The system consists of a massless ring and a bead of mass $M$. The ring rolls on the floor. Because the ring is massless, the “Center of Mass” of the entire system is simply the location of the bead. We analyze the forces acting on the system (Gravity $Mg$, Normal $N$, Friction $f$) and equate them to the motion of the bead.

Assuming pure rolling initially, the velocity of the contact point is zero. The velocity of the bead is $v$. From geometry of a rolling cycloid (or rotational kinematics about the instantaneous axis of rotation at the contact point), $v^2 = 2 (R\omega)^2 (1+\cos\theta)$.
Applying Work-Energy Theorem: $$ \text{Loss in PE} = \text{Gain in KE} $$ $$ Mg R(1 – \cos\theta) = \frac{1}{2} M [2R^2\omega^2(1+\cos\theta)] $$ $$ g(1 – \cos\theta) = R\omega^2(1+\cos\theta) $$ $$ \omega^2 = \frac{g}{R} \frac{1-\cos\theta}{1+\cos\theta} = \frac{g}{R} \tan^2\frac{\theta}{2} $$

To find the condition for slipping, we must derive the Normal force $N$ and Friction force $f$ in terms of $\theta$. Differentiating the energy equation $g(1-\cos\theta) = R\omega^2(1+\cos\theta)$ with respect to time (noting $\dot{\theta} = \omega$) yields the angular acceleration $\alpha$: $$ R\alpha = \frac{g \sin\theta}{(1+\cos\theta)^2} $$ Horizontal Equation of Motion ($f = Ma_x$): The horizontal acceleration of the bead (CM) is $a_x = R\alpha(1+\cos\theta) – R\omega^2\sin\theta$. Substituting $\alpha$ and $\omega^2$: $$ f = M \left[ \frac{g \sin\theta}{1+\cos\theta} – g \left(\frac{1-\cos\theta}{1+\cos\theta}\right)\sin\theta \right] $$ $$ f = \frac{Mg \sin\theta}{1+\cos\theta} [ 1 – (1-\cos\theta) ] = Mg \frac{\sin\theta \cos\theta}{1+\cos\theta} $$ Vertical Equation of Motion ($N – Mg = Ma_y$): The vertical acceleration is $a_y = -R\alpha\sin\theta – R\omega^2\cos\theta$. $$ N = M(g + a_y) = M \left[ g – \frac{g \sin^2\theta}{(1+\cos\theta)^2} – \frac{g(1-\cos\theta)}{1+\cos\theta}\cos\theta \right] $$ After simplifying the algebraic terms, this reduces remarkably to: $$ N = Mg \cos\theta $$ Condition for Slipping ($f = \mu N$): $$ \mu = \frac{f}{N} = \frac{Mg \frac{\sin\theta \cos\theta}{1+\cos\theta}}{Mg \cos\theta} = \frac{\sin\theta}{1+\cos\theta} $$ Using half-angle identities $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $1+\cos\theta = 2\cos^2\frac{\theta}{2}$: $$ \mu = \tan\frac{\theta}{2} $$

(a) Angle of rotation: From $\tan\frac{\theta}{2} = \mu$: $$ \theta = 2 \tan^{-1}\mu $$

(b) Speed of the centre: The speed of the center of the ring is $v_c = R\omega$. Substitute $\omega = \sqrt{\frac{g}{R}} \tan\frac{\theta}{2} = \sqrt{\frac{g}{R}} \mu$: $$ v_c = R \left( \mu \sqrt{\frac{g}{R}} \right) = \mu \sqrt{gr} $$