A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher’s hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?

Solution:

We are given the linear velocity of the ball in the pitcher’s hand, v=35.0\ \text{m/s}, and the radius of the curvature, r=0.300 \ \text{m}. Linear velocity v and angular velocity \omega are related by

v=r\omega \ \text{or} \ \omega=\frac{v}{r}

If we substitute the given values into our formula, we can solve for the angular velocity directly. That is,

\begin{align*}
\omega & = \frac{v}{r} \\
\\
\omega & = \frac{35.0 \ \text{m/s}}{0.300 \ \text{m}} \\
\\
\omega & = 116.6667 \ \text{rad/s} \\ 
\\
\omega & = 117 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The angular velocity of the forearm is about 117 radians per second.