Homework SourseA baseball pitcher pivots his extended arm about his shoulde
A baseball pitcher pivots his extended arm about his shoulder joint, applying a constant torque of 190 N · m for 0.20 s to his arm, which has a moment of inertia of 0.43 kg · m2.
(a) Find his arm\'s angular acceleration and final angular velocity, assuming it starts from rest.
(b) Find the final speed of the ball, relative to his shoulder, which is 0.77 m from the ball.
(a) Find his arm\'s angular acceleration and final angular velocity, assuming it starts from rest.
(b) Find the final speed of the ball, relative to his shoulder, which is 0.77 m from the ball.
Solution
(a)Let the arm\'s angular accelration be a. We know that t = I * a or a = (t/I) where t = 190 N.m and I = 0.43 kg.m2 or a = (190/0.43) = 441.86 rad/s2 Let the final angular velocity be w. We know that w = wo + at where wo = 0 rad/s and t = 0.2 s or w = 0 + 441.86 * 0.2 = 88.4 rad/s (b)Let the final speed of the ball, relative to his shoulder,which is 0.77 m from the ball bev.Therefore,we get v = r * w or v = 0.77 * 56.2 = 68.1 m/s
