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Rigid body rotation problem

  • #1
I am trying to solve problem number 4 part A and B from (http://people.physics.tamu.edu/kamon/teaching/phys218/exam/2003C/2003C_Exam3_Solution.pdf) but I am confused about certain aspects of it.

In part A, I understand that since we are considering the person as a cylinder, the equation for moment of inertia will include (1/2)MR^2. What I can't seem to understand is where did 2MR^2 come from and why we didn't include the moment of inertia of the arm(which is a rod).

In part B, I understand the rest of the moment of inertia equation but I am confused where did m(1l/2 + R)^2 come from?
 

Answers and Replies

  • #2
BvU
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Hello Ken, :welcome:

Please use the template and the superscript/subscript buttons. See guidelines
Apparently you have a question on moment of inertia. So isolate that problem and state it. Show the relevant equations.

And note that the solution does make a difference between M and m -- as you do not.
 
  • #3
Doc Al
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What I can't seem to understand is where did 2MR^2 come from and why we didn't include the moment of inertia of the arm(which is a rod).
The 2mR^2 is the moment of inertia of the arms -- when they are hanging down, not outstretched. (Note: m, not M.)

In part B, I understand the rest of the moment of inertia equation but I am confused where did m(1l/2 + R)^2 come from?
Here the arms are outstretched. Note that the axis of rotation is the center of the person, not the shoulder.
 
  • #4
BvU
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The 2mR^2 is the moment of inertia of the arms -- when they are hanging down, not outstretched. (Note: m, not M.)


Here the arms are outstretched. Note that the axis of rotation is the center of the person, not the shoulder.
Giving it away, eh ? not PF ! Well, let's say Ken has beginner's credit :wink:
 
  • #5
Doc Al
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Giving it away, eh ? not PF !
:wink:

Well, let's say Ken has beginners credit
First one's free! :smile:
 
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