Q. A particle moves through angular displacement $ \theta $ on a circular path of radius $r$. The linear displacement will be
Bihar CECEBihar CECE 2014
Solution:
The linear displacement, $\vec{\triangle} r =\overrightarrow{ r _{2}}-\overrightarrow{ r _{1}} ;$ where $r _{2}= r _{1}= r$
Hence, $|\Delta r |=\sqrt{ r _{2}^{2}+ r _{1}^{2}-2 r _{2} r _{1} \cos \theta}$
$
\Longrightarrow|\Delta r |=\sqrt{2 r ^{2}-2 r ^{2} \cos \theta}
$
$
\Longrightarrow|\Delta r |=\sqrt{2} r \sqrt{1-\cos \theta}
$
$
\Longrightarrow|\Delta r |=2 r \sin \left(\frac{\theta}{2}\right)
$
