I just need help solving c. the answer is supposedly 731 N

During the delivery phase of fastball pitch, the arm internally rotates at the shoulder. The angular velocity of this internal rotation peaks at 120 rad/s. At this instant, the elbow angle is 90 degrees, so the angular velocity of the forearm is also 120 rad/s. The baseball in the pitcher's hand is 35 cm from this axis of rotation through the shoulder joint. At this instant the linear velocity of the baseball is 45 m/s.

Question

I just need help solving c. the answer is supposedly 731 N

During the delivery phase of fastball pitch, the arm internally rotates at the shoulder. The angular velocity of this internal rotation peaks at 120 rad/s. At this instant, the elbow angle is 90 degrees, so the angular velocity of the forearm is also 120 rad/s. The baseball in the pitcher's hand is 35 cm from this axis of rotation through the shoulder joint. At this instant the linear velocity of the baseball is 45 m/s.

anonymous · Answer

sorry i thought i knew this but i dont sorry

summersnow8 · Answer

a) How much of the baseball's total linear velocity is due to the 120 rad/s angular velocity of the forearm? --> 120 rad/s x .35 m = 42 m/s 

b) What is the centripetal acceleration of the baseball at this instant? $$(120 rad/s)^{2} \times (.35 m) = 5040 m/s^{2}$$

c) How large is the force exerted by the pitcher on the baseball to cause this acceleration? The baseball's mass is 145 g.

anonymous · Answer

F=m * a
F = 0.145 * 5040 = 730.8N