A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.00 m and mass 6.88 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.203 rad/s^2.
1 Answer

Moment of Inertia for a rod= (MR^2)/12
(6.88*3^2)/12=5.16
I*angular acceleration= 5.16 * 0.203 = 1.05 Nm