a particle is moving in a circle of radius R with constant speed.The t

1.	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» 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 = angleR /(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

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»

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

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