A discus thrower accelerates a discus from rest to a speed of 25.7 m/s by whirli

Question

A discus thrower accelerates a discus from rest to a speed of 25.7 m/s by whirling it through 1.26 rev. Assume the discus moves on the arc of a circle 0.96 m in radius.

A discus thrower accelerates a discus from rest to a speed of 25.7 m/s by whirling it through 1.26 rev. Assume the discus moves on the arc of a circle 0.96 m in radius. Determine the magnitude of the angular acceleration of the discus, assuming it to be constant. Calculate the time interval required for the discus to accelerate from rest to 25.7 m/s.

Explanation / Answer

initial speed is u = 0

final speed is v = 25.7 m/s

the angle rotated is theta = 1.26 rev = 1.26 x (2pi) radians

radius of circle r = 0.96 m

let w be the final angular velocity

we know that

w^2 - w_o^2 = 2alpha x theta

or alpha = (w^2 - w_o^2/2theta) ------------(1)

where alpha is angular acceleration,w_o = 0 and w = (v/r)

let t be the time taken to accelerate from rest

we know that

w = w_o + alpha x t

or t = (w - w_o/alpha)

where w_o = 0 and alpha is obtained from equation (1)

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