Key to "To throw a discus, the thrower holds it..."

High School
Physics
Force, Motion and Energy
Acceleration Due to Gravity

Lennie Davis

Answered question

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

Beginner2021-12-23Added 30 answers

We have $2 r = 1.8 m$ , $t = 1 s$
The discus starts to move from the state of rest $v_{0} = 0 \frac{r a d}{s}$
And we have the expression:
$θ = ω_{0} \cdot t + \frac{1}{2} + \frac{1}{2} α t^{2}$
$θ = 0 + \frac{1}{2} α t^{2}$
Thrower performs a circular motion during movement, so $θ = 2 π$
$α = \frac{2 θ}{t^{2}}$
$α = \frac{2 \cdot 2 π}{1 s^{2}}$
$α = 4 π \frac{r a d}{s^{2}}$
$α = 4 \cdot 3.14 \frac{r a d}{s^{2}}$
$α = 12.56 \frac{r a d}{s^{2}}$
The speed of the discus at release is:
$v = r \cdot ω$
$v = 11.3 \frac{m}{s}$

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