Question 554: Centre of Mass and Rotational Motion

Question

A thin wire of length '$$L$$' and uniform linear mass density '$$m$$' is bent into a circular coil. The moment of inertia of this coil about tangential axis and in plane of the coil is

Accepted Answer

$$	herefore \quad$$ Moment of inertia of thin wire:
$$\begin{aligned}
I & =\frac{M^2}{2} \
M & =v 	imes m 	ext { and } L=2 \pi R \
R & =\frac{L}{2 \pi} \
	herefore \quad I & =\frac{L m}{2}\left(\frac{L}{2 \pi}ight)^2 \
I & =\frac{m L^3}{8 \pi^2}
\end{aligned}$$
$$	herefore \quad$$ Using Parallel axis theorem:
$$\begin{aligned}
& I^{\prime}=I+M R^2 \
& I=\frac{m L^3}{8 \pi^2}+L m\left(\frac{L}{2 \pi}ight)^2 \
& I=\frac{3 m L^3}{8 \pi^2}
\end{aligned}$$

Answer

$$\frac{3 \mathrm{~mL}^2}{5 \pi^2}$$

Answer

$$\frac{3 \mathrm{~mL}^3}{8 \pi^2}$$

Answer

$$\frac{3 \mathrm{~mL}^3}{4 \pi^2}$$

Answer

$$\frac{3 \mathrm{~mL}^2}{7 \pi^2}$$

Centre of Mass and Rotational Motion

A thin wire of length '$L' and uniform linear mass density 'm$' is bent into a circular coil. The moment of inertia of this coil about tangential axis and in plane of the coil is

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