A uniform ring is rolling without slipping on a horizontal surface such that the velocity of its cen

Question

A uniform ring is rolling without slipping on a horizontal surface such that the velocity of its center of mass is $V$. The magnitude of the net velocity of a point on the rim situated at an angular position of $60^\circ$ from the highest point (measured about the center) is

Professor PC Thomas Classes · Answer

(b) : For pure rolling, the rotational speed of the rim is $v_{rot} = R\omega = V$. The net velocity of any point on the rim is the vector sum of its translational velocity $\vec{V}_{cm}$ (horizontal) and its rotational velocity $\vec{v}_{rot}$ (tangential). For a point $60^\circ$ away from the top, the angle between $\vec{V}_{cm}$ and $\vec{v}_{rot}$ is $60^\circ$. Using vector addition: $v_{net} = \sqrt{V^2 + V^2 + 2(V)(V)\cos(60^\circ)} = \sqrt{2V^2 + 2V^2(1/2)} = \sqrt{3V^2} = \sqrt{3} V$.