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 69mph. The 0.17kg ball is 55cm 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

a = v^2 / r = 69^2 / 0.55 = 8656 m/s/s That speed can't be in m/s can it? That kind of acceleration would take your hand off. F = ma = 0.17 . 8656 = 1471.5 N You might want to add 0.17 . 9.8 = 1.666 N to balance the gravity force on the ball but I hardly think that matters.

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