A thin rod of length 4l and 4m is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point O and perpendicular to the plane of paper?
Ml23
10Ml23
Ml212
Ml224
A
10Ml23
B
Ml23
C
Ml212
D
Ml224
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Solution
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Total moment of inertia =I1+I2+I3+I4=2I1+2I2 =2(I1+I2)[I3+I1,I1=I4] Now, I2=I3=MI23 Using parallel axes theorem, we have I=ICM+Mx2 and x=√l2+l24 I1=I4=Ml212+M⎡⎣√l2+(12)2⎤⎦2 Putting all values we get Moment of inertia, I=10(MI23).
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