Problem — Pure rolling of a ring

Problem

A ring of $M = 1\;\mathrm{kg}$, $R = 0{,}20\;\mathrm{m}$, rolls without slipping on a horizontal plane with $v_\text{cm} = 5\;\mathrm{m/s}$. Calculate the total kinetic energy and verify that the two halves (translational and rotational) coincide.

Solution

For the ring $I = MR^2$ and the rolling constraint $\omega = v_\text{cm}/R$. Translational: $E_\text{transl} = \tfrac{1}{2}M v_\text{cm}^2 = \tfrac{1}{2}\cdot 1\cdot 5^2 = 12{,}5\;\mathrm{J}$. Rotational: $E_\text{rot} = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}MR^2\cdot\dfrac{v_\text{cm}^2}{R^2} = \tfrac{1}{2}M v_\text{cm}^2 = 12{,}5\;\mathrm{J}$.

For the ring the two halves are exactly equal, because all the mass sits at radius $R$ ($I = MR^2$). Total: $\ev{E_\text{tot} = 25\;\mathrm{J}\quad(\text{half translation, half rotation})}$

Links

Topics: Rotational dynamics Concepts: Rolling · Rotational kinetic energy Objects: Rotating disc