Moment of Inertia of a Circular Loop around a Horizontal Axis

The moment of inertia of a circular loop of radius R around a horizontal axis passing through its center is given by:

I = MR2

where M is the mass of the loop.

Moment of Inertia of a Circular Loop around a Parallel Axis

The moment of inertia of a circular loop of radius R around a parallel axis passing through a distance of R/2 from its center is given by:

I = 3/4 MR2

Explanation

Consider a circular loop of radius R and mass M. Let O be the center of the loop and AB be a horizontal diameter passing through O. Let P be a point on the loop such that OP = R/2. Let G be the center of mass of the loop. The distance OG is given by:

OG = AB/4

= R/2

The moment of inertia of the loop around the axis passing through P perpendicular to AB is given by:

I' = MR2 + MP2

where MP is the perpendicular distance between the axis and the line joining P and G.

By Pythagoras theorem,

MP2 = OP2 - OG2

= R2/4 - R2/4

= 0

Therefore,

I' = MR2

Now, using the parallel axis theorem, the moment of inertia of the loop around the horizontal axis passing through AB is given by:

I = I' + MD2

where D is the distance between the two parallel axes.

D = R/2

Therefore,

I = MR2 + (M(R/2)2)

= MR2 + 1/4 MR2

= 5/4 MR2

Using the perpendicular axis theorem, the moment of inertia of the loop around an axis perpendicular to its plane passing through its center is given by:

I'' = 1/2 MR2

Therefore, the moment of inertia of the loop around a parallel axis passing through a distance of R/2 from its center is given by:

I = I' + MD2

= MR2 + 1/4 MR2

= 5/4 MR2

= 1/2 I'' + MD2

= 1/2 (1/2 MR2) + (R/2)2 M

= 1/4 MR2 + 1/4 MR2

= 1/2 MR2

Therefore, the correct option is D, which gives the moment of inertia of a circular loop around a parallel axis passing through a distance of R/2 from its center as 3/4 MR2.