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3 years ago

7

The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug

h which the rotor turns. (b) What is the magnitude of the angular acceleration?

2 answers:

9966 [12]3 years ago

6 0

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

Alex3 years ago

3 0

Explanation:

Using the circular equations of motion,

dw = dθ/dt

θ = (wi - wo)*t/2

Where,

θ = angular displacement

wi = final angular velocity

= 1420 rad/s

wo = initial angular velocity

= 420 rad/s

t = time

= 5s

θ = (1420 - 420)*5/2

= 2500 rad.

2*pi rad = 1 rev

= 397.89 rev

wi = wo + αt

Where,

α = angular accelaration

1420 = 420 + 5α

α = 1000/5

= 200 rad/s^2.

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