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 75 mph. The 0.20 kg ball is 51 cm from the pivot point at her shoulder.

a)

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

Express your answer using two significant figures.

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 using two significant figures.

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?

upward, downward, left or right?

Explanation / Answer

Convert mph to m/s and cm to m

(75 mi/h)(1 hr/3600 s)(1609 m/1 mi)=33.52 m/s

51 cm= 0.51 m

a) Centripetal acceleration equals the tangent velocity squared divided by the radius. (Ac=v²/r)

Ac=v²/r
Ac=(33.52 m/s)²/.51 m
Ac=2203.23 m/s².

b) force = mg + mv²/r where m is the mass of the ball

F = 0.2x9.81 + 0.2x2203.23

F = 442.61 N

c. There is no horizontal acceleration at this point, so the force is upwards

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