Question

A point moves in the $xy$ - 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,
$v=\frac{ds}{dt}=\frac{d}{dt}[a\sin (\omega t)(\hat{i})+a\{1-\cos (\omega t)\}(\hat{j})]$
$=a\omega \cos (\omega t)(\hat{i})+a[0-\{-\omega \sin (\omega t)\}](\hat{j})$
$=a\omega \cos (\omega t)(\hat{i})+a\omega \sin (\omega t)(\hat{j})$
Acceleration vector,
$a=\frac{dv}{dt}=\frac{d}{dt}a\omega \cos (\omega t)(\hat{i})+a\omega \sin (\omega t)(\hat{j})]$
$=-a{\omega }^2\sin (\omega t)(\hat{i})+a{\omega }^2\cos (\omega t)(\hat{j})$
Now, calculation of angle between $v$ and a
$\cos \theta =\frac{v\cdot a}{|v||a|}$
$\left[a\omega \cos (\omega t)\left(\hat{i}\right)+a\omega \sin (\omega t)(\hat{j})\right]$
$\cos \theta =\frac{\cdot \left[-a{\omega }^2\sin (\omega t)\left(\hat{i}\right)+a{\omega }^2\cos (\omega t)(\hat{j})\right]}{\sqrt{[a\omega \cos (\omega t){]}^2+[a\omega \sin (\omega t){]}^2}\times \sqrt{{\left[-a{\omega }^2\sin (\omega t)\right]}^2+{\left[a{\omega }^2\cos (\omega t)\right]}^2}}$
$\cos \theta =\frac{-a^2{\omega }^3\cos (\omega t)\sin (\omega t)+a^2{\omega }^3\sin (\omega t)\cos (\omega t)}{\sqrt{a^2{\omega }^2{\cos }^2(\omega t)+a^2{\omega }^2{\sin }^2(\omega t)}\times \sqrt{a^2{\omega }^4{\sin }^2(\omega t)+a^2{\omega }^4{\cos }^2(\omega t)}}$
$\cos \theta =\frac{0}{\sqrt{a^2{\omega }^2{\cos }^2(\omega t)+a^2{\omega }^2{\sin }^2(\omega t)}\times \sqrt{a^2{\omega }^4{\sin }^2(\omega t)+a^2{\omega }^4{\cos }^2(\omega t)}}$
$\cos \theta =0$
$\Rightarrow \theta =\frac{\pi }{2}$