In the given figure,a ring of mass m is kept | vedclass

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(A) Let $a_1$ be the acceleration of the centre of mass of the ring and $a_2$ be the acceleration of the hanging mass. Let $T$ be the tension in the s

Vedclass · Accepted Answer

Statement $(i)$ and $(ii)$ only

Vedclass · Answer

(A) Let $a_1$ be the acceleration of the centre of mass of the ring and $a_2$ be the acceleration of the hanging mass. Let $T$ be the tension in the string and $f$ be the frictional force acting on the ring.For the hanging mass: $mg - T = ma_2$  --- $(1)$For the ring (translational motion): $T + f = ma_1$  --- $(2)$For the ring (rotational motion about the centre of mass): $	au = I\alpha \Rightarrow (T - f)R = (mR^2)\alpha \Rightarrow T - f = ma_R\alpha$  --- $(3)$Since the ring rolls without slipping,$a_1 = R\alpha$. Also,the acceleration of the point on the ring where the string is attached is $a_1 + R\alpha = 2a_1$. Thus,$a_2 = 2a_1$.Substituting $a_2 = 2a_1$ and $\alpha = a_1/R$ into the equations:From $(3)$,$T - f = ma_1$.Adding $(2)$ and $(3)$: $2T = 2ma_1 \Rightarrow T = ma_1$.Substitute $T = ma_1$ into $(1)$: $mg - ma_1 = m(2a_1) \Rightarrow mg = 3ma_1 \Rightarrow a_1 = g/3$.Then $a_2 = 2a_1 = 2g/3$.From $(2)$,$f = ma_1 - T = ma_1 - ma_1 = 0$.Since the calculated frictional force is $0$,none of the statements $(iii)$ or $(iv)$ are correct. Thus,only statements $(i)$ and $(ii)$ are correct.

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