State an expression for the moment of intertia of a solid uniform disc, rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration:

i. about a tangent in the plane of the disc, and

ii. about a tangent perpendicular to the plane of the disc.

#### Solution

The M. I of a thin uniform disc about an axis passing through its centre and perpendicular to its plane is given by,

I_{C} = (1/2)MR^{2}

i. According to theorem of parallel axis,

I_{T} = I_{d} + Mh^{2} = I_{d} + MR^{2} [∵ h = R]

But I_{d} = MR^{2}/4

`therefore I_T=(MR^2)/4+MR^2`

`therefore I_T=5/4 MR^2`

Now , radius of gyration is given by,

`K=sqrt(I/M)`

`therefore K=sqrt((5MR^2)/(4M))`

`thereforeK=sqrt5/2R`

ii. Applying theorem of parallel axis,

I_{T} = I_{O} + Mh^{2} = I_{O} + MR^{2} [∵ h = R]

But I_{O} = MR^{2}/2

`I_T=(MR^2)/2+MR^2=3/2MR^2`

Now, radius of gyration is given by,

`K=sqrt(I/M)`

`thereforeK=sqrt((3MR^2)/(2M))`

`thereforeK=sqrt(3/2)R`