[SOLVED] Magnitude of the angular acceleration about the pivot point? 1. The problem statement, all variables and given/known data A uniform horizontal rod of mass 2.7 kg and length 2.4 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I=ml^2/12 If 4 N force at an angle of 64 degrees to the horizontal acts on the rod as shown, what is the magnitude of the resulting angular acceleration about the pivot point? The acceleration of gravity is 9:8 m/s^2 : Answer in units of rad/s^2 2. Relevant equations I=(2.7*(2.4)^2)/12 F=4N=2.7*a 3. The attempt at a solution I=(2.7*(2.4)^2)/12=1.296 F=4N=2.7*a How do I figure out the magnitude of a? or [tex]\alpha[/tex]? I just cant understand how to relate the F and the I, I would think the magnitue of a was sin(64).