Rigid Body Kinetics: Angular Acceleration of Rod w/ Friction Torque

[gus_lyon]

Homework Statement

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?

Homework Equations

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.

 

[Doc Al]

Moment of inertia for a rod = (1/12)*m*(l^2)

That's the moment of inertia about what axis?

And when you find the net torque, be sure to include the torque due to friction.