2. During a tennis serve, a racket is given an angular acceleration of magnitude 155 rad/s². At the top of the serve, the racket has an angular speed of 17.0 rad/s. If the distance between the top of the racket and the shoulder is 1.40 m, find the magnitude of the total acceleration of the top of the racket. m/s²
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
![### Example Problem: Calculating Total Acceleration During a Tennis Serve
**Problem Statement:**
2. During a tennis serve, a racket is given an angular acceleration of magnitude 155 rad/s². At the top of the serve, the racket has an angular speed of 17.0 rad/s. If the distance between the top of the racket and the shoulder is 1.40 m, find the magnitude of the total acceleration of the top of the racket.
\[ \_\_\_\_\_\_\_\_\_ \text{m/s}^2 \]
**Concepts to Apply:**
1. **Angular Acceleration (\( \alpha \))**: The rate of change of angular velocity. Given as \( 155 \text{ rad/s}^2 \).
2. **Angular Speed (\( \omega \))**: The rate at which the racket rotates. Given as \( 17.0 \text{ rad/s} \).
3. **Distance (r)**: The length from the top of the racket to the shoulder. Given as \( 1.40 \text{ m} \).
4. **Total Acceleration**: The sum of both tangential and centripetal accelerations.
**Key Equations:**
- **Tangential Acceleration (\( a_t \))**:
\[ a_t = \alpha \times r \]
- **Centripetal Acceleration (\( a_c \))**:
\[ a_c = \omega^2 \times r \]
- **Total Acceleration (\( a \))**:
\[ a = \sqrt{a_t^2 + a_c^2} \]
**Solution Steps:**
1. **Calculate Tangential Acceleration (\( a_t \)):**
\[
a_t = \alpha \times r = 155 \, \text{rad/s}^2 \times 1.40 \, \text{m} = 217 \, \text{m/s}^2
\]
2. **Calculate Centripetal Acceleration (\( a_c \)):**
\[
a_c = \omega^2 \times r = (17.0 \, \text{rad/s})^2 \times 1.40 \, \text{m} = 404.6 \, \text{m/s}^2
\]
3. **Calculate Total Acceleration (\( a \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5ebd1cdc-7ac8-46e8-9a0a-3be05b72e4f6%2F371e9b86-8d09-4f46-af6e-39106e85b596%2Fvmd1hb6_processed.jpeg&w=3840&q=75)
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