1. The problem statement, all variables and given/known data A 2.0-m measuring stick of mass 0.175 kg is resting on a table. A mass of 0.500 kg is attached to the stick at a distance of 74.0 cm from the center. Both the stick and the table surface are frictionless. The stick rotates with an angular speed of 5.30 rad/s. (a) If the stick is pivoted about an axis perpendicular to the table and passing through its center, what is the angular momentum of the system? (b) If the stick is pivoted about an axis perpendicular to it and at the end that is furthest from the attached mass, and it rotates with the same angular speed as before, what is the angular momentum of the system? 2. Relevant equations L=Iω I=MR^2 I=(1/12)ML^2 I=(1/3)ML^2 3. The attempt at a solution Okay so for the first question I used the moment of inertia formula and treated the mass as a point mass. (0.500*0.740^2)=0.27 Plus the moment of inertia for the stick(rod) since it passing through the center I used I=(1/12)ML^2 (1/12)(0.175*2^2)=0.058 Then I added and multiplied by the angular velocity 0.27+0.058=0.33*5.3=1.7 kg*m/s Now this answer was right however when I did the second one I repeated the same process but used I= (1/3)ML^2 because it at the end of the rod. So I did (0.500*0.740^2)=0.27 Then (1/3)(0.175*2^2)=0.23 Then I added and multiplied by the angular velocity 0.27+0.23=0.50*5.3=2.65 rounded to 2.7 kg*m/s It is saying both 2.65 and 2.7 were wrong. Where did I go wrong?