Q.Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.

Total rolling KE splits as Translational : Rotational : Total = $1 : \frac{K^2}{R^2} : \left(1+\frac{K^2}{R^2}\right)$, a ratio fixed purely by the object's shape (via $K^2/R^2$), never by its mass or size. …

For an object of mass M, radius R and radius of gyration K, rolling uniformly with centre-of-mass speed v (so $\omega=v/R$ for pure rolling), the total kinetic energy is the SUM of a translational part (from the centre-of-mass motion) and a rotational part (from spinning about the centre of mass): $E=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2=\frac{1}{2}Mv^2+\frac{1}{2}(MK^2)\left(\frac{v}{R}\right)^2=\frac{1}{2}Mv^2\left(1+\frac{K^2}{R^2}\right)$ Reading off the coefficients of $\frac{1}{2}Mv^2$ in each term gives the ratio $\text{Translational K.E.}:\text{Rotational K.E.}:\text{Total K.E.}=1:\frac{K^2}{R^2}:\left(1+\frac{K^2}{R^2}\right)$ Since $K^2/R^2$ is a pure NUMBER determined entirely by the object's SHAPE (equal to 1 for a ring/hollow cylinder, $\frac{1}{2}$ for a disc/solid cylinder, $\frac{2}{5}$ for a solid sphere, $\frac{2}{3}$ for a thin hollow sphere), this three-way split is completely fixed once the shape is known, and is entirely INDEPENDENT of the object's mass M, its radius R, or its actual rolling speed v (all of which cancel out of the ratio). For …

Write the total rolling kinetic energy as the sum $\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$, substitute $I=MK^2$ and $\omega=v/R$, and read the three-way ratio …

Believing the translational/rotational split depends on the object's actual speed or size, rather than realising v, M and R all cancel out, leaving a …