During the delivery phase of fastball pitch, the arm internally rotates at the s
ID: 1409639 • Letter: D
Question
During the delivery phase of fastball pitch, the arm internally rotates at the shoulder. The angular velocity of this internal rotation peaks at 120rad/s. At this instant, the elbow angle is 90 degrees, so the angular velocity of the forarm is also 120rad/s. The baseball in the pitcher's hand is 35cm from this axis of rotation through the shoulder joint. At this instant the linear velocity of the baseball is 45m/s.
a.) how much of the baseball's total linear velocity is due to the 120rad/s angular velocity of the forarm?
b.) what is the centripetal acceleration of the baseball at this instant?
c.) how large is the force exerted by the pitcher on the baseball to cause this acceleration? the baseball's mass is 145g
Explanation / Answer
a)
Vt = w*r
Vt = 120 rad/sec x 0.35
Vt = 42 m/s
x / 100 = 42 / 45
x = 4200 / 45
x = 93 % of total linear velocity is due to the 120 rad/sec angular velocity of forearm.
b)
centripetal acceleration = w^2*r = (120 rad/sec)*(0.35 m) = 5040 m/s^2
c)
The baseball mass is 145 grams
F = ma
F = 0.145 kg x 5040 m/s^2
F = 730.8 N
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