Question

In class, I have finished except for the last step in deriving a general expression for the angular frequency ω for a physical pendulum. Given that the moment of inertia is I , the mass of the physical pendulum m and the distance from the pivot point to the centre of mass of the physical pendulum to be dcm , the expression you obtained for ω is?

Answer 1

we have

mgLsin$\theta$ = I$\alpha$

for small angles

mgL$\theta$ = I$\alpha$

so,

$\alpha$ = mgL$\theta$ / I

so,

$\alpha$ = -$\omega$²$\theta$

Therefore,

$\omega$ = sqrt ( mgL/I)

so,

in terms of given variables,

$\omega$ = (mgd_cm / I)^1/2

Proper format is

$\omega = \sqrt{\frac{mgd_{cm}}{I}}$