Lennie Davis

2021-12-22

To throw a discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discuss 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?

Jimmy Macias

Expert

We have
The discus starts to move from the state of rest ${v}_{0}=0\frac{rad}{s}$
And we have the expression:
$\theta ={\omega }_{0}\cdot t+\frac{1}{2}+\frac{1}{2}\alpha {t}^{2}$
$\theta =0+\frac{1}{2}\alpha {t}^{2}$
Thrower performs a circular motion during movement, so $\theta =2\pi$
$\alpha =\frac{2\theta }{{t}^{2}}$
$\alpha =\frac{2\cdot 2\pi }{1{s}^{2}}$
$\alpha =4\pi \frac{rad}{{s}^{2}}$
$\alpha =4\cdot 3.14\frac{rad}{{s}^{2}}$
$\alpha =12.56\frac{rad}{{s}^{2}}$
The speed of the discus at release is:
$v=r\cdot \omega$
$v=11.3\frac{m}{s}$

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