Question

A point moves in the $x y$ - plane according to the following equation $x=a \sin \omega t, y=a(1-\cos \omega t)$, where $a, \omega$ are positive constants. Find the angle between the point's velocity and acceleration vectors.

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\pi$
• D. $2 \pi$

Answer 1

Answer: (A)

Displacement vector,
$s=x(\hat{i})+y(\hat{j})$
$=a \sin (\omega i)(\hat{i})+a\lfloor 1-\cos (\omega t) \hat{j})$
Velocity vector,
$v=\frac{d s}{d t}=\frac{d}{d t}[a \sin (\omega t)(\hat{i})+a\{1-\cos (\omega t)\}(\hat{j})]$
$=a \omega \cos (\omega t)(\hat{i})+a[0-\{-\omega \sin (\omega t)\}](\hat{j})$
$=a \omega \cos (\omega t)(\hat{i})+a \omega \sin (\omega t)(\hat{j})$
Acceleration vector,
$a=\frac{d v}{d t}=\frac{d}{d t}a \omega \cos (\omega t)(\hat{i})+a \omega \sin (\omega t)(\hat{j})]$
$=-a \omega^{2} \sin (\omega t)(\hat{i})+a \omega^{2} \cos (\omega t)(\hat{j})$
Now, calculation of angle between $v$ and a
$\cos \theta=\frac{v \cdot a}{|v||a|}$
$\left[a \omega \cos (\omega t)\left(\hat{i}\right)+a \omega \sin (\omega t)(\hat{j})\right]$
$\cos \theta=\frac{\cdot\left[-a \omega^{2} \sin (\omega t)\left(\hat{i}\right)+a \omega^{2} \cos (\omega t)(\hat{j})\right]}{\sqrt{[a \omega \cos (\omega t)]^{2}+[a \omega \sin (\omega t)]^{2}} \times \sqrt{\left[-a \omega^{2} \sin (\omega t)\right]^{2}+\left[a \omega^{2} \cos (\omega t)\right]^{2}}}$
$\cos \theta=\frac{-a^{2} \omega^{3} \cos (\omega t) \sin (\omega t)+a^{2} \omega^{3} \sin (\omega t) \cos (\omega t)}{\sqrt{a^{2} \omega^{2} \cos ^{2}(\omega t)+a^{2} \omega^{2} \sin ^{2}(\omega t)} \times \sqrt{a^{2} \omega^{4} \sin ^{2}(\omega t)+a^{2} \omega^{4} \cos ^{2}(\omega t)}}$
$\cos \theta=\frac{0}{\sqrt{a^{2} \omega^{2} \cos ^{2}(\omega t)+a^{2} \omega^{2} \sin ^{2}(\omega t)} \times \sqrt{a^{2} \omega^{4} \sin ^{2}(\omega t)+a^{2} \omega^{4} \cos ^{2}(\omega t)}}$
$\cos \theta=0 $
$\Rightarrow \theta=\frac{\pi}{2}$