Question

A fast pitch softball player does a "windmill" pitch moving her hand through a vertical circular arc to pitch a ball at 72mphmph. The 0.19 kg ball is 53 cm from the pivot point at her shoulder.

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

2. 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?

3. 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?

Answer #1

Part A.

Centripetal acceleration is given by:

ac = V^2/R

V = Speed of ball = 72 mph = (72 mi/hr)*(1609.34 m/1mi)*(1 hr/3600 sec) = 72*1609.34/3600 = 32.1868 m/s

R = radius of circular arc = 53 cm = 0.53 m

So,

**ac = (32.1868 m/s)^2/(0.53 m) = 1954.7 m/s^2**

Part B.

Net force at the lowest point of circle will be:

F_net = Fc - W

W = Weight of ball = m*g

Fc = m*ac

m = mass of ball = 0.19 kg

F_net = 0.19*1954.7 - 0.19*9.81

**F_net = 369.5
N**

Part C.

Since at the lowest point of circle, force will be towards the
circle, So at that moment force will be **towards the shoulder of
player.**

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