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Four rods of equal mass m each , Ab , BC , CA , DB are placed as shown. The moment of inertia of the system about B perpendicular to the plane ABC is (AB=a)

Physics
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|dw:1360169550988:dw|
Obviously break down the problem rod by rod.

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Other answers:

And if you're really stuck, you're just going to integrate rod by rod.
I (BC) = ma^2/6 ryt ?
Its basically a maths ques, you just need to find rod lengths, after doing that you just have to calculate moment of inertia of each rod about B and add all
Yup . i Have Calculated the length of rods BC = a/sqrt(2) , AC = a/sqrt(2) , DB = (sqrt2) * a / sqrt3
I (BC) = ma^2/6 ryt ?
yep
I (DB) = ma^2/18 ?
won' that be (2/9)ma^2
Hw ?
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I (DB) = M L^2 / 3 * sin^2 30 ryt ?
why sin ?
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I don;t think angle matters
For calculation of I,
Because the axis is perpendicular to plane, the rod will make 90 degrees with the axis, doesn't matter how it is placed.
Do you have the final ans ?
Yup 25 M a^2 / 18
So DB = 2M a^2 /9
And AC = ma^2/6 + 3ma^2/6 = 2ma^2 /3
So sum = ma^2 (2/3 + 2/9 + 1/6) = 19/18ma^2
hmm, okay doesn't match
i think that angle matters
Wait I missed AB
AB = ma^2 /3 Sum = 25/18
Got it! B|
But Hw Did u Claculate I (AC) ??
calculate about C, then B
Ok...got it....:)
"Angle will not matter here" ;)

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