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What will be the moment of inertia of a ring of mass 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 \(m r^2\).

Note by Kushal Patankar
1 year, 9 months ago

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

i am not sure too..............are getting the same??? Yash Sharma · 1 year, 8 months ago

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@Yash Sharma This is wrong, putting \(\theta = \pi/2\) does not yield \(\frac {mR^2} {2}\) Raghav Vaidyanathan · 1 year, 8 months ago

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@Yash Sharma You can check by putting θ=0 Kushal Patankar · 1 year, 8 months ago

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@Kushal Patankar what are you getting ? Karan Shekhawat · 1 year, 8 months ago

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@Karan Shekhawat I have a guess \[I= mr^2 \cos^2 \theta +\frac{mr^2}{2} \sin ^2 \theta \] Kushal Patankar · 1 year, 8 months ago

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did you solved it ... ? Karan Shekhawat · 1 year, 9 months ago

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@Karan Shekhawat Yep but i m not sure about the answer. Kushal Patankar · 1 year, 9 months ago

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