JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 8)
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 :
$${1 \over 4}{{M{L^2}} \over {{\pi ^2}}}$$
$${1 \over 2}{{M{L^2}} \over {{\pi ^2}}}$$
$${2 \over 5}{{M{L^2}} \over {{\pi ^2}}}$$
$${{M{L^2}} \over {{\pi ^2}}}$$
Giải thích
_18th_March_Evening_Shift_en_8_2.png)
$$ \therefore $$ From figure,
L = $$\pi$$R
$$ \Rightarrow $$ R = $${L \over \pi }$$
Moment of inertia about center O,
I = MR2 = M$${\left( {{L \over \pi }} \right)^2}$$ = $${{{M{L^2}} \over {{\pi ^2}}}}$$