Question 190: Centre of Mass and Rotational Motion

Question

Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the center is :

Accepted Answer

$$ 	herefore $$ From figure,
L = $$\pi$$R
$$ \Rightarrow $$ R = $${L \over \pi }$$
Moment of inertia about center O,
I = MR
2
= M$${\left( {{L \over \pi }} ight)^2}$$ = $${{{M{L^2}} \over {{\pi ^2}}}}$$

Answer

$${1 \over 4}{{M{L^2}} \over {{\pi ^2}}}$$

Answer

$${1 \over 2}{{M{L^2}} \over {{\pi ^2}}}$$

Answer

$${2 \over 5}{{M{L^2}} \over {{\pi ^2}}}$$

Answer

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

Centre of Mass and Rotational Motion

Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the center is :

Choose the correct answer: