Question

A uniform thin rod of length 0.812 m is hung from a horizontal nail passing through a small hole in the rod located 0.043 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

The basic equation of motion is

I d^2ϕ/dt^2 = τ

where ϕ is the angle, I the moment of inertia and τ the torque. Let h denote the distance between the hole and the top of the rod (h<L/2).

The moment of inertia of a uniform rod about its end is 1/3 M L^2. Invoking the Parallel Axes Theorem therefore gives us 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)

As usual the small angle approximation uses sin(ϕ) ~ ϕ. The equation of motion therefore becomes

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

which can be written as

ϕ“ + 3g(L-2h)/(2L^2-6h^2) ϕ = 0

From this we read off

ω^2 = 3g(L-2h)/(2L^2-6h^2)

and so a period

T = 2π/ω

= 2π √︎ ( (2L^2-6h^2)/(3g(L-2h)) )
= 2*3.14*SQRT((2*.812^2-6*0.043^2)/3g(0.812-2*.043)))
Here it calculates to

T = 1.55s

Hope this helps you.

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