Question
A thin wire of length ' $L$ ' and linear mass density ' $m$ ' is bent into a circular ring (in $x - y$ plane) with centre ' $C$ ' as shown in figure. The moment of inertia of the ring about an axis $yy^{'}$ will be:
Options
Solution
Sol. [Key Concept: Moment of inertia of a ring] Mass of thin wire $= ($ Linear mass density $) \times ($ Length $)$
$$\begin{matrix} & \ \Rightarrow \ M = mL \\ & L = 2\pi r,\text{~}\text{where}\text{~}r = \text{~}\text{radius of circular ring}\text{~} = \frac{L}{2\pi} \end{matrix}$$
Using parallel axis theorem,
$$\begin{matrix} & I_{yy^{'}} = I_{CM} + Mr^{2} = \frac{Mr^{2}}{2} + Mr^{2} = \frac{3}{2}Mr^{2} \\ & \ \Rightarrow \ I_{yy^{'}} = \frac{3}{2} \times (mL) \times \left( \frac{L}{2\pi} \right)^{2} = \frac{3mL^{3}}{8\pi^{2}} \end{matrix}$$
Create a free account to view solution
View Solution Free