The main objective of this question is to find the** speed** of the **disc** when it is **released**.

This question uses the concept of **circular motion**. In a circular motion, the motion **direction** is **tangential** and **constantly changing**, but the velocity is** constant**.

The force necessary to vary the **velocity** is always **perpendicular** to the motion and **directed** toward the **circle center**.

## Expert Answer

We are **given**:

\[ \space 2r \space = \space 1.8 \space m \]

\[ \space t \space = \space 1 \space s \]

The** disc** starts to **move** from **rest** **position**, so:

\[ \space v_o \space = \space 0 \space \frac{rad}{s} \]

By **applying kinematics**, we result in:

\[ \space \theta \space = \space w_o \space . \space t \space + \space \frac{1}{2} \space + \space +\frac{1}{2} \alpha t^2 \]

\[ \space \theta \space = \space 0 \space + \space \frac{1}{2} \alpha t^2 \]

We** know** that:

\[ \space \theta \space = \space 2 \pi \]

\[ \space \alpha \space = \space \frac{2 \theta}{t^2} \]

\[ \space \alpha \space = \space \frac{2 \space . \space 2 \pi}{1s^2} \]

\[ \space \alpha \space = \space 4 \pi \frac{rad}{s^2} \]

\[ \space \alpha \space = \space 4 \space \times \space 3.14 \frac{rad}{s^2} \]

\[ \space \alpha \space = \space 12.56 \frac{rad}{s^2} \]

The **speed** is given as:

\[ \space v\space = \space r \space . \space w \]

\[ \space v\space = \space 0.9 \space m \space . \space 4 \pi \]

\[ \space v\space = \space 11.3 \space \frac{m}{s} \]

## Numerical Answer

The **speed** of the **disc** when it is **released** is:

\[ \space v\space = \space 11.3 \space \frac{m}{s} \]

## Example

The **thrower holds** the discus with an **arm fully** extended while releasing it.

He starts to **turn at rest** with a **steady angular acceleration** and releases the handle after **one full rotation**, if discus moves in a **circle** that is **approximately** $ 2 $ meters in **diameter** and it takes the thrower $ 1 $ second to **make** one turn from **rest**, what is the** speed** of discus when it is **thrown**?

We are **given** that:

\[\space 2r \space = \space 2 \space m \]

\[ \space t \space = \space 1 \space s \]

The** disc** starts to **move** from **rest position**, so:

\[ \space v_o \space = \space 0 \space \frac{rad}{s} \]

By **applying kinematics**, we results in:

\[ \space \theta \space = \space w_o \space . \space t \space + \space \frac{1}{2} \space + \space +\frac{1}{2} \alpha t^2 \]

\[ \space \theta \space = \space 0 \space + \space \frac{1}{2} \alpha t^2 \]

We **know** that:

\[ \space \theta \space = \space 2 \pi \]

\[ \space \alpha \space = \space \frac{2 \theta}{t^2} \]

\[ \space \alpha \space = \space \frac{2 \space . \space 2 \pi}{1s^2} \]

\[ \space \alpha \space = \space 4 \pi \frac{rad}{s^2} \]

\[ \space \alpha \space = \space 4 \space \times \space 3.14 \frac{rad}{s^2} \]

\[ \space \alpha \space = \space 12.56 \frac{rad}{s^2} \]

The **speed** is given as:

\[ \space v\space = \space r \space . \space w \]

\[ \space v\space = \space 1 \space m \space . \space 4 \pi \]

\[ \space v\space = \space 12.56\space \frac{m}{s} \]