# Q. The moment of inertia of semi-circular plate of radius `R`and mass `M`about axis AB in its plane passing through centre:- A) `(MR^2)/2` B) `(MR^2)/4cos^2theta` C) `(MR^2)/2sin^2theta` D)...

Q. The moment of inertia of semi-circular plate of radius `R`and mass `M`about axis AB in its plane passing through centre:-

A) `(MR^2)/2`

B) `(MR^2)/4cos^2theta`

C) `(MR^2)/2sin^2theta`

D) `(MR^2)/4`

### 1 Answer | Add Yours

Suppose you have a full plane disc on mass 2M as in the figure below. The moment of inertia of the full disc, with respect to **any axis in the plane of figure passing through its center O** is

`I_("disc") = ((2M)*R^2)/4 = (M*R^2)/2`

We can write the moment of inertia of the full disc also as

`I_("disc") = int(r^2*(2dm)) =2*int(r^2*dm) =2*I_("halfdisk") =2*I_(AB)`

no matter how we take the half disk position, as long as `r` is perpendicular to the infinitesimal mass element `dm` . (This happens because of symmetry considerations). See the figure attached below.

Otherwise said, the moment of the inertia of the full disc is twice the moment of inertia of a half a disk, no matter how the half disk is cut from the entire disk.

Therefore, the moment of inertia of the half disc in the figure is just

`I_(AB) =I_("disc")/2 = (M*R^2)/4`

**The correct answer is D)** `(M*R^2)/4`