homework help 12763

##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)##