PH0101 UNIT 1 LECTURE 2

PH0101 UNIT 1 LECTURE 2 1

PH0101 UNIT 1 LECTURE 2

• Shafts • Torsion Pendulum-Theory and Uses• Worked Problem


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.


• 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

Requirements for a good shaft


• 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.


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 (r2

2 – r12)

r2 = r22 – r1

2


In the case of a solid cylinder, twisting couple per unit twist,

lrrN

lrrN

C2

)r - (r)(2

)( 21

22

21

22

41

421

lrrN

lrNC

2)(

2

221

22

4

For a hollow cylinder, twisting couple per unit twist

Dividing C1 by C

21

22

21

22

1

rrrr

CC


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.


Disc

Torsionally flexible elastic wire

Fixed End

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.


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,


2

2

dtdI

C θ =

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.

This couple acting on the disc produces in it an angular acceleration given by equation.


IC

dtd

2

2

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


T = 2π

or T = 2 π

on AcceleratintDisplaceme

IC

2

I/C

The time period of the vibration will be given by


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.


• 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

CI

CII 1

CII 2

CI1

24

CI 2

24

we haveT1

2 – T2 =

and T2

2 – T2 =


2

1

22

12

222

221

/4/4

II

CICI

TTTT

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.

221

222

12 TTTT

II


(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.


T = 2 π

and T1= 2 π

where I is the moment of inertia of the disc and I1 the moment of inertia of the regular body placed on the disc.

C I

C

I II

If T is the time period of the pendulum in first case and T1 in the second case, then we have


T12 – T2 =

or

For a wire of modulus of rigidity N, length l and radius r, we have

CI1

24

221

124TTI

C

lNrC2

4


lNr

TTI

24 4

221

12

4221

1

)(8

rTTlI

N

Thus, the value of N can be determined.


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.

42

8rTIN


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


40.59 2 380 10 2 102 4NT r 2

48 I 8 0.14382 10

5.531ml

SRM


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PH0101 UNIT 1 LECTURE 2