Question

a fast - pitch softball player does a \windmill\ pitch moving her hand through a vertical circular arc to pitch a ball at 71 mph. the 0.20 kg ball is 55 cm from the pivot point at her shoulder. for the steps and strategies involved in solving a similar problem, you may view a video tutor solution. part a just before the ball leaves her hand, what is its centripetal acceleration? express your answer in meters per second squared. 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 in newtons.

Explanation:

Step1: Convert speed to m/s

The speed of the ball is $v = 71\ mph$. We know that $1\ mph=\frac{1609.34}{3600}\ m/s$. So $v = 71\times\frac{1609.34}{3600}\ m/s\approx 31.7\ m/s$. The radius of the circular - arc $r = 55\ cm=0.55\ m$.

Step2: Calculate centripetal acceleration (Part A)

The formula for centripetal acceleration is $a_c=\frac{v^{2}}{r}$. Substitute $v = 31.7\ m/s$ and $r = 0.55\ m$ into the formula: $a_c=\frac{(31.7)^{2}}{0.55}=\frac{1004.89}{0.55}\approx1827.07\ m/s^{2}$.

Step3: Analyze forces at the lowest point (Part B)

At the lowest point of the vertical circle, the net - force towards the center provides the centripetal force. The forces acting on the ball are the tension $T$ in the hand and the weight $mg$ of the ball. According to Newton's second law, $F_{net}=T - mg=ma_c$. So $T=m(a_c + g)$. Given $m = 0.20\ kg$, $a_c\approx1827.07\ m/s^{2}$, and $g = 9.8\ m/s^{2}$. Then $T=0.20\times(1827.07 + 9.8)=0.20\times1836.87 = 367.374\ N$.

Answer:

Part A: $a_c\approx1827.07\ m/s^{2}$
Part B: $T\approx367.4\ N$