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:
\[ \space v\space = \space r \space . \space w \]
\[ \space v\space = \space 1 \space m \space . \space 4 \pi \]
\[ \space v\space = \space 12.56\space \frac{m}{s} \]