Afast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 65 mph. the 0.19 kg ball is 50 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?

A)  The centripetal acceleration just before the ball leaves her hand is;

α = 1693.6 m/s²

B) The magnitude of the force on her right hand that is exerted on the ball gives; F = 321.8 N

C) The direction of the force her hand exerts on the ball at the point;

Upward direction

We are given;

Mass of ball; m = 0.19 kg

Distance of ball from the pivot at her shoulder; r = 50 cm = 0.5 m

Speed; v = 65 m/h = 29.1 m/s

A) Formula for centripetal acceleration is;

α = v²/r

Thus;

α = 29.1²/0.5

α = 1693.6 m/s²

B) At the lowest point of the circle where the ball has reached maximum speed, magnitude of force which is the centripetal force is gotten from;

F = mv²/r

F = 0.19 * 1693.6

F = 321.8 N

C) Since the ball is now at the lowest point, it means that balls motion will be towards the center and as such the force will be acting in an upward direction.

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force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2