A uniform rod of mass 0.62 kg is 6 m long. The rod is pivoted about a horizontal, fric- tionless pin at the end of a thin extension (of negligible mass) a distance 6 m from the cen- ter of mass of the rod. Initially the rod makes an angle of 48◦ with the horizontal. The rod is released from rest at an angle of 48◦ with the horizontal, as shown in the figure. What is the angular speed of the rod at the instant the rod is in a horizontal position? The acceleration of gravity is 9.8m/s2 and the moment of inertia of the rod about its center of mass is I = (1/12)mL2 Answer in units of rad/s. 2. Relevant equations U = mgh KE = .5Iw2 3. The attempt at a solution So, I set U = KE mgh = .5Iw2 Then I plugged in the moment of inertia mgh = .5(1/12)(mass of rod)(length of rod)2w2 Cancelled out the m and solved for the height of the center of mass g(6sin(48degrees)) = .5(1/12)mL2w2 and finally, I plugged in values and solved. (9.8)(6sin(48 degrees)) = .5(1/12)(6)2w2 And solved for w, which I found to equal 5.397339822. This is wrong, but I have no idea where my error is. Can you guys help me spot it??