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`

Asked on by user8235304

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valentin68 | College Teacher | (Level 3) Associate Educator

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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`

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