A particle is moving in a circle of radius R with constant speed . The time period of particle is T=1 Second in a time t = T/6 , If the difference between average speed and magnitude of average velocity of the partical is 2m/sec, find the radius of the circle(in meteres)

Dear Student,

Please find below the solution to the asked query:

Given that the time period of the particle is T. Then, the angular velocity is,

$\omega =\frac{2\pi }{T} \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
So, in time t = T/6, the particle will be having the angular displacement,

$\theta =\omega t=\frac{2\pi }{T}\times \frac{T}{6}=\frac{\pi }{3}$
Therefore, the distance travelled by the body is,

$dis\tan ce =\frac{{\displaystyle \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2\pi }\times 2\pi R\Rightarrow dis\tan ce=\frac{\pi R}{3}$

Displacement of the body is,

$dispacement=R$

So, the magnitude of velocity is,

$\left|V\right|=\frac{Displacement}{time}=\frac{R}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}=\frac{6R}{T}$

The magnitude of speed is,

$\left|S\right|=\frac{Dis\tan ce}{time}=\frac{{\displaystyle \raisebox{1ex}{$\pi R$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\displaystyle \raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}=\frac{2\pi R}{T}$

Therefore, the difference of the magnitude of the speed and the magnitude of the velocity is,

$$\begin{gathered} Difference = \frac{2\pi R}{T}-\frac{6R}{T}\Rightarrow \frac{\left(2\pi -6\right)R}{T}=2 \\ \Rightarrow R=\frac{2}{2\pi -6}\Rightarrow R=\frac{2}{0.28}=\frac{50}{7} \\ \Rightarrow R=7.14 m \end{gathered}$$


 

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