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:

let the angular speed of the particle be 'w'

1 rev in one second = 2πrad/s

angular speed v = wR = 2πR m/s

t = T/6 = 1/6

average speed = total distance / time  = angle*R /(1/6)

in T/6 angle swept is 2π/6 = π/3 (60°)

Avg speed = (π/3)R/(1/6) = 2πR m/s(same as angular speed)

Avg velocity = displacement/ time

in T/6 time the particle will make 60° at the centre of the circle. if you draw a triangle with initial, fianl points and the centre you'll see the displacement is R.

Avg velocity = R/(1/6) = 6R m/s

Given,

Avg speed - Avg velocity = 2m/s

2πR - 6R = 2

6.282R - 6R = 2

0.282R = 2

R = 2/0.282 = 7.1m