Question

plz give sol
ans is a ​

Answer 1

The moment of Inertia of a disc about it's centre of mass = mr^2/2

According to the perpendicular axis theorem,
Iz = Ix + Iy
Now a disc is symmetrical about x y plane
So,
Iz = 2Ix = 2Iy
Therefore,
Ix = Iy = 1/2*(mr^2/2) = mr^2/4
Now, according to the Parallel axis theorem ,

I = I(cm) + md^2
where d is the distance from the axis passing through centre of mass to the new axis where the moment of Inertia has to be calculated

So
I = mr^2/4 + mr^2 = 5mr^2/4

I have taken d = r as we want the axis passing through the tangent to the disc, that means the distance from centre to tangent should be r .

hope this helps you !

Answer 2

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