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"In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a fast pitch the ball leaves the hand with a speed of 139 km/hr. (Assume that the pitching motion originates with the arm straight back and terminates with the arm straight down. Do not ignore gravity.) (a) Find the rotational kinetic energy (in J) of the pitcher's arm given its moment of inertia is 0.720 kg · m2 and the ball leaves the hand at a distance of 0.604 m from the pivot in the shoulder. (Enter the rotational kinetic energy of the pitcher's arm just after the ball is released.)"(b) What force (in N) did the muscles exert to cause the arm to rotate if their effective perpendicular lever arm is 4.00 cm and the ball is 0.187 kg? (Enter the magnitude.) XN

Respuesta :

Answer:

Explanation:

139 km /h = 38.61 m / s.

the rotational kinetic energy (in J) of the pitcher's arm

= (1/2) I ω²

Where I is moment of inertia of the arm and ω is its angular velocity

Given , I = 0.72 kg m²

ω = vR , v is linear velocity and R is length of arm ( given .604 m )

ω = 38.61  x .604 = 23.32 rad / s

(1/2) I ω²

= .5 x 0.72 x (23.32)²

= 195.77 J

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