1. The problem statement, all variables and given/known data A uniform rod of mass M and lenght L is free to rotate about a pivot at the left end. It is released from rest in the horizontal position ( Thetha=90 degrees). What is the torque on the rod when it makes and angle (theta) with the vertical? What is the downward linear acceleration of the right end of the rod when it is first released (at Thetha=90degrees)? 2. Relevant equations T=Ialpha, t=+-rFsintheta 3. The attempt at a solution For the first part of the question i manged to solve by with -rFsin(theta) my radius was half of the lenght of the rod, my force was found inthe center of mass of the rod which was mg and multiply that by sin(theta) according to the formula i get -(Lmg sin(theta)/2). The second part is thepart where i'm having trouble with. I know that it has to do with t=I(alpha) since we are looking for linear acceleration. But i don't seem to know how to start up with because I no no units. Any help on this one??