Torsional Pendulum

OBJECTIVE

INTRODUCTIONWhen a torque is applied to an object, torsion is produced. Torsion is the strain experienced by a length of wire or rod when a twisting force is applied to one end while the other end is fixed. It can be measured by observing how much an object twists due to a given torque. Engineering materials employed in rotating machine parts, such as engine crankshafts and ship-propeller shafts, must resist the torsional stresses induced by the twisting loads.

OBJECTIVE

This experiments objective is to study the behavior of a torsional pendulum.

EQUIPMENT

A collet which can grip shafts of various diameters firmly is attached to a wall-bracket.The vertical shafts supports a large body of revolution by a similar collet at its lower end.This body can be made to rotate tortionally.

THEORY

A fine wire whose shear modulus is to be determined is suspended from a fixed support A and the other end B is fixed to the center of a heavy disc (collet) whose moment of inertia is known. The disc is allowed to oscillate with small amplitude(, from the equilibrium position A. The torsion of the wire will try to restore the disc to the position A, but due to inertia it will continue to move beyond A to C till it swings back and in the process executes simple harmonic motion. The equation of motion is given by:

I d2( + G( = 0

dt2Therefore the time period T which can be measured by means of a stopwatch is given by :

T = 2(( I/G

Where, I = moment of inertia of disc I = Mr2Where, M is mass of the disc and

r is the radius of disc.

Also, G = shear modulus.

G = n(a4 (a4 = J (second moment of area of the shaft)

2l 2

Here, a is radius of shaft and

l is the length of shaft.

Support

Shaft Length (L)

d

Cylindrical body

PROCEDURE

Choose a shaft and length L and determine T by timing a number of vibrations. Repeat several times and average your results. Move a shaft through the collet and repeat for a range of lengths. Repeat for shafts of the same material and different diameters. Determine the variation of T with d. Repeat for shafts of other materials. Determine G, and compare with published values.

RESULTS

Material: Mild steel

Diameter of shaft: 6.3mm

No of oscillation: 20 m = 0.0053 s2/cm

L/cmt1/st2/sAve.time/sT/sT2/s2

55.0010.8011.0010.900.5450.297

60.0010.7011.0010.850.5430.295

65.00 12.0011.8011.900.5950.354

70.0012.2012.4012.300.6150.378

75.0012.5012.6012.550.6280.394

Tab.1




L/cmt1/st2/sAve.time/sT/ST2/S2

55.0015.4015.6015.500.7750.601

60.0016.0016.2016.100.8050.648

65.0016.8016.6016.700.8350.697

70.0017.4017.4017.400.8700.757

75.0018.2018.2018.200.9100.828

Tab.2





55.0037.8037.9037.851.8933.583

60.0039.6039.2039.401.9703.881

65.0041.2040.6040.902.0454.182

70.0042.4042.2042.302.1154.473

75.0043.8043.8043.802.1904.796

Tab.3

Material: Copper


No of oscillations: 20 m = 0.0097 s2/cm


55.0014.8014.8014.800.7400.548

60.0015.2015.4015.300.7650.585

65.0015.8015.8015.800.7900.624

70.0016.4016.6016.500.8250.681

75.0016.6017.0016.800.8400.706

Tab.4

Material: copper




55.0022.8022.8022.801.1401.300

60.0023.8024.0023.901.1951.428

65.0024.8024.8024.801.2401.538

70.0025.8025.8025.801.2901.664

75.0026.2026.4026.301.3151.729

Tab.5

Material: Copper



L/cmt1/st2/sAve.time/sT/S T2/S2

55.0065.0064.8064.903.24510.530

60.0067.6067.8067.703.38511.458

65.0070.2069.8070.003.50012.250

70.0072.8072.8072.803.64013.250

75.0074.4074.8074.603.73013.913

Tab.6

DETERMINATION OF G,

From theory,

Therefore from graph,

Slope (m) =

( G = 4(2I ....... eqn * Where, I = Mr2, J = (d4 and m = slope

mJ 32

G for the various diameters of mild steel shaft:

6.3mm diameter,

I = 0.1387029

J = 1.546(10-10

m = 0.53, substituting into eqn * gives: G = 66.9 GPa5.2mm diameter I = 0.1387029

