Question

A connecting rod of mass 5.5 kg is placed on a horizontal platform whose mass is 1.5 kg. It is suspended by three equal wires, each 1.25 m long, from a rigid support. The wires are equally spaced round the circumference of a circle of 125 mm radius. When the c.g. of the connecting rod coincides with the axis of the circle, the platform makes 10 angular oscillations in 30 seconds. Determine the mass moment of inertia about an axis through its c.g.

(a) 0.198 kg-m^2

(b) 1.198 kg-m^2

(c) 2.198 kg-m^2

(d) 3.198 kg-m^2

This question was posed to me in an interview.

I want to ask this question from Closely-coiled Helical Spring and Compound Pendulum topic in section Simple Harmonic Motion of Machine Kinematics

Answer 1

The correct option is (a) 0.198 kg-m^2

Explanation: Given : m1 = 5.5 kg ; m2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125 m

Since the platform makes 10 angular oscillations in 30 s, therefore frequency of oscillation,

n = 10/30 = 1/3 Hz

Let kG = Radius of gyration about an axis through the c.g.

We know that frequency of oscillation (n)

1/3 = r/2πkG √g/l = 0.125/2πkG√9.81/1.25 = 0.056/kG

kG = 0.056 x 3 = 0.168 m

and mass moment of inertia about an axis through its c.g.,

I = mk^2G = (m1 + m2)k^2G

  = (5.5 + 1.5) (0.168)^2 kg-m^2

  = 0.198 kg-m^2.