A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v₀. It encounters an inclined plane at angle $$\theta$$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as;

I₁ = M.I. of thin circular ring about its diameter,

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

Then :

The linear mass density of a thin rod AB of length L varies from A to B as
$$\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)$$, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :

Four point masses, each of mass m, are fixed at the corners of a square of side $$l$$. The square is rotating with angular frequency $$\omega $$, about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is :