Question

A uniform thin rod of length 0.778 m is hung from a horizontal nail passing through a small hole in the rod located 0.057 m from the rod's end. When the rod is set swinging about the nail at small amplitude, what is the period of oscillation?

Answer 1

Solution-
Here we use the basic equation of motion

I d^2ϕ/dt^2 = τ

here ϕ = the angle
I =the moment of inertia
τ= the torque
Let us assume h be the distance between the top of the rod and hole (h<L/2).

The moment of inertia of a uniform rod about its end
= 1/3 M L^2.
Using the the Parallel Axes Theorem
We have that the moment of inertia for rotation about the hole s 1/3ML^2 - Mh^2.

The net torque from gravity is::

τ = -M(1-h/L)g (L-h)/2 sin(ϕ) + M h/L g h/2 sinϕ)

= -1/2 Mg/L * ( (L-h)^2 - h^2 ) sin(ϕ)

= -1/2 M g (L-2h) sin(ϕ)

Since angle is small approximation uses sin(ϕ) ~ ϕ.
Therefore the equation of motion becomes

(1/3ML^2 - Mh^2) ϕ“ + 1/2 Mg(L-2h) ϕ = 0

Therefore

ϕ“ + 3g(L-2h)/(2L^2-6h^2) ϕ = 0

it can be written as

ω^2 = 3g(L-2h)/(2L^2-6h^2)

Therefore the period

T = 2π/ω

= 2π √︎ ( (2L^2-6h^2)/(3g(L-2h)) )

Now just putting the value
2*3.14 sqrt ( (2* 0.778^2-6*0.057^2)/(3*9.8(0.778-2*0.057)) )
=1.55 s