A uniform thin straight rod of mass m = 2.31 kg and length L = 0.85 m can oscill

Question

A uniform thin straight rod of mass m = 2.31 kg and length L = 0.85 m can oscillate freely in a verticle plane about one of its end O, as a compound pendulum.

Using parallel axis theorem, calculate the moment of inertia of the rod I in kg m2, about an axis through the end O and perpendicular to the rod. Give your answer to 3 decimal places.Provide working and solution.

(Note : The moment of inertia of a uniform rod about a perpendicular axis passing through its centre of mass is given by I = (1/12)mL2.)

Explanation / Answer

moment of inertia of rod about its center of mass, Icm = m*L^2/12

moment of inertia of rod about one end(Using parallel axis theorm),

I = Icm + m*d^2 (d is the distance from center to end)

= m*L^2/12 + m*(L/2)^2

= m*L^2/12 + m*L^2/4

= (m*L^2 + 3*m*L^2)/12

= m*L^2/3

= 2.31*0.85^2/3

= 0.556 kg.m^2 <<<<<<<<<<<<------------Answer

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