Question

A uniform thin rod of length 0.984 m is hung from a horizontal nail passing through a small hole in the rod located 0.081

m from the rod's end. When the rod is set swinging about the nail at small amplitude, what is the period of oscillation?

period of oscillation:

Answer 1

Solution:

Given:

length of a rod is 0.984 m

the time period is, T = 2 sqrt [ I / mgD]

here, D = x

D = L / 2 - 0.081 = (0.984 / 2) - 0.081 = 0.411 m

And, the moment of inertia is : I = I_cm + MD²

I = (1/12) M L² + Mx²

Therefore : T = 2 sqrt [ { (1/12) ML² + Mx²)} / m g D ]

T = 2 sqrt [ { (1/12) L² + x² } / g D ]

T = 2 sqrt [ { (1/12)(0.984)² + (0.411)²} / (9.81 x 0.411) ]

T = 1.563 s