## Aim:

To determine:

- The acceleration
*g* of gravity** **using a compound pendulum.

- The radius of gyration k
_{G }of the compound pendulum about an axis perpendicular to the plane of oscillation and passing through its centre of mass.

- The moment of inertia I
_{G }of the compound pendulum about an axis perpendicular to the plane of oscillation and passing through its centre of mass.

## Theory:

In Fig. 1,** **O is the point of suspension of the compound pendulum and G is its centre of mass; we consider the force of gravity to be acting at G. If *h* is the distance from O to G, the equation of motion of the compound pendulum is

Where *I*_{0}is the moment of inertia of the compound pendulum about the point O.

Comparing to the equation of motion for a simple pendulum

We see that the two equations of motion are the same if we take

It is convenient to define the radius of gyration *k*_{0 }of the compound pendulum such that if all the mass *M* were at a distance *k*_{0} from O,

the moment of inertia about O would be *I*_{0}, which we do by writing *I*_{0 }=_{ }M*k*_{0}^{2}

Substituting this into (1) gives us

The point O′, a distance *l* from O along a line through G, is called the center of oscillation. Let *h*′ be the distance from G to O′, so that *l=h+h'. *Substituting this into (2), we have

If *I*_{G} is the moment of inertia of the compound pendulum about its centre of mass, we can also define the radius of gyration *k*_{G} about the centre of mass by writing *I*_{G}_{ }=_{ }M*k*_{G}^{2}.

The parallel axis theorem gives us

Comparing to (3), we have,

If we switch *h* with *h*′, equation (4) doesn’t change, so we could have derived it by suspending the pendulum from O′. In that case, the center of oscillation would be at O and the equivalent simple pendulum would have the same length *l*. Therefore the period would be the same as when suspended from O. Thus if we know the location of G, by measuring the period *T* with suspension at O and at various points along the extended line from O to G, we can find O′ and thus *h*′.

Then using equation (4), we can calculate *k*_{G }and *I*_{G}_{ }=_{ }M*k*_{G}^{2}.

Knowing *h*′ gives us *l* = *h* + *h*′, and since for small angle oscillations the period

We can calculate *g* using

The minimum period *T*_{min}** **, corresponds to the minimum value of *l*. Recall that *l* = *h* + *h*′ and that *k*_{G}^{2 }_{= }hh' is a constant, depending only on the physical characteristics of the pendulum.

Thus, *l=h+k*_{G}^{2}/h, and the minimum *I* occurs when,

i.e, when *h*^{2}=k_{G}^{2},* h=h'* and *l=2h=2k*_{G.}

Thus, at *T*_{min}, *l=2k*_{G}.