Question

The period of oscillation of a 7.49-kg body is suspended at a point 0.22 m from its center of mass is 6.45 s. Find the moment of inertia about an axis through the pivot point of the body. Round off your final answer to two decimal places.

Answer 1

$$T=2\pi\sqrt{\dfrac{I}{mgd}},$$ $$T^2=4\pi^2\dfrac{I}{mgd},$$ $$I=\dfrac{mgdT^2}{4\pi^2},$$ $$I=\dfrac{7.49\ kg\times9.8\ \dfrac{m}{s^2}\times0.22\ m\times(6.45\ s)^2}{4\pi^2}=17\ kg\times m^2.$$