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(D) For a ring rolling without slipping,the velocity of the point of contact $A$ with the ground is zero,so $v_A = u - R\omega = 0$,which implies $u =

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All of the above

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(D) For a ring rolling without slipping,the velocity of the point of contact $A$ with the ground is zero,so $v_A = u - R\omega = 0$,which implies $u = R\omega$.The velocity of the top point $B$ is $v_B = u + R\omega = 2u$.Thus,the range of speeds for any point $P$ on the ring is $0 \leq v \leq 2u$.Let $P$ be a point such that the radius $CP$ makes an angle $	heta$ with the horizontal. The velocity vector of $P$ is the vector sum of the translational velocity $\vec{u}$ and the rotational velocity $\vec{v}_{rot} = \vec{\omega} 	imes \vec{r}$.The magnitude of velocity is $v_P = \sqrt{u^2 + (R\omega)^2 + 2u(R\omega)\cos(	heta + 90^\circ)}$.Since $u = R\omega$,$v_P = \sqrt{u^2 + u^2 + 2u^2(-\sin	heta)} = u\sqrt{2(1 - \sin	heta)}$.For option $B$: If $CP$ is horizontal,$	heta = 0^\circ$,then $v_P = u\sqrt{2(1 - 0)} = u\sqrt{2}$. This is correct.For option $C$: If $	heta = 30^\circ$ and $P$ is below the horizontal,then $v_P = u\sqrt{2(1 - \sin 30^\circ)} = u\sqrt{2(1 - 0.5)} = u\sqrt{2(0.5)} = u$. This is also correct.Therefore,all statements are correct.

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

$A$ ring rolls without slipping on the ground. Its centre $C$ moves with a constant speed $u$. $P$ is any point on the ring. The speed of $P$ with respect to the ground is:

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