Question

A physical pendulum consists of a body of mass contained in the xy-plane. The moment of inertia of the object about an axis perpendicular to the plane and passing through the object's center of mass is I_cm .

The object oscillates in the xy-plane about the point S a distance d from the center of mass as shown. What is the period of the pendulum for small angle oscillations where sin(theta )neraly =theta ?

Answer 1

I assume your mass is m and moment of inertia about the center of mass is Ic

Then the total moment of inertia about the point S is I = (Ic + m d^2)

If the xy plane is vertical and this is a physical pendulum hanging from point S then the gravity force down is mg at the center of mass. At angle theta, which I call Th, this has components mg cos theta along the rod and mg son Theta toward the center. The torque about S is therefore m g d sin Th trying to bring the pendulum back to center
Then m g d sin Th = -I alpha
where alpha is the angular acceleration
If Th = x sin 2 pi f t (harmonic motion)
then V =dTh/dt 2 pi f x cos 2 pi f t
and alpha = dv/dt = -(2 pi f)^2 sin 2 pi f t
which is -(2pif)^2 Th
then
m g d sin Th = I ( 2 pi f)^2 T
now for small angles sin Th = Th in radians so
m g d = I (2 pi f)^2
or
m g d = (Ic + m d^2)( 2 pi f)^2
solve for f, the frequency
then
period = 1/f

Answer 2

Thank you very much Damon