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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)##