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A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 55 mph. The 0.19 kg ball is 45 cm 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?
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

Respuesta :

Answer:

Explanation:

a ) 55 mph = 55 x 1.6 x 10³ / 60 x 60 m / s

= 24.44 m / s

centripetal acceleration

= v² / r

= 24.44² / .45

= 1327.36 m / s²

b )

velocity at lowest point v

1/2 m v² = 1/2 m x 24.44² + mgh

v² = 24.44² + 2 gh

= 24.44² + 2 x 9.8 x .45

= 597.3 + 8.82

= 606.12

v = 24.62 m / s

If R be the force by hand

R - mg = mv² /r

R = mg + mv²/r

= m ( g + v²/r )

= .19 ( 9.8 + 24.62²/.45)

= 257.78 N . Ans

c ) Direction of force by hand will be towards the center of circular path or

towards shoulder joint.

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