Question 613: Centre of Mass and Rotational Motion

Question

A thin metal wire of length 'L' and uniform linear mass density '$\rho$' is bent into a circular coil with 'O' as centre. the moment of inertia of a coil about the axis XX' is

Accepted Answer

Key Idea Moment of inertia of a thin circular coil about its diameter,
$$I=\frac{M R^2}{2}$$
Moment of inertia of a thin circular coil,
$$I=\frac{M R^2}{2}$$
Now, moment of inertia of a ring about axis $X X^{\prime}$ as in figure below,
$I_{X x^{\prime}}=\frac{M R^2}{2}+M R^2=\frac{3}{2} M R^2\quad 	ext{.... (i)}$
(Using theorem of parallel axis)
Given, $L=$ length of wire of ring
and $ho=$ linear mass density
Then, mass of the ring $=$ linear density $	imes$ length
$$\begin{array}{ll}
\Rightarrow & M=ho L \quad	ext{... (ii)}\
	ext { and } & L=2 \pi R \
\Rightarrow & R=\frac{L}{2 \pi}\quad	ext{... (iii)}
\end{array}$$
Now, putting the value from Eqs. (ii) and (iii) in (i), we get
$$I_{x x^{\prime}}=\frac{3}{2}(ho L) \frac{L^2}{4 \pi^2} \Rightarrow \frac{3 ho L^3}{8 \pi^2}$$
Hence, option a is correct.

Answer

$\frac{3 \rho L^3}{8 \pi^2}$

Answer

$\frac{\rho L^3}{4 \pi^2}$

Answer

$\frac{3 \rho L^2}{4 \pi^2}$

Answer

$\frac{\rho L^3}{8 \pi^2}$

Centre of Mass and Rotational Motion

A thin metal wire of length 'L' and uniform linear mass density '\rho' is bent into a circular coil with 'O' as centre. the moment of inertia of a coil about the axis XX' is

Choose the correct answer: