How can i calculate the radius of gyration K of a compound pendulum?

##k=sqrt(I/M)=(gh)/omega^2=sqrt(I/M-h^2)=sqrt((T^2gh)/(4pi^2)-h^2)##
For an object of mass M and I about an arbitrary axis the radius of gyration k is that distance from the same axis of rotation where a mass M needs to be placed to yield the same moment of inertia as the object.
ie ##I=Mk^2 => k=sqrt(I/m)##
The period of a compound pendulum is given by
##T=(2pi)/omega=2pisqrt(I/(Mgh)##
where h is the distance from the axis of rotation to the centre of mass and I is the moment of inertia about the centre of mass.
Thus the angular velocity of the compound pendulum may be given by
##omega=sqrt((Mgh)/I)=sqrt((Mgh)/(Mk^2))=sqrt((gh)/k^2)##
We may now apply the parallel axis theorem for moments of inertia to write
##I=I_(com)+ML^2##
##=Mk^2+Mh^2##
##=M(k^2+h^2)##
##thereforeT=2pisqrt((M(k^2+h^2))/(Mgh))=2pisqrt((k^2+h^2)/(gh))##
You may now solve for k in any of these above equations depending on what information is given to find the radius of gyration.
##k=sqrt(I/M)=(gh)/omega^2=sqrt(I/M-h^2)=sqrt((T^2gh)/(4pi^2)-h^2)##