A thin heavy disc can turn freely, about an axis in its own plane, and this revolves horizontally with a uniform angular velocity $\omega$ about a fixed point on itself. Show that the inclination $\theta$ of the plane of the disc to the vertical is $$ \cos^{-1} ( \frac{gh} {k^2 w^2})$$ where h is the distance of the Centre of Inertia of the disc from the axis and k is the radius of gyration of the disc about the axis.
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What did you try, how far did you get, what do you know in general about this topic? Please mark the question if you decide to move it to the more appropriate http://physics.stackexchange.com forum. – Lutz Lehmann Mar 23 '19 at 12:58
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Here I know that the Moment of inertia of a thin disc about an axis passing through the center and perpendicular to its plane is ma^2/2 . Further we can somehow apply D' Alembert's principle of reverse effective force and take moment about the fixed point to get some equations of motion of the disc. – Soumya Kanti Das Mar 23 '19 at 16:05