College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Problem 3: Analyzing the Total Acceleration in a Tennis Serve**
In a tennis serve, a racket is subjected to an angular acceleration of 180 rad/s². At the top of the serve, the racket exhibits an angular speed of 13.0 rad/s. Given that the distance from the shoulder to the top of the racket is 1.10 m, we are tasked with calculating the magnitude of the total acceleration at the top of the racket.
**Solution:**
To find the total acceleration of the top of the racket, we need to consider both the tangential and centripetal components of acceleration. The tangential acceleration \(a_t\) is given by the product of the angular acceleration \(\alpha\) and the radius \(r\), while the centripetal acceleration \(a_c\) is determined by the square of the angular speed \(\omega\) times the radius:
1. **Tangential Acceleration (\(a_t\)):**
\[
a_t = \alpha \times r
\]
2. **Centripetal Acceleration (\(a_c\)):**
\[
a_c = \omega^2 \times r
\]
3. **Total Acceleration (\(a\)):**
\[
a = \sqrt{a_t^2 + a_c^2}
\]
**Given:**
- Angular acceleration \(\alpha = 180 \text{ rad/s}^2\)
- Angular speed \(\omega = 13.0 \text{ rad/s}\)
- Radius/distance \(r = 1.10 \text{ m}\)
By applying these formulas, we can solve for the total acceleration, ensuring we take into account both components to accurately describe the motion experienced by the racket during the serve. The final answer will be in meters per second squared (\( \text{m/s}^2 \)).](https://content.bartleby.com/qna-images/question/88cd9656-9419-481b-bc0b-e9aadd0312cb/2d150f0c-bca8-41e4-9521-72a9c34bffce/g3uu1pr_processed.png)
Transcribed Image Text:**Problem 3: Analyzing the Total Acceleration in a Tennis Serve**
In a tennis serve, a racket is subjected to an angular acceleration of 180 rad/s². At the top of the serve, the racket exhibits an angular speed of 13.0 rad/s. Given that the distance from the shoulder to the top of the racket is 1.10 m, we are tasked with calculating the magnitude of the total acceleration at the top of the racket.
**Solution:**
To find the total acceleration of the top of the racket, we need to consider both the tangential and centripetal components of acceleration. The tangential acceleration \(a_t\) is given by the product of the angular acceleration \(\alpha\) and the radius \(r\), while the centripetal acceleration \(a_c\) is determined by the square of the angular speed \(\omega\) times the radius:
1. **Tangential Acceleration (\(a_t\)):**
\[
a_t = \alpha \times r
\]
2. **Centripetal Acceleration (\(a_c\)):**
\[
a_c = \omega^2 \times r
\]
3. **Total Acceleration (\(a\)):**
\[
a = \sqrt{a_t^2 + a_c^2}
\]
**Given:**
- Angular acceleration \(\alpha = 180 \text{ rad/s}^2\)
- Angular speed \(\omega = 13.0 \text{ rad/s}\)
- Radius/distance \(r = 1.10 \text{ m}\)
By applying these formulas, we can solve for the total acceleration, ensuring we take into account both components to accurately describe the motion experienced by the racket during the serve. The final answer will be in meters per second squared (\( \text{m/s}^2 \)).
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