J = 7.179(10-11m = 1.09, substituting into eqn * gives: G = 70.0 GPa3.2mm diameter I = 0.1387029

J = 1.0296(10-11m = 6.43 substituting into eqn * gives: G = 82.8 GPa

Average of the Gs of mild steel, G = 73.2 GPa

Standard value of G for mild steel, G = 75.0 GPa( Error = 75.0 73.2 ( 100(

75.0

= 2.4(G for the various diameters of copper shaft,

6.3mm diameter, I = 0.1387029

J = 1.546(10-10m = 0.97 substituting into eqn * gives: G = 36.5 GPa5.1mm diameter,

I = 0.1387029

J = 7.17909(10-11m = 2.35 substituting into eqn * gives: G = 32.5 GPa3.0mm diameter,

I = 0.1387029

J = 7.953(10-12m = 18.87 substituting into eqn * gives: G = 36.5 GPaAverage of the Gs of copper shaft, G = 35.2 GPaStandard value of G for copper, G = 40.0 GPa( Error = 40.0 35.2 ( 100 ( 40.0

= 12(VARIATION OF (T) WITH (d),

The period T is inversely proportional to the diameter d for both mild steel and copper. As d increases, T decreases and as d decreases T increases.EXERCISE: Show from analysis that the period of the torsional pendulum is

T=

Where, I = moment of inertia and

J = Second moment of area.

G = Shearing modulus

DEDUCTION A torsional pendulum consists of a rigid body supported on a shaft as shown below.

When the pendulum is twisted through a small angle ( and released, a simple harmonic angular motion results from the stresses developed in the shaft. Assume that a vibratory motion of body B in diagram above is started by twisting the assembly through the small angle(, as shown, and then releasing the pendulum. If the limit of proportionality of the material in a solid circular shaft is not exceeded, the moment necessary to twist the shaft is proportional to the angle of twist and can be obtained from:

T = JG( L

=K(

(Second moment of area)G = Shearing modulus

( = Angle of twist in radians

L = Length of shaft

The moment exerted by the shaft on the body B is equal in magnitude and opposite in sense to the moment,T,required to twist the shaft.The moment exerted by the shaft on B has a sense opposite to the sense of the angular displacement of the lower end of the shaft.

The equation of motion for body B, in scalar form, is:

(M= I( or T=-K(

= I(The angular acceleration is, =

This equation proves that body B has simple angular motion, and the natural circular frequency of the motion is:

The frequency is given by:

Hence Period,

Where,

Therefore,

CONCLUSION

From the results obtained, the period T is dependent on the length L and diameter d of the shaft used. As length L of shaft is increased the period T also increases, whiles as the diameter d of the shaft is increased the period T decreases. As diameter d decreases the period T increases.

Also from the experimental values obtained, the square of the period T can be plotted against length L of the shaft, to give a straight line. From this line the shear modulus G of the shaft material can be determined from its slope. Experimental values not equal to theoretical values due to the following: I. The shaft not perfectly straight. II. Angle of twist ( of the shaft not uniform throughout the experiment.III. Unbalance forces due to air resistance.IV. Loss of some of the shafts elasticity due to the years it has being used.

For the inaccuracies to be corrected, all the assumptions behind the theory must be obeyed and followed when performing the experiment to achieve very accurate results.

The assumptions behind the theory are:

Deformation the angle of twist ( must be small.

Geometry the shaft must be straight.

Material behaviour the shaft should not be stressed beyond its elastic limit.

Equilibrium - all external loading should be such that the internal force resulting at any section along the axis of the bar is a torque or couple parallel to the axis of the bar.

APPLICATIONS OF EXPERIMENT

- It can be used to determine material behaviour (ie; strength, stiffness, etc).

- It is used to determine material properties.

- It is used to solve statically indeterminate torsional problems.

- It can be used to determine the moment of inertia of a body.

PRECAUTIONS

1) We made sure the angular displacement ( of the material was small.

2) We made sure the time taken for 20 oscillations was done twice and the average of taken.

3) We made sure that the shafts did not swing as it was oscillating.

REFERENCES

1) Engineering Mechanics by Archie Higdon.

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Torsional Pendulum

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