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 1.9 m?

Answer 1

Complete 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 1.9 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:

11.94 m/s will be the speed of the discus at release.

Explanation:

Given data:

Time taken to complete one revolution = 1.0 s

Diameter of the circle, D = 1. 9 m

The radius of the circle is the half of the diameter. So,

        $r=\frac{D}{2}=\frac{1.9}{2}=0.95 \mathrm{m}$

One revolution is equal to 360 degree or 2 pi or 6.28 radians. Average speed is the ratio of total distance to the time. It can be expressed as

        $\text {average speed, } v_{a v}=\frac{\text {total distance}}{\text {total time}}=\frac{2 \times \pi \times r}{1}$

$2 \times \pi \times r$ – The distance reached by the thrower to make one revolution

         $v_{a v}=2 \times 3.14 \times 0.95=5.966 \mathrm{m} / \mathrm{s}$

Now, we need to the final velocity (speed of the discus at release). This can be done as below

        $v_{a v}=\frac{1}{2} \times\left(v_{i}+v_{f}\right)$

By taking the average of combining both initial and final velocity, we get average velocity. Here, initial velocity is zero.

         $5.966=\frac{1}{2} \times\left(0+v_{f}\right)$

         $v_{f}=5.966 \times 2=11.932 \mathrm{m} / \mathrm{s}$