A ring of mass m and radius r has a particle of mass m attached to it at a point A. The ring can rotate about a smooth horizontal axis which is tangential to the ring at a point B diametrically opposite to A. The ring is released from rest when A B is horizontal. Find the angular velocity and the angular acceleration of the body when A B has turned through an angle $\frac{π}{3}$ .

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(1) $\sqrt{\frac{3 g}{4 r}}, \frac{2 g}{10 r}$

(2) $\sqrt{\frac{6 \sqrt{3} g}{11 r}}, \frac{3 g}{11 r}$

(3) $\sqrt{\frac{2 \sqrt{3} g}{13 r}}, \frac{7 g}{12 r}$

(4) $\sqrt{\frac{6 g}{11 r}}$ $\frac{6 g}{13 r}$

answer is B.

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Detailed Solution

Angular velocity From conservation of mechanical energy,

decrease in gravitational PE

= increase in rotational KE

$or mgrsin 60 ° + mg (2 rsin 60 °) = \frac{1}{2} [\frac{3 {mr}^{2}}{2} + m (2 r)^{2}] ω^{2} 2 = \frac{3 \sqrt{3} mgr}{4} {mr}^{2} ω^{2} ∴ ω = \sqrt{\frac{6 g \sqrt{3}}{11 r}}$

Angular acceleration

$α = \frac{τ}{1} = \frac{mgrcos 60 ° + mg (2 rcos 60 °)}{[\frac{3 {mr}^{2}}{2} + m (2 r)^{2}]} = \frac{\frac{3 mgr}{2}}{\frac{11}{2} {mr}^{2}} = \frac{3 g}{11 r}$