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.

270 m/s²

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