Moment of Inertia of a 4 rod system

  • #1
Homework Statement:
Calculate the Moment of Inertia of each and determine the greatest:
Relevant Equations:
MOI=mr2
This was the question
20211125_163051.jpg

20211125_163147.jpg

(The line below is probably some translation of upper line in different language)

For disc it was ma^2/2
For ring it was ma^2
For square lamina it was 2ma^2/3
For rods
It was different
20211125_163250.jpg

Please explain

Thank You🙏
 

Answers and Replies

  • #2
TSny
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I don't know what was being calculated by this equation
1637853924455.png


The hand-written calculation is a little messy and hard to follow, but it looks correct for the 4 rods making a square.
 
  • #3
It probably calculated the MOI of the system (because it was the solution given in the booklet)

Also my MOI came out to be
(m/4)×(8a^2)÷12 + 2×(m/4)×a^2
[About x and y axes]

&&

(ma^2)+(ma^2)/3


{About z axis}
 
  • #4
TSny
Homework Helper
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Also my MOI came out to be
(m/4)×(8a^2)÷12 + 2×(m/4)×a^2
[About x and y axes]
Ok, you are saying that ##I_{\rm x\, axis} = I_{\rm y\, axis} = \frac{m}{4} \frac{8a^2}{12} + 2 \frac{m}{4}a^2##
That looks correct.
It would be nice to simplify the result.
&&

(ma^2)+(ma^2)/3


{About z axis}
Here you are using ##I_{\rm z\,axis} = I_{\rm x\,axis}+I_{\rm y\,axis} = 2I_{\rm x\,axis} = ma^2 + \frac{ma^2}{3}##.
This looks correct. Of course you can simplify this result to just one term.
 

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