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A discus thrower accelerates a discus from rest to a speed of 25.4 m/s by whirling it through 1.21 rev. Assume the discus moves on the arc of a circle 0.95 m in radius. A discus thrower moving in a circle as he prepares to throw the discus. (a) Calculate the final angular speed of the discus. rad/s (b) Determine the magnitude of the angular acceleration of the discus, assuming it to be constant. rad/s2 (c) Calculate the time interval required for the discus to accelerate from rest to 25.4 m/s. s

Respuesta :

Answer:

Explanation:

Given:

vo = 0

v = 25.4 m/s

r = 0.95 m

n = 1.21 rev

A.

r*ω = v

ω = v/r

= 25.4/.95

= 26.74 rad/s

B.

θ = 2πn

= 2π × (1.21)

= 7.6 m

Using equation of motion,

ωf^2 = ωi^2 + 2αθ

(26.74)^2 = 2 × α × (7.6)

α = 715.03/15.2

= 47.04 rad/s^2

C.

Using equation of motion,

θ = ωi*t + 1/2*α*t^2

θ = 0 + 1/2*α*t^2

7.6 = 47.04/2 × t^2

= sqrt(0.323)

= 0.57 s

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