UNIVERSITY OF STRATHCLYDE DEPARTMENT OF MECHANICAL ENGINEERING Engineering Mechanics 2 (Dynamics) 16232 PRIMARY OBJECTIVE The Trifilar Suspension The trifilar suspension is used to determine the moments of inertia of objects.

This apparatus is employed for objects of irregular shape, which make calculation methods impractical, or when CAD modelling packages are not available. The primary objective is to measure the moment of inertia of an IC engine connecting rod about an axis passing through its centre of gravity and normal to its in-service plane of motion. This is the quantity required by designers for analysing dynamic force actions and thus enabling bearing sizes etc to be determined. Two methods are proposed: 1. By making a compound pendulum of the connecting rod by suspending it in a suitable way and then measuring the period of the pendulum motion 2. The trifilar suspension, the validity of which should first be checked by measuring the moment of inertia of an object of simple shape for which the moment of inertia is already known. A "thin?" rod is provided for this purpose. 1 EXPERIMENTAL WORK 1. 2. Pendulum Using a stop-watch, measure the time taken for a large number of small oscillations of the connecting rod, suspended from point A and from point B (see diagram in Theory). The connecting rod mass and the length dimensions AG and BG are given (see data sheet in lab) where G is the centre of mass. The theory section explains how the period and frequency of the pendulum motion are related to the moment of inertia of the connecting rod about G. Note the use of the parallel axis theorem. Trifilar Suspension Calculate the theoretical values of the moments of inertia of the table and the rectangular bar using the formulas given in the CALCULATIONS section of this hand-out. Using a stop-watch, determine the period of (i) the table alone, (ii) the table plus rectangular bar with the table and bar centres of gravity coincident and (iii) table plus connecting rod with their centres of gravity coincident In each case, find the period by measuring the time taken for a large number of small oscillations. Use these values to determine the moments of inertia of the table, rectangular bar and con-rod, by making use of the expression for the period given in the theory and compare with theoretical values were applicable. 2 THEORY 1. The Trifilar Suspension 0 Mg 3 Displacement AA' re=lo r •· p=²0 Eq.1. 9: If weight is equally distributed, forces at A' are as shown in the following page: The force actions tending to restore the table to its equilibrium position are applied by the tension in the wires and table weight. Restoring Force at Point A' Vector sum of applied forces T resultant restoring force Tension in wire T Mg 3 Forces applied at attachment point A' Ф A' .. period T₁ = 2π @ n Igl = 2π₂ √ Mgr² From vector sum of forces, resultant restoring force Mg 3 (if o is small) tan o 4 Mg 3 = seconds Total restoring torque for 3 wires Mgr² = Mgor= 1 -P Now from Newton's 2nd Law, Σ torques = 10 -0 (from Eq.1.) ..— — (restoring torque) = 1,Ö Mgr.² 1 Mg 3 i.e. 2nd order Differential Equation describing Simple Harmonic Motion (SHM) of natural circular frequency on rad/s or Hz (see lectures on 2π this) :.Ö + Mgr² Ial or 0 + ²0 = 0 -0 = IGÖ W₁ r= = radius from G to attachment point of a wire. g = acceleration of gravity = 9.81 m/s². 0=0 where T₁ = the period which is the time, in seconds, for one oscillation. IG = moment of inertia about G of table plus object (if any). M = mass of table plus object (if any). 7 = length of trifilar wires. = 2. Compound Pendulum +ve B where G Mg 0 G' .. period T = 2π @n i.e. the same 2nd order Differential Equation describing SHM of natural n circular frequency rad/s or @ Hz. on 2π = M = mass of connecting rod. (AG) = Length A to G. g= acceleration of gravity. Force tending to restore con-rod to equilibrium position = Mg I₁ = Ig+M(AG)² Moment arm of restoring force = G'G=(AG) sin 0 = (AG)0 (assuming small 0) = 2π, Restoring torque about A = Mg (AG) 0 From Newton's 2nd Law, Now, from the Parallel Axis Theorem: 5 Σ torques = 16 :.- Mg(AG)0 = I ¸Ö ·Ö + 0 T₁= the period which is the time, in seconds, for one oscillation. IA = moment of inertia of connecting rod about A. Mg(AG)0 IA or + ²0 = 0 IA Mg(AG) seconds :.IĠ = I₁ - M(AG)² G