Question

A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre of gravity. If the rod makes 20 oscillations in 40 seconds, find the mass moment of inertia of the rod about a vertical axis through the centre of gravity.

(a) 0.014 kg-m^2

(b) 0.015 kg-m^2

(c) 0.016 kg-m^2

(d) 0.017 kg-m^2

I have been asked this question in homework.

This key question is from Closely-coiled Helical Spring and Compound Pendulum in chapter Simple Harmonic Motion of Machine Kinematics

Answer 1

Right option is (d) 0.017 kg-m^2

For explanation I would say: Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m

Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,

n = 20/40 = 0.5 Hz

Let kG = Radius of gyration of the connecting rod.

We know that frequency of oscillation (n),

0.5 = 1/2πkG √gxy/l = 1/2πkG √9.81 x 0.12 x 0.12/1.25 = 0.0535/k

kG = 0.0535/0.5 = 0.107 m = 107 mm

We know that mass moment of inertia,

I = mk^2G = 1.5(0.107)^2 = 0.017 kg-m^2.