√ω1. The problem statement, all variables and given/known data A uniform rod of mass 1.2 kg and length 1.2 m is pivoted in the horizontal position as shown (black point). The rod is at rest and then released. The acceleration due to gravity is g = 9.8 m/s2. Part 1. Find the rotational inertia of the rod relative to the axis perpendicular to the screen and passing through the pivot point (in kgm2). Part 2. As the rod swings downward, its angular speed velocity and the magnitude of its angular acceleration, respectively are (increasing, decreasing, staying the same) and (increasing, decreasing, staying the same). Part 3. What is the angular speed (in rad/s) of the rod as it passes through the vertical position (when end marked B is at the bottom)? Part 4. What is the linear speed of the bottom end of the rod (marked B) when it passes the vertical position?(in m/s) 2. Relevant equations Icm=1/12ml2 Ip=Icm+md2 Ug=mgh Ui+Ki+Wnoncon=Uf+Kf Ktrans=1/2mv2 Krot=1/2Iω2 t=Frsinθ K=Ktrans+Krot v=ωR 3. The attempt at a solution Picture/FBD is attached. Part 1 Ip=1/12m(l)2+m(l/4)2=1/12*1.2*1.22+1.2*.32=.252 Part 2 increasing and decreasing Part 3 0=1/2mv2+1/2Iω2+mgh where h=-l/4(distance from pivot to cm which will be directly below pivot when the rod is at the bottom of its swing) mgh=1/2mv2+1/2Iω2=mgh=1/2mω2R2+1/2Iω2=1/2mω2R2+1/2Iω2 1.2(9.8)(.3)=(1/2(1.2)(.3)2+1/2*.252)ω2=(.18)ω2 ω=√(3.528/.18)=4.42718872424 I am not sure if this is correct, but I used l/4 (distance from pivot to center of mass) as R. Part 4 mgh=1/2mv2+1/2Iω2 2mgh=mv2+Iv2/R2 2(1.2)(9.8)(.3)=(1.2+.252/.32)v2 v=√(7.056/4)=1.32815661727 or v=ωR2=4.42718872424(.3)^2=0.398446985181 Since these two values for v don't match up, I think I'm doing something wrong.