Question

During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket has an angular speed of 12.0 rad/s. If the distance between the top of the racket and the shoulder is 1.30 m, find the magnitude of the total acceleration of the top of the racket.

Answer 1

The magnitude of the total acceleration of the top of the racket during the serve, considering both tangential and radial components, is approximately 270.5 m/s^2.

Step-by-step explanation:

To find the magnitude of the total acceleration of the top of the racket, we need to consider both tangential and radial (centripetal) acceleration components. The tangential acceleration (at) is directly provided by the angular acceleration (α), and is calculated by multiplying the angular acceleration by the radius (r), so at = α × r. Here, α = 150 rad/s2 and r = 1.30 m, giving at = 195 m/s2. The radial acceleration (ar), also known as centripetal acceleration, depends on the angular speed (ω) and the radius (r), and is calculated with the formula ar = ω2 × r. Given ω = 12.0 rad/s and r = 1.30 m, we find ar = 187.2 m/s2. Finally, the total acceleration (a) is the square root of the sum of the squares of at and ar, resulting in a = √(at2 + ar2). This gives us a = √(1952 + 187.22) m/s2, and by calculating, we find the magnitude of the total acceleration to be approximately 270.5 m/s2.

Answer 2

270 m/s²

Step-by-step explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²