Lennie Davis

Answered

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?

Answer & Explanation

Jimmy Macias

Expert

2021-12-23Added 30 answers

We have $2r=1.8\text{}m$, $t=1\text{}s$

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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