Asked by Anonymous

At the top of the serve, the racket has an angular speed of 23.7 rad/s. If the distance between the top of the racket and the shoulder is 1.39 m, find the magnitude of the total acceleration of the top of the racket.
Answered by bobpursley
assuming the racket velocity is constant, and not changing, then the centripetal acceleration downward is 23.7^2 * 1.39 m/sec^2
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