A small pine tree has a mass of 15kg. Its center of mass is located at .72m from the ground. Its trunk is sawed through at ground level, causing the tree to fall, with the severed trunk acting as the pivot point. At the instant the falling tree makes a 17° angle with the vertical, the angular acceleration fo the tree is 2.4 rad/s2. What is the moment of inertia of the tree?
cm=.72m (from the ground)
(The correct answer is 12.9 kg⋅m2 according to the answer key of this study guide, what I want to know is how one gets to that answer.)
Kinda just listing all the ones I thought might help here but:
I=Icm+Mh2 (parallel axis theorem, not sure this is actually relevant.)
(angular momentum)=m(r x v)
(angular momentum of solid about an axis)=Iω
3. The Attempt at a Solution
I made many attempts at a solution such that, it would take way too long to put them all down here but first I tried just using parallel axis theorem, which didn't work. Then I tried using conservation of angular momentum which, was also a bad plan. Then I tried using the torque somehow which didn't pan out. I'm just not sure of the correct steps when supplied with the information that I have. The falling at an angle with an angular acceleration seems to be throwing me for a loop, what am I supposed to do with that? I already know the answer, but can anyone help with the steps I should take to get there?