question_answer

A particle moves around a circular path of radius with a constant speed u. If particle describes angle \[\theta \]in 4 s, then average acceleration of the particle is

A) \[\frac{{{v}^{2}}}{r}\cdot \sin \left( \frac{2v}{R} \right)\]

B) \[\frac{2{{v}^{2}}}{R}\cdot \sin \left( \frac{\theta }{2} \right)\]

C) \[\frac{2{{v}^{2}}}{R\theta }\cdot \cos \left( \frac{\theta }{2} \right)\]     

D) \[\frac{{{v}^{2}}}{2}\cdot \sin \left( \frac{2v}{R} \right)\]

Correct Answer: D

Solution :

As particle moves by an angle\[\theta .\]

Then, change in velocity is \[\Delta v={{v}_{2}}-{{v}_{1}}.\]

Magnitude of change in velocity is

\[|\Delta v|=\sqrt{{{({{v}_{2}}-{{v}_{1}})}^{2}}}\]

\[\Rightarrow \]   \[|\Delta v|=\sqrt{v_{1}^{2}+v_{2}^{2}-2{{v}_{1}}{{v}_{2}}\cos \theta }\]

Here, \[{{v}_{1}}={{v}_{2}}=v\]

\[\Rightarrow \]   \[|\Delta v|=\sqrt{{{v}^{2}}+{{v}^{2}}-2{{v}^{2}}\cos \theta }\]

\[=\sqrt{2{{v}^{2}}(1-\cos \theta })\]

\[=\sqrt{2{{v}^{2}}\left[ 1-\left( 1-2{{\sin }^{2}}\frac{\theta }{2} \right) \right]}\]

\[=\sqrt{2{{v}^{2}}\left( 1-1+2{{\sin }^{2}}\frac{\theta }{2} \right)}\]

\[=\sqrt{2{{v}^{2}}\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)}=\sqrt{4{{v}^{2}}{{\sin }^{2}}\frac{\theta }{2}}\]

\[\therefore \]\[|\Delta v|=2v\sin \frac{\theta }{2}\]

Time duration of motion is given 4s

So, \[\Delta t=4=\frac{R\theta }{v}\]\[\Rightarrow \]\[\theta =\frac{4v}{R}\]

So, average acceleration of the particle is

\[|{{a}_{avg}}|=\frac{|\Delta v|}{\Delta t}=\frac{2v\cdot \sin \frac{\theta }{2}}{4}=\frac{v}{2}\sin \left( \frac{2v}{R} \right)\]