Energy Distribution In Rolling

Question

A solid cylinder rolls without slipping along a flat horizontal surface at a steady speed. What fraction of its total kinetic energy is contained in its rotation about the central axis?

RankGuru · Accepted Answer

A rolling body carries kinetic energy in two forms: translation of the centre of mass and rotation about that centre, with $KE_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. For a solid cylinder $I = \frac{1}{2}MR^2$, and rolling without slipping enforces the relation $\omega = v/R$. The rotational part then becomes $\frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{4}Mv^2$, while the translational part is $\frac{1}{2}Mv^2$. The total is $\frac{3}{4}Mv^2$, 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.

Energy Distribution In Rolling

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