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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
2 years, 10 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???

- 2 years, 10 months ago

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

- 2 years, 10 months ago

You can check by putting θ=0

- 2 years, 10 months ago

what are you getting ?

- 2 years, 10 months ago

I have a guess $I= mr^2 \cos^2 \theta +\frac{mr^2}{2} \sin ^2 \theta$

- 2 years, 10 months ago

did you solved it ... ?

- 2 years, 10 months ago