Trifilar Suspension

Introduction: The mass moment of inertia ( I ) is the measure of an object’s resistance to changes in rotation direction about a specific axis. The higher the object’s mass moment of inertia, the lower is its angular acceleration. The mass moment of inertia plays the role of mass in rotational dynamics How is Mass Moment of inertia calculated? For a point mass about a specific axis the calculation of mass moment of inertia is simple and easy. Its calculated by adding the masses together times the square of a distance perpendicular to the reference axis. I = m r2 I = moment of inertia M = mass R = distance between the axis and rotational mass.

AIM:  Derive the formula for the theoretical data to show the relation between the time period and mass moment of inertia. This is done by equating the equations of motion together.

 Comparing the values that we got from the experiment (Experimental Results) and the values that we calculate (theoretical) in this report.

 Have a better understanding of how the trifilar system works?

 Discussing and concluding why do experimental and theoretical values differ.

Theory: The trifilar suspension is hung by three chains from the roof. Once all the weights are kept on the circular wooden board it’s displaced sideways from its resting equilibrium position. When released, the force due to gravity will accelerate it back to its original position. However the force combined with free weight causes it to oscillate beyond the original resting position. Thus

it swings back and forth. The time for one complete swing (Left and right) is known as the time period. The value of the time period can be used to calculate the Mass moment of inertia of the objects on this suspension.

1. We begin with the energy Equation. KE + PE + Work Done = 0 Where KE = Rotation Kinetic Energy ½ Iω2 PE = Potential Energy Mgh

2. Double differential is performed on both the equation. KINETIC ENERGY ½ Iω2 ½ I(θ’)2 ½(2)I(θ’’)(θ’) POTENTIAL ENERGY Mgh Where, h = L - LcosΦ Φ = Rθ/L Φ’ =R/L

mgL(1-cosΦ) mgl(sinΦ)θ’(R/L) Small angle sinΦ = Φ mgL(Φ)θ’(R/L) mgL(Rθ/L)θ’(R/L) (θ’)mgR2θ/L 3. It is then put back in the original energy equation. ½(2)I(θ’’)(θ’) + (θ’)mgR2θ/L = 0 4. The whole equation is then divided by θ’ I(θ’’) + mgR2θ/L = 0 5. We have to now make sure that the time period can be integrated into the above equation.

Assumption of the general solution for θ = θ sin (ωt) Therefore θ’ = ωθ cos (ωt) θ’’ = -ω2θsint(ωt)

6. Substitute it into the equation I(θ’’) + mgR2θ/L = 0 I (-ω2 θ sint(ωt)) + mgR2/L(θ sin (ωt)) = 0

7. Divide the above equation by (θ sin (ωt) I (-ω2) + mgR2/L= 0 -I (ω2) = -mgR2/L ω2 =-mgR2/IL ω =(mgR2/IL)1/2 8. ω= 2πf is to be substituted in the above equation.

2πf=(mgR2/IL)1/2 9. f = 1/τ (τ = time period) is to be substituted in the above equation. 1/τ =*(mgR2/IL)1/2+/2πf

2 1/2

τ = 2π (IL/ mgR ) I = Mass Moment of Inertia

R = Distance from centre of the mass to the reference axis. L = Length of Chain from board to the roof.

TIME VERSUS MASS MOMENT OF INERTIA

Equipment:  Stop Watch – To measure the timings for the oscillations.  Weighing Machine – Weighing the three masses. RECTANGULAR, HOLLOW CYLINDER & SOLID CYLINDER.  Measuring tape – Calculating the lengths of chain, radius etc.  A trifilar suspension system – 3 chains, hung off a bracket on the roof, holding a circular, wooden board.

A.

B. Hollow

C. Solid

Rectangular

Cylinder

cylinder

Mass (kg)

2.238

2.35

8.865

Dimension

Thickness .006

Inner Diameter: 0.078

Diameter: 0.126

Outer Diameter: 0.098

(m)

Procedure:  Start off by measuring the chains of the suspension and the diameters of the weights and weighing the weights.  The three objects have to align in such a way that they are exactly in the middle of the marked positions on the circular platform.  The distances are to be measure once again. This insures that the objects are not off the center.  The circular platform is held and pulled towards one side. It is then set free.  The oscillations are then counted and timed using the provided stop watch.  10 oscillations are timed with the weight and then 10 oscillations without the weight. Both of them are done thrice for accuracy.  20 oscillations are done twice without the weight and then a third one is done for 21 oscillations.

Results: WITH WEIGHTS 10 oscillations - w/ weights 3 TRIALS Trial 1 2 3 AVERAGE

Time 28.2 28.28 28.55 Time28.34

Period 2.82 2.828 2.855 Period2.834

21 oscillations - w/ weights Trial Time Period 1 56.72 2.700

WITHOUT WEIGHTS 10 oscillations - w/o weights 3 TRIALS Trial Time Period 1 28 2.8 2 28.09 2.809 3 28 2.8 AVERAGE Time- Period28.03 2.803

20 oscillations – w/ weight Trial Time Period 1 56.09 2.804

Calculations: Theoretical Time: With Weights: √

√

WHERE L = 2.12 m Without Weights: √

√

Percentage Error - with weights

Percentage Error - without Mass