Center of Mass and Moment of Inertia

  • #1

Homework Statement


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?
m=15kg
cm=.72m (from the ground)
θ=17°
∝=2.4 rad/s2
Itree=?

(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.)

Homework Equations


Kinda just listing all the ones I thought might help here but:
I=Icm+Mh2 (parallel axis theorem, not sure this is actually relevant.)
∝=at/r
Tnet=I∝
(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?
 

Answers and Replies

  • #3
Please post that attempt.
I didn't really know what I was doing, so I tried out multiplying the mass by acceleration for a force
F=(15)(2.4)=36
then I did T=(.72)(36sin(17))=7.6
7.6=I(2.4)
I=3.2
which is, not correct. And I'm also not clear as to the thought process.
 
  • #4
haruspex
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I didn't really know what I was doing, so I tried out multiplying the mass by acceleration for a force
F=(15)(2.4)=36
then I did T=(.72)(36sin(17))=7.6
7.6=I(2.4)
I=3.2
which is, not correct. And I'm also not clear as to the thought process.
The 2.4 is an angular acceleration, not a linear acceleration, so multiplying it by a mass does gives anything useful.
You quoted two equations relating to angular acceleration. (The correct symbol is α, the Greek letter alpha, not ∝.)
Which one looks appropriate here?
 
  • #5
The 2.4 is an angular acceleration, not a linear acceleration, so multiplying it by a mass does gives anything useful.
You quoted two equations relating to angular acceleration. (The correct symbol is α, the Greek letter alpha, not ∝.)
Which one looks appropriate here?
My guess is the one with tangential acceleration. And I only use that symbol because that's what my professor uses. So you're saying if I do what I did but using the correct form of acceleration I should get the correct answer?
 
  • #6
No, there's more. I'm going to take a minute to try out a different method of solving.
 
  • #8
Tnet=I∝ you mean this one? But then what is the net torque? I know torque is usually force multiplied by how far away from the axis you are so then would the force be the force of gravity at this angle multiplied by .72m?
 
  • #9
haruspex
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Tnet=I∝ you mean this one? But then what is the net torque? I know torque is usually force multiplied by how far away from the axis you are so then would the force be the force of gravity at this angle multiplied by .72m?
Not quite. The torque a force exerts about an axis is the magnitude of the force multiplied by the distance from the axis to the line of action of the force.
Draw a diagram. Show the force of gravity as an arrow. What is the nearest that arrow (extended to infinity) gets to the axis?
 
  • #10
T=.72m*15kg*9.8m/s2*sin(17)=30.94
30.94=I(2.4rad/s2)
so I=12.9
 
  • #11
12.9 is actually the number I was looking for so I think I found the correct process for what my professor wanted. Thanks for your help.
 
  • #12
haruspex
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12.9 is actually the number I was looking for so I think I found the correct process for what my professor wanted. Thanks for your help.
You are welcome.
Regarding the symbol, whichever way your professor happens to write it, I'm sure it is intended to be the Greek lowercase alpha, not the "is proportional to" symbol. ∝ might look more like what your professor wrote, but the correct one to use in typing is definitely α. Of course, there are different fonts. Maybe ##\alpha## looks better.
 

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