In the case of pure rolling,what will be the | vedclass

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(B) In pure rolling,the velocity of any point on the ring is the vector sum of the translational velocity of the center of mass $(v_{cm})$ and the tan

Vedclass · Accepted Answer

$\sqrt{2} v_{cm}$

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(B) In pure rolling,the velocity of any point on the ring is the vector sum of the translational velocity of the center of mass $(v_{cm})$ and the tangential velocity due to rotation $(v_{rot} = \omega R = v_{cm})$.For point $A$,the translational velocity is directed horizontally to the right $(v_{cm})$,and the rotational velocity is directed vertically upwards $(v_{rot} = v_{cm})$.Since these two velocity vectors are perpendicular to each other,the net velocity $v_{net}$ is given by:$v_{net} = \sqrt{v_{cm}^2 + v_{rot}^2} = \sqrt{v_{cm}^2 + v_{cm}^2} = \sqrt{2 v_{cm}^2} = \sqrt{2} v_{cm}$

In the case of pure rolling,what will be the velocity of point $A$ of the ring of radius $R$?

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