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Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in cartesian coordinates $\vec{A}=A_x \hat{i}+A_y \hat{j}$, 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 $A$. Motion can also be studied by expressing vectors in circular polar coordinates as $\vec{A}=A_r \hat{r}+A_\theta \hat{\theta}$, where $\hat{r}=\frac{\vec{r}}{r}=\cos \theta \hat{i}+\sin \theta \hat{j}$ 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 $\frac{d}{d t}(\hat{r})=\omega \hat{\theta}$, where $\omega=\frac{d \theta}{d t}$ and $\frac{d}{d t}(\hat{\theta})=-\theta \hat{r}$
(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.
(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 $\frac{d}{d t}(\hat{r})=\omega \hat{\theta}$, where $\omega=\frac{d \theta}{d t}$ and $\frac{d}{d t}(\hat{\theta})=-\theta \hat{r}$
(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.
Solution:
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Verified Answer
8. (a) As given that, unit vector
$$
\begin{aligned}
&\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j} \\
&\hat{\theta}=-\sin \theta \hat{i}+\cos \theta \hat{j}
\end{aligned}
$$
By multiplying Eq. (i) by $\sin \theta$ and Eq. (ii) with $\cos \theta$ and then adding both of them.
$$
\begin{aligned}
\hat{r} \sin \theta+\hat{\theta} \cos \theta=& \sin \theta \cdot \cos \theta \hat{i} \\
&+\sin ^2 \theta \hat{j}+\cos ^2 \theta \hat{j} \\
&-\sin \theta \cdot \cos \theta \hat{i} \\
\hat{r} \sin \theta+\hat{\theta} \cos \theta=& \hat{j}\left(\cos ^2 \theta+\sin ^2 \theta\right) \\
&\left(\because \cos ^2 \theta+\sin ^2 \theta=1\right)
\end{aligned}
$$
So, $\hat{r} \sin \theta+\hat{\theta} \cos \theta=\hat{j}$
Now, by multiplying equation (i) by $\cos \theta$ and eq. (ii) with $\sin \theta$ and then subtracting both of them.
$$
\hat{r} \cos \theta-\hat{\theta} \sin \theta=\left(\cos ^2 \theta+\sin ^2 \theta\right) \hat{i}+0
$$
So, $(\hat{r} \cos \theta-\hat{\theta} \sin \theta)=\hat{i}$
(b) From (i), (ii) by dot product
$$
\begin{aligned}
\hat{r} \cdot \hat{\theta} &=(\cos \theta \hat{i}+\sin \theta \hat{j}) \cdot(-\sin \theta \hat{i}+\cos \theta \hat{j}) \\
&=-\cos \theta \cdot \sin \theta+\sin \theta \cdot \cos \theta=0 \\
|\hat{r}| \cdot|\hat{\theta}| \cos \theta=0 \Rightarrow \cos \theta=0
\end{aligned}
$$
So, angle between $\hat{r}$ and $\hat{\theta}=90^{\circ}=\frac{\pi}{2}$.
(c) Given that,
$$
\begin{aligned}
&\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j} \\
&\frac{d \hat{r}}{d t}=\frac{d}{d t}(\cos \theta \hat{i}+\sin \theta \hat{j})=-\sin \theta \cdot \frac{d \theta}{d t} \hat{i}+\cos \theta \cdot \frac{d \theta}{d t} \hat{j} \\
&\frac{d \hat{r}}{d t}=\omega[-\sin \theta \hat{i}+\cos \theta \hat{j}] \quad\left[\because \omega=\frac{d \theta}{d t}\right]
\end{aligned}
$$
(d) Given $\vec{r}=a \hat{\theta} \hat{r}$, here, writing dimensions
$$
[\vec{r}]=[a][\theta][\hat{r}]
$$
dimension of $[a]$ is constant $=\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^0\right]$
(e) As given that, $a=1$ unit
$$
r=\theta \hat{r}=\theta[\cos \theta \hat{i}+\sin \theta \hat{j}]
$$
Velocity,
$$
\begin{aligned}
v &=\frac{d r}{d t}=\frac{d \theta}{d t} \hat{r}+\theta \frac{d}{d t} \hat{r} \\
&=\frac{d \theta}{d t} \hat{r}+\theta \frac{d}{d t}[(\cos \theta \hat{i}+\sin \theta \hat{j})] \\
