PH0101 UNIT 1 LECTURE 2

1. PH0101 UNIT 1 LECTURE 2 • Shafts • Torsion Pendulum-Theory and Uses • Worked Problem PH0101 UNIT 1 LECTURE 2

2. Shafts • Any rotating member which is transmitting torque is called shaft. • It is an arrangement for the transmission of a couple applied at one end to appear at the other end without any appreciable twist in it. • For example, in automobile such as buses, lorries and vans, the driving shaft (axle) transmits the power torque form engine to the wheel. PH0101 UNIT 1 LECTURE 2

3. Requirements for a good shaft • Shaft must transmit the couple without any appreciable twist in it • The twist in the shaft should be very small when the large couple is applied to it. • It is found that the efficiency of a shaft varies as πNr4 2l where N - rigidity modulus of the material r - radius of shaft l - length of shaft PH0101 UNIT 1 LECTURE 2

4. Hence, it is better to use shafts of large radius and the material is of high rigidity modulus. • Hollow shaft is stronger and better than the solid shaft of the same length, mass and material. PH0101 UNIT 1 LECTURE 2

5. Note:Hollow shaft is better than solid shaft • If two cylinders of the same length and same mass and of the same material are made such that one cylinder is solid of radius r and the other cylinder is hollow of inner radius r1 and outer radius r2 then, • Mass of solid cylinder = Mass of hollow cylinder πr2 = π (r22 – r12) r2 = r22 – r12 PH0101 UNIT 1 LECTURE 2

6. In the case of a solid cylinder, twisting couple per unit twist, For a hollow cylinder, twisting couple per unit twist Dividing C1 by C PH0101 UNIT 1 LECTURE 2

7. i.e., C1 > C • Thus twisting couple for a hollow cylinder is greater than that for a solid cylinder of the same mass, length and material. • Due to this reason, hollow shaft is better than solid shaft. • Driving shafts (axles) in automobiles are of hollow tubes and not of solid rods. PH0101 UNIT 1 LECTURE 2

8. Fixed End Torsionally flexible elastic wire Disc Torsion Pendulum A torsion pendulum is an oscillator for which the restoring force is torsion • The device consists of a disc or other body of large moment of inertia mounted on one end of a torsionally flexible elastic rod wire whose other end is held fixed; • If the disc is twisted and released, it will undergo simple harmonic motion, provided the torque in the rod is proportional to the angle of twist. PH0101 UNIT 1 LECTURE 2

9. Theory • When the disc is rotated in a horizontal plane so as to twist the wire, the various elements of the wire undergo shearing strains. • Restoring couples, which tend to restore the unstrained conditions, are called into action. • Now when the disc is released, it starts executing torsional vibrations. • If the angle of twist at the lower end of the wire is θ, then the restoring couple is C θ, where C is the torsional rigidity of the wire, PH0101 UNIT 1 LECTURE 2

10. This couple acting on the disc produces in it an angular acceleration given by equation. where I is the moment of inertia of the disc about the axis of the wire. The minus sign indicates that the couple C θ tends to decrease the twist. C θ = PH0101 UNIT 1 LECTURE 2

11. The above relation shows that the angular acceleration is proportional to the angular displacement θ and is always directed towards the mean position. Hence the motion of the disc is simple harmonic motion PH0101 UNIT 1 LECTURE 2

12. The time period of the vibration will be given by T = 2π or T = 2 π PH0101 UNIT 1 LECTURE 2

13. Uses of Torsion Pendulum (1)For determining the moment of inertia of an irregular body • First, the time period of pendulum is determined when it is empty and then the time period of the pendulum is determined after placing a regular body on the disc and after this the time period is determined by replacing the regular body by the irregular body whose moment of inertia is to be determined. • It is ensured that the body is placed on the disc such that the axes of the wire pass through the centre of gravity of the body placed on the disc. PH0101 UNIT 1 LECTURE 2

14. If I, I1 and I2 are the moments of inertia of the disc, regular body and irregular body and T, T1 and T2 are the time periods in the three cases respectively, then • T = 2 π • T1 = 2 π • T2 = 2 π we have T12 – T2 = and T22 – T2 = PH0101 UNIT 1 LECTURE 2

15. or The moment of inertia of the regular body I1 is determined with the help of the dimensions of the body, thus the moment of inertia of the irregular body is calculated. PH0101 UNIT 1 LECTURE 2

16. (2) Determination of Torsional Rigidity For determining the modulus of rigidity N, the time period of the pendulum is found (i) when the disc is empty, and (ii) when a regular body is placed on the disc with axis of wire passing through the centre of gravity of the body. PH0101 UNIT 1 LECTURE 2

17. If T is the time period of the pendulum in first case and T1 in the second case, then we have T = 2 π and T1= 2 π where Iis the moment of inertia of the disc and I1 the moment of inertia of the regular body placed on the disc. PH0101 UNIT 1 LECTURE 2

18. T12 – T2 = or For a wire of modulus of rigidity N, length l and radius r, we have PH0101 UNIT 1 LECTURE 2

19. Thus, the value of N can be determined. PH0101 UNIT 1 LECTURE 2

20. Worked Problem A torsion pendulum is made using a steel wire of diameter 0.5mm and sphere of diameter 3cm. The rigidity modulus of steel is 80 GPa and density of the material of the sphere is 11300 kg/m3. If the period of oscillation is 2 second, find the length of the wire. PH0101 UNIT 1 LECTURE 2

21. For sphere, I = 2/5 MR2 M = volume × density M = 4/3π (3/2 × 10-2)3× 11300 = 0.1598 kg I = 2/5 × 0.1598× (3/2 × 10-2 )2 = 0.14382 × 10-4 kgm2 PH0101 UNIT 1 LECTURE 2

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23. THANK YOU THANK YOU PH0101 UNIT 1 LECTURE 2