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 68mph . The 0.20kg ball is 51cmfrom the pivot point at her shoulder.

Part A: Just before the ball leaves her hand, what is its centripetal acceleration? Express your answer using two significant figures.

Part B: 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? Express your answer using two significant figures.

Explanation / Answer

Here ,

v = 68 mph

v = 30.4 m/s

A)

centripetal acceleration = v^2/r

centripetal acceleration = 30.4^2/.51

centripetal acceleration = 1811.2 m/s^2

centripetal acceleration of the ball is 1811.2 m/s^2

B)

at the lowest point

A = ac + g

FOrce = m*(g + ac)

Force = .20 * (9.8 + 1811.2)

Force = 364.2 N

the force on the ball by her hand is 364.2 N

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