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a rod of mass m and length l lies in the axis, with its centre at the origin. what is its moment of inertia about the line x=y?
dear reshab ,
first take elemental part dx anywhere on the rod & let the distance of this part from origin is x ..
angle bw rod & axis is @ = 45o (in this case)
perpendicular distance bw that part & axis is r = xsin@
mass of this element = m/l (dx)
dI (inertia of this small part) = (dm) r2
= (m/l) (xsin@)2 dx
I = (msin2@/l) x2dx lim from -l/2 to l/2
after integrating & putting limit we get
I = ml2 / 24
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