A baseball pitcher pivots his extended arm about his shoulderjoint, applying a c

Question

A baseball pitcher pivots his extended arm about his shoulderjoint, applying a constant torque of 151 N· m for 0.19 s to his arm, whichhas a moment of inertia of 0.51 kg· m2. (a) Find his arm's angular acceleration andfinal angular velocity, assuming it starts from rest. angular acceleration 1rad/s2 final angular velocity 2 rad/s
(b) Find the final speed of the ball, relative to his shoulder,which is 0.85 m from the ball.
3 m/s (a) Find his arm's angular acceleration andfinal angular velocity, assuming it starts from rest. angular acceleration 1rad/s2 final angular velocity 2 rad/s
(b) Find the final speed of the ball, relative to his shoulder,which is 0.85 m from the ball.
3 m/s angular acceleration 1rad/s2 final angular velocity 2 rad/s

Explanation / Answer

(a)Let the arm's angular accelration be . We know that = I * or = (/I) where = 151 N.m and I = 0.51 kg.m2 or = (151/0.51) = 296 rad/s2 Let the final angular velocity be w. We know that w = wo + t where wo = 0 rad/s and t = 0.19 s or w = 0 + 296 * 0.19 = 56.2 rad/s (b)Let the final speed of the ball, relative to his shoulder,which is 0.85 m from the ball bev.Therefore,we get v = r * w or v = 0.85 * 56.2 = 47.7 m/s Let the final angular velocity be w. We know that w = wo + t where wo = 0 rad/s and t = 0.19 s or w = 0 + 296 * 0.19 = 56.2 rad/s (b)Let the final speed of the ball, relative to his shoulder,which is 0.85 m from the ball bev.Therefore,we get v = r * w or v = 0.85 * 56.2 = 47.7 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.