&=\frac{d \theta}{d t} \hat{r}+\theta\left[(-\sin \theta \hat{i}+\cos \theta \hat{j}) \frac{d \theta}{d t}\right] \\
v &=\frac{d \theta}{d t} \hat{r}+\theta \hat{\theta} \omega=\omega \hat{r}+\omega \theta \hat{\theta}
\end{aligned}
$$
Acceleration,
$$
\begin{aligned}
&\vec{a}=\frac{d}{d t}[\omega \hat{r}+\omega \theta \hat{\theta}]=\frac{d}{d t}\left[\frac{d \theta}{d t} \hat{r}+\frac{d \theta}{d t}(\theta \hat{\theta})\right] \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\frac{d \theta}{d t} \cdot \frac{d \hat{r}}{d t}+\frac{d^2 \theta}{d t^2}(\theta \hat{\theta})+\frac{d \theta}{d t} \frac{d}{d t}(\theta \hat{\theta}) \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\omega \frac{d}{d t}(\cos \theta \hat{i}+\sin \theta \hat{j}) \\
&+\quad+\frac{d^2 \theta}{d t^2}(\theta \cdot \hat{\theta})+\omega \frac{d}{d t}(\theta \hat{\theta}) \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\omega\left[-\sin \theta \hat{i} \frac{d \theta}{d t}+\cos \theta \hat{j} \frac{d \theta}{d t}\right] \\
&\text { From (ii) }+\frac{d^2 \theta}{d t^2}(\theta \hat{\theta})+\frac{\omega d}{d t}(\theta \hat{\theta}) \\
&\vec{a}=\frac{d^2 \theta}{d t^2} \hat{r}+\omega(\hat{\theta}) \omega+\frac{d^2 \theta}{d t^2}(\theta \hat{\theta})+\omega\left[\frac{d \theta}{d t} \hat{\theta}+\theta \frac{d \hat{\theta}}{d t}\right] \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\omega^2 \hat{\theta}+\frac{d^2 \theta}{d t^2}(\theta \cdot \hat{\theta})+\omega\left[\omega \hat{\theta}+\theta \frac{d}{d t} \frac{\vec{\theta}}{|\vec{\theta}|}\right] \\
&\vec{a}=\frac{d^2 \theta}{d t^2} \hat{r}+\omega^2 \hat{\theta}+\frac{d^2 \theta}{d t^2} \times(\theta \hat{\theta})+\omega^2 \hat{\theta}+\omega \omega^2 \theta(-\hat{r}) \\
&\vec{a}=\left(\frac{d^2 \theta}{d t y^2}-\omega^2\right) \hat{r}+\left(2 \omega^2+\frac{d^2 \theta}{d t^2} \theta\right) \hat{\theta}
\end{aligned}
$$
$$
\begin{aligned}
&\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j} \\
&\hat{\theta}=-\sin \theta \hat{i}+\cos \theta \hat{j}
\end{aligned}
$$
By multiplying Eq. (i) by $\sin \theta$ and Eq. (ii) with $\cos \theta$ and then adding both of them.
$$
\begin{aligned}
\hat{r} \sin \theta+\hat{\theta} \cos \theta=& \sin \theta \cdot \cos \theta \hat{i} \\
&+\sin ^2 \theta \hat{j}+\cos ^2 \theta \hat{j} \\
&-\sin \theta \cdot \cos \theta \hat{i} \\
\hat{r} \sin \theta+\hat{\theta} \cos \theta=& \hat{j}\left(\cos ^2 \theta+\sin ^2 \theta\right) \\
&\left(\because \cos ^2 \theta+\sin ^2 \theta=1\right)
\end{aligned}
$$
So, $\hat{r} \sin \theta+\hat{\theta} \cos \theta=\hat{j}$
Now, by multiplying equation (i) by $\cos \theta$ and eq. (ii) with $\sin \theta$ and then subtracting both of them.
$$
\hat{r} \cos \theta-\hat{\theta} \sin \theta=\left(\cos ^2 \theta+\sin ^2 \theta\right) \hat{i}+0
$$
So, $(\hat{r} \cos \theta-\hat{\theta} \sin \theta)=\hat{i}$
(b) From (i), (ii) by dot product
$$
\begin{aligned}
\hat{r} \cdot \hat{\theta} &=(\cos \theta \hat{i}+\sin \theta \hat{j}) \cdot(-\sin \theta \hat{i}+\cos \theta \hat{j}) \\
&=-\cos \theta \cdot \sin \theta+\sin \theta \cdot \cos \theta=0 \\
|\hat{r}| \cdot|\hat{\theta}| \cos \theta=0 \Rightarrow \cos \theta=0
\end{aligned}
$$
So, angle between $\hat{r}$ and $\hat{\theta}=90^{\circ}=\frac{\pi}{2}$.
