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A baseball pitcher pivots his extended arm about his shoulder joint, applying a constant torque of 170 N · m for 0.15 s to his arm, which has a moment of inertia of 0.53 kg · m2.
(a) Find his arm's angular acceleration and final angular velocity, assuming it starts from rest.
angular acceleration
final angular velocity

(b) Find the final speed of the ball, relative to his shoulder, which is 0.76 m from the ball.

Answer 1

Newton’s 2 Law for rotation:
M=I• ε,
angular acceleration is
ε = M/I=170/0.53=320.8 rad/s^2
Final angular velocity (if initial velocity is zero) is
ω = εt = 320.8•0.15= 48.1 rad/s.
v= ωR = 48.1•0.76= 36.6 m/s