Oscillations physics question

Question

Consider a physical pendulum in which the distance from the axis of rotation to the center of mass is 22.5 cm. The formula for the pendulum's moment of inertia goes as k·mr² and its radius is 31.5 cm. If the angular frequency is 4.22 rad/s what is the value of k?

Answer 1

The equation of SHM of the given physical pendulum is:

(kmr²)θ'' = - mgd sinθ

For small oscillations, θ'' = - (gd/kr²) θ

Comparing with standard equation of an SHM, y'' = -ω²y, where ω is the angular frequency:

gd/kr²= ω² => k = gd/(ωr)²

Now plug in the values k = (9.8 ms^-2)x(.225m)/(0.315 m x 4.22 s^-1)²=1.25 [ three significant figures].