To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.8 m.If the thrower takes 1.0 s to complete one revolution, starting from rest, what will be the speed of the discus at release?

Respuesta :

Answer:

v = 11.3 m / s

Explanation:

This rotation exercise can be treated using the rotation kinematics. Remember that all these equations work in radians

Let's look for angular acceleration

    θ = w₀ t + ½ α t²

They tell us that it takes t = 1.0 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)

    θ = 0 + ½ α t²

    α = 2θ / t²

   α= 2 2π / 1²

    α = 4π = 12.57 rad / s²

Let's calculate the angular velocity at this point

    w = wo + α t

    w = 0 + α t

    w = 12.57 1

    w = 12.57 rad / s

The relationship between linear and angular velocity is

    r = d / 2

    r = 1.8 / 2 = 0.90 m

   v = w r

   v = 12.57 .90

   v = 11.3 m / s