Question

Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 23.7 m/s (about 53.0 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 982 m/s 2 . What is the length of her arm from the pivot point at her shoulder

Answer 1

length of her arm from the pivot point at her shoulder = 0.572m

Step-by-step explanation:

Formula for radial acceleration = V²/r

From the question, radial acceleration = 982 m/s²

While speed(V) = 23.7m/s

And r is the radius of motion and in this case her arm is doing the motion, and thus r it's the length of her arm.

Thus; 982 = 23.7²/r

Making r the subject;

r = 23.7²/982 = 561.69/982 = 0.572m

Answer 2

0.572 m

Step-by-step explanation:

Assume that the ball is subjected to radial acceleration only and there's no tangential acceleration. Then the speed 23.7 m/s of the ball leaving her hand is the result of the centripetal acceleration a = 982 m/s2. We can use the following equation of motion to calculate the radius of motion, or her arm length:

`a = (v^2)/(r)`

`r = (v^2)/(a) = (23.7^2)/(982) = 0.572 m`