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Q6.1-5PE

Expert-verifiedFound in: Page 221

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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 ${\mathbf{35}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{\hspace{0.33em}m/sec}}}$ and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?**

The angular velocity of the forearm is $166.66\text{\hspace{0.33em}rad/s}$.

**The distance acts as the radius for the circular track followed by the pitcher's hand before pitching.**

- The velocity of the ball in the pitcher’s hand is $35.0\text{\hspace{0.33em}m/sec}$.
- The distance between the ball and the elbow joint is 0.300 m.

The relationship between the angular velocity and linear velocity is given as follows:

Before pitching, the ball is in the hand of the baseball pitcher.

Therefore, the angular velocity of the forearm is equal to the angular velocity of the ball.

The angular velocity of the forearm can be calculated as:

$\omega =\frac{v}{r}$

Putting the values in the above equation, we get:

$\begin{array}{c}\omega =\frac{35}{0.300}\\ =166.66\text{\hspace{0.33em}rad/s}\end{array}$

Therefore, the angular velocity is $166.66\text{\hspace{0.33em}rad/s}$.

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