}=\omega \,\hat{\theta }\] and \[\frac{d\theta }{dt}=\,\omega \]                 \[\therefore \] \[\vec{\upsilon }=\,\omega \,\theta \,\hat{\theta }\,+\,\omega \,\hat{r}=\,\omega \,(\hat{r}\,+\,\theta \,\,\hat{\theta })\]                 Acceleration, \[\vec{a}\,\,=\,\frac{d\vec{\upsilon }}{dt}=\frac{d}{dt}\,(\omega \,\hat{r}+\,\omega \theta \,\,\hat{\theta })\]                 \[=\,\hat{r}\,\frac{d\omega }{dt}+\omega \,\frac{d\hat{r}}{dt}+\,\omega \,\frac{d\theta }{dt}\,\hat{\theta }\]           \[+\,\frac{d\omega }{dt}\,\theta \,\hat{\theta }+\,\omega \theta \,\frac{d\hat{\theta }}{dt}\]                 Since \[\frac{d\theta }{dt}=\omega ,\,\,\frac{d\hat{r}}{dt}=\omega \,\hat{\theta }\] and \[\frac{d\hat{\theta }}{dt}\,=\,-\omega \,\hat{r}\]                 \[\therefore \]\[\,\vec{a}=\,\hat{r}\,\frac{d\omega }{dt}+\,{{\omega }^{2}}\hat{\theta }+\,{{\omega }^{2}}\,\hat{\theta }+\,\frac{d\theta }{dt}\,\theta \,\hat{\theta }\,-\,{{\omega }^{2}}\,\theta \,\hat{r}\]                 \[=\,\hat{r}\,\,\left( \frac{d\omega }{dt}-\,{{\omega }^{2}}\theta  \right)\,+\,\hat{\theta }\,\left( \theta \,\frac{d\omega }{dt}+\,2{{\omega }^{2}} \right)\]                 Now \[=\,\hat{r}\,\,\left( \frac{d\omega }{dt}-\,{{\omega }^{2}}\theta  \right)\,+\,\hat{\theta }\,\left( \theta \,\frac{d\omega }{dt}+\,2{{\omega }^{2}} \right)\]                 \[\therefore \]\[\vec{a}=\,\hat{r}\left( \frac{{{d}^{2}}\theta }{d{{t}^{2}}}-{{\omega }^{2}}\theta  \right)\,\,+\hat{\theta }\,\left( \theta \frac{{{d}^{2}}\theta }{d{{t}^{2}}}+2{{\omega }^{2}} \right)\]