Question

In: Physics

A uniform thin rod of length 0.812 m is hung from a horizontal nail passing through...

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?

Solutions

Expert Solution

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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