1. The problem statement, all variables and given/known data A long straight rod has the following: mass = 0.016 kg length = 0.16 m It is being pulled on by a force on its right end. The force pulls down on the rod, making an angle of 22.5 degrees with the vertical. The magnitude of the force is 3.2E-2 N. Give the angular acceleration of the rod at the instant shown. 2. Relevant equations 3. The attempt at a solution I chose the center of mass to be the rotation axis, I hope that's okay. Then the torque due to this force is cos(22.5)(L/2)F = cos(22.5)(0.16 m/2)(3.2E-2 N) = 0.00237 Nm. Torque = I * angular acceleration, so angular acceleration = torque/I I = (1/12)mL2 = 3.41E-5 kgm2 0.00237 Nm / 3.41E-5 kgm2 = 69.29 rad/s2 But this is wrong, the answer should be 115 rad/s2 Can someone help?