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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.

Correct Answer: 0.017 kg-m2
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 = mk2G = 1.5(0.107)2 = 0.017 kg-m2.

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 radius of gyration of the rod about a vertical axis through the centre of gravity.

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