Question 491: Centre of Mass and Rotational Motion

Question

A thin metal wire of length ' L ' and mass ' M ' is bent to form semicircular ring as shown. The moment of inertia about $\mathrm{XX}^{\prime}$ is

Accepted Answer

Both the semi-circular arc of a ring and the complete ring have the same moment of inertia because the mass in both cases is distributed at the same radial distance R from the central axis. Thus, the moment of inertia is determined solely by the distance R and the total mass M , not by the shape of the distribution along the ring.
M.I. of ring about diameter $\mathrm{I}=\frac{\mathrm{MR}^2}{2} \ldots$. (i)
$$ \because \quad \mathrm{L}=\pi \mathrm{R} \Rightarrow \mathrm{R}=\mathrm{L} / \pi $$
$	herefore \quad$ From equation (i), $\mathrm{I}=\frac{\mathrm{ML}^2}{2 \pi^2}$

Answer

$\frac{\mathrm{ML}^2}{4 \pi^2}$

Answer

$\frac{2 M L^2}{\pi^2}$

Answer

$\frac{\mathrm{ML}^2}{2 \pi^2}$

Answer

$\frac{\mathrm{ML}^2}{\pi^2}$

Centre of Mass and Rotational Motion

A thin metal wire of length ' L ' and mass ' M ' is bent to form semicircular ring as shown. The moment of inertia about \mathrm{XX}^{\prime} is

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