A fast pitch softball player does a \"windmill\" pitch, moving her hand through

Question

A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 66 mph. The 0.19 kg ball is 54 cm from the pivot point at her shoulder. Just before the ball leaves her hand, what is its centripetal acceleration?At the lowest point of the circle the ball has reached its maximum speed. What is the magnitude of the force her hand exerts on the ball at this point? At the lowest point of the circle the ball has reached its maximum speed. What is the direction of the force her hand exerts on the ball at this point?

Explanation / Answer

Centripetal force = mv^2/R

Centripetal acc. = v^2/R

v = 66mph = 29.50 mt/sec

R = 54cm = 0.54 mt

Centripetal acc. = (29.5)^2/0.54 = 1611.57 m/sec^2

Force = mass*Centripetal acc. = 0.19*1611.57 = 306.2 Newton

At the lowest point gravity is acting downwards so the force of her hand will be directed upward. The force necessary to maintain an object in constant circular motion is directed perpendicular to the circumference of the circle.

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