Question

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 . Part A If the thrower takes 1.2 s to complete one revolution, starting from rest, what will be the speed of the discus at release

Answer 1

Answer:

9.42 m/s

Explanation:

a) Using Newton's law of motion formula:

$\theta=\frac{(\omega+\omega_o)}{2}t\\\\where \ \theta=angular\ displacement=1\ rev =2\pi, w_o=initial\ velocity\ of\ discus\\=0\ rad/s, \omega=angular\ speed\ of\ discus\ at\ release,t=time\ = 1.2\ s.\\\\Hence:\\\\2\pi=\frac{(0+\omega)}{2}(1.2)\\\\\omega=\frac{2*2\pi}{1.2} \\\\\omega=10.47\ rad/s$

The speed of the discus at release (v) is:

v = ωr; where r = radius of discus

diameter = 1.8 m, r = diameter / 2= 1.6 / 2 = 0.9 m

v = ωr = 10.47 * 0.9

v = 9.42 m/s