A uniform meter stick is suspended through a small pin hole at the 10 cm mark. Find the time period of small oscillation about the point of suspension.

Explanation

Let the mass of the stick be m. The moent of inertia of the stick about the axis of rotation through the point of suspension is
`I=(ml^2)/12+md^2, `
where l=1 m and d=40cm

The seperation between the centre of mass of the stick and the point of suspension is d=40 cm. The time period of this physical pendulum is
`T=2pi sqrt(I/(mgd))`
`=2pi sqrt(((ml^2)/12+md^2)/(mgd))`
`=2pi[sqrt(1/12+0.16)/4])s=1.55s`