Question

During a tennis serve, a racket is given an

angularacceleration of magnitude 160 rad/s2. At thetop
of the serve, the racket has an angular speed of 14 rad/s. Ifthe
distance between the top of the racket and the shoulder is1.5m,
find the magnitude of the total acceleration of the top ofthe
racket.

Answer 1

379.521 m/s²

Step-by-step explanation:

r = Radius = 1.5 m

`\alpha` = Angular acceleration = 160 rad/s²

`\omega` = Angular speed = 14 rad/s

Tangential acceleration is given by

`a_t=r\alpha\\\Rightarrow a_t=1.5* 160\\\Rightarrow a_t=240\ m/s^2`

Centripetal accelration is given by

`a_c=r\omega^2\\\Rightarrow a_c=1.5* 14^2\\\Rightarrow a_c=294\ m/s^2`

The resultant acceleration is given by

`a=√(a_t^2+a_r^2)\\\Rightarrow a=√(240^2+294^2)\\\Rightarrow a=379.521\ m/s^2`

The magnitude of the total acceleration of the top of the racket is 379.521 m/s²