A uniform slender bar AB of the length of / has an angular velocity of omega and its center of gravity has a velocity of vG as shown. Calculate the angular momentum of the bar about A, B, and the center of gravity. Solution a) angular momentum about center of gravity = m*v g + I*omega = m*v g + ml^2*omega/12 moment of inertia of a slender rod about center of gravity = ml^2/12 b) angular momentum about A = m*v g + I*omega = m*v g + (ml^2/12+ ml^2/4)*omega = m*v g + ml^2*omega/3 c) angular momentum about B = m*v g + I*omega = m*v g + (ml^2/12+ ml^2/4)*omega = m*v g + ml^2*omega/3 .