*m* and radius *r* in xy plane about an axis making angle \(\theta\) with the z-axis given that moment of inertia about z axis is \(m r^2\).

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## Comments

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TopNewesti am getting 2m(r^2)(ln(R)/(pi) - m((cos(theta))^2)(R^2)/(2pi)

i am not sure too..............are getting the same???

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This is wrong, putting \(\theta = \pi/2\) does not yield \(\frac {mR^2} {2}\)

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You can check by putting θ=0

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what are you getting ?

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did you solved it ... ?

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Yep but i m not sure about the answer.

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