A long slender rod has a mass of 0.66 kg and a length of 648.0 mm. The friction torque at the hinge is 0.41 Nm.
What is the angular acceleration of the rod when it is 38 degrees below the horizontal?
The Attempt at a Solution
Torque = moment of inertia x angular acceleration
Moment of inertia for a rod = (1/12)*m*(l^2)
= (1/12)*0.66*(0.648^2) = 0.0231 Kgm^2
Torque = Radius * Force
= (0.648*cos(38))*(9.81*0.66) = 3.306 Nm
Angular acceleration = Torque/Moment of inertia
= 3.306/0.0231 = 143.12
Which is wrong.