# Angular Acceleration of a Fixed Rod

1. The problem statement, all variables and given/known data
Consider a thin 14 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 3.2 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is Irod =1/3mL^2 and the moment of inertia of the sphere about its center of mass is Isphere =2/5mr^2 . What is the angular acceleration of the rod immediately after it is released from its initial position of 35◦ from the vertical? The acceleration of gravity g = 9.8 m/s^2

If needed a graphical representation of the question can be found at the following link:
http://ift.tt/OpPkbU

2. Relevant equations
τ = Iα
I = (1/3)MR2
τ = FRsin(Θ)

3. The attempt at a solution
Finding the position of the center of mass (x meters from pivot):
(x-7) = (14+3.2-x)
x=12.1 meters

Using xcm = 12.1 m, we can solve for α:
I = (1/3)(18 kg)(12.1 m)2 ≈ 878.46 kg·m2
τ = (18 kg)(9.8 m/s2)(12.1 m)sin(35°) ≈ 1224.264489 N·m
α = τ/I ≈ 1.39361 rad/s2

This answer is taken as incorrect by my online program. Any suggestions or assumptions that I have made in my calculations would be helpful. Thanks. One thing that I was worried about was that this method didn’t use the moments of inertia that I was given.

http://ift.tt/1hka7dP