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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)
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|dw:1360169550988:dw|
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@JamesJ
JamesJ
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Obviously break down the problem rod by rod.
JamesJ
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And if you're really stuck, you're just going to integrate rod by rod.
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I (BC) = ma^2/6 ryt ?
shubhamsrg
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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
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Yup . i Have Calculated the length of rods BC = a/sqrt(2) , AC = a/sqrt(2) , DB = (sqrt2) * a / sqrt3
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I (BC) = ma^2/6 ryt ?
shubhamsrg
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yep
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I (DB) = ma^2/18 ?
shubhamsrg
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won' that be (2/9)ma^2
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Hw ?
shubhamsrg
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|dw:1360246728479:dw|
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I (DB) = M L^2 / 3 * sin^2 30 ryt ?
shubhamsrg
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why sin ?
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|dw:1360246891927:dw|
shubhamsrg
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I don;t think angle matters
shubhamsrg
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For calculation of I,
shubhamsrg
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Because the axis is perpendicular to plane, the rod will make 90 degrees with the axis, doesn't matter how it is placed.
shubhamsrg
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Do you have the final ans ?
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Yup 25 M a^2 / 18
shubhamsrg
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<Lets see if we get that by my method>
shubhamsrg
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So DB = 2M a^2 /9
shubhamsrg
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And AC = ma^2/6 + 3ma^2/6 = 2ma^2 /3
shubhamsrg
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So sum =
ma^2 (2/3 + 2/9 + 1/6) = 19/18ma^2
shubhamsrg
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hmm, okay doesn't match
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i think that angle matters
shubhamsrg
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Wait I missed AB
shubhamsrg
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AB = ma^2 /3
Sum = 25/18
shubhamsrg
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Got it! B|
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But Hw Did u Claculate I (AC) ??
shubhamsrg
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calculate about C, then B
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Ok...got it....:)
shubhamsrg
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"Angle will not matter here" ;)