JEE MAIN - Physics (2005 - No. 67)
The moment of inertia of a uniform semicircular disc of mass $$M$$ and radius $$r$$ about a line perpendicular to the plane of the disc through the center is
$${2 \over 5}M{r^2}$$
$${1 \over 4}Mr$$
$${1 \over 2}M{r^2}$$
$$M{r^2}$$
Explanation
Let mass of the semi circular disc = M
Now assume a disc which is combination of two semi circular parts. Let $$I$$ be the moment of inertia of the uniform semicircular disc. So $$2I$$ will be the moment of inertia of the full circular disc and 2M will be the mass.
$$ \Rightarrow 2I = {{2M{r^2}} \over 2}$$
$$ \Rightarrow I = {{M{r^2}} \over 2}$$
Now assume a disc which is combination of two semi circular parts. Let $$I$$ be the moment of inertia of the uniform semicircular disc. So $$2I$$ will be the moment of inertia of the full circular disc and 2M will be the mass.
$$ \Rightarrow 2I = {{2M{r^2}} \over 2}$$
$$ \Rightarrow I = {{M{r^2}} \over 2}$$
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