Friday, April 10th, 2009 | Author:

Vertical Windmill

What is centripetal acceleration & what is the magnitude of force.?

A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 72mph. The 0.18kg ball is 50cm from the pivot point at her shoulder.
a)Just before the ball leaves her hand, what is its centripetal acceleration?
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?

We are given v=72mi/hr and we have to convert that to meters/second:
72mi/hr*1.609km/mi*1hr/3600s*1000m/km
=32.18m/s
CA=v^2/r=(32.18^2)/.5=2071.1m/s^2 a)
For b assume the speed at the lowest point of the circle is 72mph=32.18ms/s and the centripedal force is m(v^2)/r
=372.8N and add weight +9.81*.18kg
= 374.6N b)

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