Question

A ring or radius $3a$ is fixed rigidly on a table. A small ring whose mass is $m$ and radius a, rolls without slipping inside it as shown in the figure. The small ring is released from position $A.$ When it reaches at the lowest point, the speed of the centre of the ring at that time would be :-

• A. $\sqrt{2ga}$
• B. $\sqrt{3ga}$
• C. $\sqrt{6ga}$
• D. $\sqrt{4ga}$

Answer 1

The correct option is A $\sqrt{2ga}$

$\begin{array}{c}\text{Given-}\ast \text{mass of the}\mathrm{Ring}=m\text{and readius (a)}\\ \text{we know that Kinetic energy of a rolling ring is given as -}\\ kE=\frac{1}{2}m{v}^2+\frac{1}{2}\text{Iw (Rotational kinetic Energy)}\end{array}$

$\begin{array}{cc}I& =m{a}^2\text{(moment of Inertia of Ring)}\\ \Rightarrow kE& =\frac{1}{2}m{v}^2+\frac{1}{2}m{a}^2{w}^2=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2(v=aw)\text{circular motion}\\ kE& =m{v}^2\end{array}$

$\begin{array}{c}\text{Foom conservation of energy -}\\ \text{Initial potential energy = Final kinetic energy}\\ \Rightarrow mgh=m{v}^2\\ \Rightarrow mg(3a-a)=m{v}^2(h=3a-a)\text{change in height.}\\ \Rightarrow v=\sqrt{2ag}\end{array}$