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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 1.7 m.

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Khoso123

Incomplete question.The complete question is here

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 1.7 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?

Answer:

[tex]v=10.676m/s[/tex]

Explanation:

Radius r=1.7m/2 =0.85 m

The angular displacement in time t is:

θ=1/2αt²

where α is angular acceleration

Given

θ=2π

t=1.0s

So

α=2θ/t²

[tex]\alpha =\frac{2*2\pi }{(1.0s)^{2} } \\\alpha =4\pi rad/s^{2}[/tex]

Angular speed after 1.0s is:

[tex]w=\alpha t=4\pi rad/s^{2} *1.0s\\w=4\pi rad/s[/tex]

Speed of discus is given by:

[tex]v=rw\\v=(0.85)(4\pi )\\v=10.676m/s[/tex]

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