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(A) For a wheel rolling on a plane surface,the velocity of the highest point $B$ is $V_B = 2v$,where $v$ is the velocity of the center of mass.Given $

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$4 \sqrt{2} \ m/s$

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(A) For a wheel rolling on a plane surface,the velocity of the highest point $B$ is $V_B = 2v$,where $v$ is the velocity of the center of mass.Given $V_B = 8 \ m/s$,we have $2v = 8 \ m/s$,which implies $v = 4 \ m/s$.The point $A$ at the bottom is the instantaneous center of rotation.The velocity of any point $P$ on the rim is given by $V_P = \omega r_P$,where $r_P$ is the distance from the instantaneous center $A$.For a point $P$ at the same level as the center $C$,the distance $AP = \sqrt{R^2 + R^2} = R\sqrt{2}$,where $R$ is the radius of the wheel.Since $v = \omega R$,we have $\omega = v/R$.Thus,$V_P = (v/R) 	imes (R\sqrt{2}) = v\sqrt{2}$.Substituting $v = 4 \ m/s$,we get $V_P = 4\sqrt{2} \ m/s$.

$A$ wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is $8 \ m/s$. The speed of the particle on the rim of the wheel at the same level as the centre of the wheel will be

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