(c) Given that,
$$
\begin{aligned}
&\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j} \\
&\frac{d \hat{r}}{d t}=\frac{d}{d t}(\cos \theta \hat{i}+\sin \theta \hat{j})=-\sin \theta \cdot \frac{d \theta}{d t} \hat{i}+\cos \theta \cdot \frac{d \theta}{d t} \hat{j} \\
&\frac{d \hat{r}}{d t}=\omega[-\sin \theta \hat{i}+\cos \theta \hat{j}] \quad\left[\because \omega=\frac{d \theta}{d t}\right]
\end{aligned}
$$
(d) Given $\vec{r}=a \hat{\theta} \hat{r}$, here, writing dimensions
$$
[\vec{r}]=[a][\theta][\hat{r}]
$$
dimension of $[a]$ is constant $=\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^0\right]$
(e) As given that, $a=1$ unit
$$
r=\theta \hat{r}=\theta[\cos \theta \hat{i}+\sin \theta \hat{j}]
$$
Velocity,
$$
\begin{aligned}
v &=\frac{d r}{d t}=\frac{d \theta}{d t} \hat{r}+\theta \frac{d}{d t} \hat{r} \\
&=\frac{d \theta}{d t} \hat{r}+\theta \frac{d}{d t}[(\cos \theta \hat{i}+\sin \theta \hat{j})] \\
&=\frac{d \theta}{d t} \hat{r}+\theta\left[(-\sin \theta \hat{i}+\cos \theta \hat{j}) \frac{d \theta}{d t}\right] \\
v &=\frac{d \theta}{d t} \hat{r}+\theta \hat{\theta} \omega=\omega \hat{r}+\omega \theta \hat{\theta}
\end{aligned}
$$
Acceleration,
$$
\begin{aligned}
&\vec{a}=\frac{d}{d t}[\omega \hat{r}+\omega \theta \hat{\theta}]=\frac{d}{d t}\left[\frac{d \theta}{d t} \hat{r}+\frac{d \theta}{d t}(\theta \hat{\theta})\right] \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\frac{d \theta}{d t} \cdot \frac{d \hat{r}}{d t}+\frac{d^2 \theta}{d t^2}(\theta \hat{\theta})+\frac{d \theta}{d t} \frac{d}{d t}(\theta \hat{\theta}) \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\omega \frac{d}{d t}(\cos \theta \hat{i}+\sin \theta \hat{j}) \\
&+\quad+\frac{d^2 \theta}{d t^2}(\theta \cdot \hat{\theta})+\omega \frac{d}{d t}(\theta \hat{\theta}) \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\omega\left[-\sin \theta \hat{i} \frac{d \theta}{d t}+\cos \theta \hat{j} \frac{d \theta}{d t}\right] \\
&\text { From (ii) }+\frac{d^2 \theta}{d t^2}(\theta \hat{\theta})+\frac{\omega d}{d t}(\theta \hat{\theta}) \\
&\vec{a}=\frac{d^2 \theta}{d t^2} \hat{r}+\omega(\hat{\theta}) \omega+\frac{d^2 \theta}{d t^2}(\theta \hat{\theta})+\omega\left[\frac{d \theta}{d t} \hat{\theta}+\theta \frac{d \hat{\theta}}{d t}\right] \\
&=\frac{d^2 \theta}{d t^2} \hat{r}+\omega^2 \hat{\theta}+\frac{d^2 \theta}{d t^2}(\theta \cdot \hat{\theta})+\omega\left[\omega \hat{\theta}+\theta \frac{d}{d t} \frac{\vec{\theta}}{|\vec{\theta}|}\right] \\
&\vec{a}=\frac{d^2 \theta}{d t^2} \hat{r}+\omega^2 \hat{\theta}+\frac{d^2 \theta}{d t^2} \times(\theta \hat{\theta})+\omega^2 \hat{\theta}+\omega \omega^2 \theta(-\hat{r}) \\
&\vec{a}=\left(\frac{d^2 \theta}{d t y^2}-\omega^2\right) \hat{r}+\left(2 \omega^2+\frac{d^2 \theta}{d t^2} \theta\right) \hat{\theta}
\end{aligned}
$$
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