Question

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 particle is 2 m/sec,find the radius of circle

Answer 1

Answer:

R = 7.06 m

Explanation:

As we know that time period of the particle is T = 1 s

so the angle covered by it in t = T/6 seconds is given as

$\theta = \frac{2\pi}{T}t$

$\theta = \frac{2\pi}{1}\times \frac{1}{6}$

$\theta = \frac{\pi}{3}$

now we know that total distance moved by the particle will be

$d = R\theta = R\times \frac{\pi}{3}$

now average speed is given as

$v = \frac{R\times \frac{\pi}{3}}{\frac{1}{6}}$

$v = 2\pi R$

now for displacement of the particle we know that

$\vec d = \sqrt{R^2 + R^2 - 2(R)(R)cos\frac{\pi}{3}}$

$\vec d = R$

so average velocity is given as

$\vec v = \frac{R}{\frac{1}{6}} = 6R$

now we have

$v - \vec v = 2\pi R - 6 R = 2 m/s$

$R = \frac{2}{2\pi - 6} = 7.06 m$

Answer 2

at t=T/6, total distance covered= 2*pi*r/6  and  displacement= r

 difference of them = 2*t
so, 2*pi*r/6- r =2(1/6)=1/3

r=1/(pi-3)=7.062