Energy Distribution In Rolling

A rolling body carries kinetic energy in two forms: translation of the centre of mass and rotation about that centre, with $K{E}_{total}=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$. For a solid cylinder $I=\frac{1}{2}M{R}^2$, and rolling without slipping enforces the relation $\omega =v/R$. The rotational part then becomes $\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{v}{R}\right)}^2=\frac{1}{4}M{v}^2$, while the translational part is $\frac{1}{2}M{v}^2$. The total is $\frac{3}{4}M{v}^2$, so the rotational fraction is $\frac{1/4}{3/4}=\frac{1}{3}$. The value 1/2 wrongly assumes an equal split between the two forms. The value 2/3 reports the translational fraction instead of the rotational one. The value 2/5 belongs to a solid sphere, not a cylinder. This is the NCERT energy partition for rolling bodies. As a check, since the cylinder's translational energy clearly exceeds its rotational energy, the rotational fraction must be below one-half, and 1/3 satisfies that requirement. This one-third share is fixed by the cylinder's geometry alone and does not depend on its speed, mass, or radius. A hollow pipe, by contrast, stores a full one-half of its energy in rotation, which is exactly why solid bodies reach the bottom of an incline ahead of hollow ones of the same mass and radius.