Ncert Exercises & Solutions

4.36. Motion in two dimensions, in a plane, can be studied by expressing position, velocity, and acceleration as a vector in Cartesian coordinates $A = A_{x}^i + A_{y}^j$ $A = A_{x} \hat{i} + A_{y} \hat{j}$ where $ˆ i$ $\hat{i}$ and $ˆ j$ $\hat{j}$ are unit vectors along x and y directions, respectively and $A_{x}$ $A_{x}$ and $A_{y}$ $A_{y}$ are corresponding components of A. Motion can also be studied by expressing vectors in circular polar coordinates as $A = A_{r}^r + A_{θ}^θ$ $A = A_{r} \hat{r} + A_{θ} \hat{θ}$ where $^r = \frac{r}{r} = cos θ^i + sin θ^j$ $\hat{r} = \frac{r}{r} = \cos θ \hat{i} + \sin θ \hat{j}$ $and^θ = - sin θ^i + cos θ^j$ $and \hat{θ} = - \sin θ \hat{i} + \cos θ \hat{j}$ are unit vectors along the direction in which r and θ are increasing.

a) express $^i and^j$ $\hat{i} and \hat{j}$ in terms of $^r and^θ .$ $\hat{r} and \hat{θ} .$

b) show that both $^r and^θ$ $\hat{r} and \hat{θ}$ are unit vectors and are perpendicular to each other

c) show that $\frac{d}{d t} (\hat{r}) = ω \hat{θ}, where ω = \frac{d θ}{d t} and \frac{d}{d t} (\hat{θ}) = - θ \hat{r}$

d) for a particle moving along a spiral given by $r = a θ \hat{r}$ where a = 1 find dimensions of ‘a’

e) find velocity and acceleration in polar vector representation for a particle moving along spiral described in d) above

Hint: Velocity, v =

$\frac{d \overset{⇀}{r}}{dt}$ and acceleration, a =

$\frac{d \vec{v}}{dt} .$

(a)Step 1: Express $\hat{i} and \hat{j}$ in terms of $\hat{r} and \hat{θ} .$

Given, unit vector

$\begin{array}{l} \hat{r} = \cos θ \hat{i} + \sin θ \hat{j} . . . (i) \\ \hat{θ} = - \sin θ \hat{i} + \cos θ \hat{j} . . . (ii) \end{array}$

Multiplying Eq. (i) by sin $θ$ and Eq. (ii) with cos $θ$ and adding

$\begin{array}{r} \hat{r} \sin θ + \hat{θ} \cos θ = \sin θ \cdot \cos θ \hat{i} + \sin^{2} θ \hat{j} + \cos^{2} θ \hat{j} - \sin θ . \cos θ \hat{i} \\ = \hat{j} (\cos^{2} θ + \sin^{2} θ) = \hat{j} \end{array}$
$\begin{array}{l} \Rightarrow \hat{r} \sin θ + θcos θ = \hat{j} \\ By Eq. (i) \times \cos θ - Eq. (ii) \times \sin θ \\ (\hat{r} \cos θ - \hat{θ} \sin θ) = \hat{i} \end{array}$

Step 2: Use dot product to find the angle between $\hat{r} and \hat{θ} .$

(b)

$\hat{r} . \hat{θ} = (\cos θ \hat{i} + \sin θ \hat{j}) \cdot (- \sin θ \hat{i} + \cos θ \hat{j}) = - \cos θ \cdot \sin θ + \sin θ \cdot \cos θ = 0$
$\Rightarrow θ = 90^{\circ} Angle between \hat{r} and \hat{θ} .$

Step 3: Find velocity.

(c)

$Given, \hat{r} = \cos θ \hat{i} + \sin θ \hat{j}$
$\begin{array}{l} \frac{d \hat{r}}{dt} = \frac{d}{dt} (\cos θ \hat{i} + \sin θ \hat{j}) = - \sin θ \cdot \frac{dθ}{dt} \hat{i} + \cos θ \cdot \frac{dθ}{dt} \hat{j} \\ = ω [- \sin θ \hat{i} + \cos θ \hat{j}] \end{array} [∵ θ = \frac{dθ}{dt}]$

Step 4: Find the dimension of a using homogeneity principle.

(d)

$\begin{array}{l} Given, r = aθ \hat{r}, here, writing dimensions [r] = [a] [θ] [\hat{r}] \\ \Rightarrow [L] = [a] \Rightarrow [a] = [L] = [M^{0} L^{1} T^{0}] \end{array}$

Step 5: Find velocity and acceleration on the spiral path.

(e)

Given, a= 1 unit $r = θ \hat{r} = θ [\cos θ \hat{i} + \sin θ \hat{j}]$
Velocity, $v = \frac{dr}{dt} = \frac{dθ}{dt} \hat{r} + θ \frac{d \hat{r}}{dt} = \frac{dθ}{dt} \hat{r} + θ \frac{d}{dt} [(\cos θ \hat{i} + \sin θ \hat{j})]$

$\begin{array}{l} = \frac{dθ}{dt} \hat{r} + θ [(- \sin θ \hat{i} + \cos θ \hat{j}) \frac{dθ}{dt}] \\ = \frac{dθ}{dt} \hat{r} + θ \hat{θ} ω = ω \hat{r} + ωθ \hat{θ} \end{array}$

Acceleration, $a = \frac{d}{dt} [ω \hat{r} + ωθ \hat{θ}] = \frac{d}{dt} [\frac{dθ}{dt} \hat{r} + \frac{dθ}{dt} (θ \hat{θ})]$

$\begin{array}{l} = \frac{d^{2} θ}{{dt}^{2}} \hat{r} + \frac{dθ}{dt} \cdot \frac{d \hat{r}}{dt} + \frac{d^{2} θ}{{dt}^{2}} θ \hat{θ} + \frac{dθ}{dt} \frac{d}{dt} (θ \hat{θ}) \\ = \frac{d^{2} θ}{{dt}^{2}} \hat{r} + ω [- \sin θ \hat{i} + \sin θ \hat{j}] + \frac{d^{2} θ}{{dt}^{2}} θ \hat{θ} + ω \frac{d}{dt} (θ \hat{θ}) \\ = \frac{d^{2} θ}{{dt}^{2}} \hat{r} + ω^{2} \hat{θ} + \frac{d^{2} θ}{{dt}^{2}} \times θ \hat{θ} + ω^{2} \hat{θ} + ω^{2} θ (- \hat{r}) \\ (\frac{d^{2} θ}{{dt}^{2}} - ω^{2} θ) \hat{r} + (2 ω^{2} + \frac{d^{2} θ}{{dt}^{2}}) \hat{θ} \end{array}$

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