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 72 mph. The 0.20 kg ball is 50 cm from the pivot point at her shoulder.

Part A

Just before the ball leaves her hand, what is its centripetal acceleration?

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?

Part C

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

Here,

velocity , v = 72 mph

v = 32.18 m/s

mass , m = 0.20 kg

radius , r = 0.50 m

part A)

for the centripetal acceleration

centripetal acceleration = m * v^2/r

centripetal acceleration = 0.20 * 32.18^2/0.50

centripetal acceleration = 414.2 m/s^2

part B)

let the magnitude of force of hand is Fnet

Fnet = m * g + m * a

Fnet = 0.2 * (9.8 + 414.2)

Fnet = 811.9 N

the lowest point of the circle has reached its maximum speed is 811.9 N

part C)

as the force is centripetal

the direction of force is in upwards dircetion

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