A uniform rod of length 1.15m is attached to a frictionless pivot at one end. It is released from rest an angle theta=17.0 degrees above the horizontal.
Find the magnitude of the initial acceleration of the rod's center of mass
torque=moment of inertia*angular acceleration
The Attempt at a Solution
I set the torque about pivot equal to the moment of inertia about pivot times the angular acceleration to find the angular acceleration. Then I found the linear acceleration by multiplying the found angular acceleration by the length of rod. So my work was....
12.22 rad/s^2= (angular acceleration)
a=(angular acceleration)*L= 14.05 m/s^2
However the answer I got is wrong. (Unfortunately, I do not have access to the correct answer) What step did I miss?