Introduction:Torsion is the twisting of a metallic rod shaped object, when a torque is applied on two sides’ perpendicular to the radius of a uniform cross-sectional bar.

Objective:Determining the natural frequency of a system undergoing tortional vibration.

Theory:Using Newton’s second law of tortional system.

( [pic] …………………. ( Equation 1 )

where Io = mass moment of inertia of the disk

Hence, [pic] ……..……... ( Equation 2 )

where k = torsional stiffness of the shaft

Rearrange Equation 2

( [pic] .………..……... ( Equation 3 )

where natural frequency of the system,

[pic] …..…….…..……... ( Equation 4 )

From Simple Theory of Torsion, [pic]

where T = Applied torqueJ = Polar second moment of area

[pic] = Shear stressR = Radius of shaft

G = Shear modulus[pic] = Angle of twist

L = Length of shaft

As torsional stiffness [pic], it can be determined through [pic] ………….. ( Equation 5 )

Apparatus:

• One solid circular disk with mass = 4.536kg, diameter = 150mm and thickness = 30mm.

• One annular circular disk with mass 1.89kg, outer diameter 150mm, inner diameter = 110mm and thickness = 30mm.

• Two chucks; one steel rod; one stopwatch.

Procedure:

1. The diameter of the provided rod is measured at three different locations to get the average diameter of the rod.

2. The anchor is chucked tightly to the solid circular disk.

3. The length of the rod or the distance between the two chucks is initially kept 30cm.

4. The disk is displaced slightly, so that the rod can be twisted.

5. The disk is released and the stopwatch is switched on simultaneously.

6. The time taken is recorded according 10, 20, 30, 40 and 50 cycles of the disk.

7. From step 3 to step 6 is repeated by increasing the length between the two chucks from 35 cm to 40 cm.

8. The whole procedure is repeated by attaching the annular circular disk on top of the solid disk.

Results:

| |Time, (s) | |Length |0.3m |0.4m |0.5m |0.3m(With Annular |0.4m(With Annular |0.5m(With Annular | | | | | |disk) |disk) |disk) | |Number of cycles | | | | | | | |10 |4.74 |4.89 |5.08 |4.53 |5.55 |7.24 | |20 |8.93 |10.08 |10.55 |11.25 |11.62 |13.33 | |30 |13.27 |14.64 |15.7 |16.77 |17.77 |19.55 | |40 |18.33 |19.67 |21.14 |22.08 |24.24 |25.4 | |50 |23.05 |25.73 |26.45 |27.49 |30.37 |31.71 |

Given Information:

Mass of the circular disk = 4.536kg.

Diameter of the circular disk = 150mm.

Thickness of the circular disk = 20mm.

Mass of the annular disk = 1.86kg.

Outer diameter of the annular disk = 150mm.

Inner diameter of the annular disk = 110mm.

Thickness of the annular disk = 30mm.

G ( Shear module ) = 80GPa.

Sample calculation:

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