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 \(m r^2\).

*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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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??? – Yash Sharma · 1 year, 6 months ago

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– Raghav Vaidyanathan · 1 year, 6 months ago

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

– Kushal Patankar · 1 year, 6 months ago

You can check by putting θ=0Log in to reply

– Karan Shekhawat · 1 year, 6 months ago

what are you getting ?Log in to reply

– Kushal Patankar · 1 year, 6 months ago

I have a guess \[I= mr^2 \cos^2 \theta +\frac{mr^2}{2} \sin ^2 \theta \]Log in to reply

did you solved it ... ? – Karan Shekhawat · 1 year, 6 months ago

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– Kushal Patankar · 1 year, 6 months ago

Yep but i m not sure about the answer.Log in to reply