dynamics of mahinery lab manual_cycle 2

Exp. No: 07

Date :

Balancing of rotating masses Aim:

To study the static and dynamic balancing of the given setup.

Procedure:

1. Fix the known weigts to slot 1 & 4.

2. Measure the radius r1 & r4.

3. Draw the couple poly gon and force polygon.

4. Find out unknown radius r2 &r3.

5. Fix the remaining known weights to the slot 2 & 3 at the radius equal to r2 &

r3.

6. Check whether the system is in balancing condition.

Observations:

M1 = ___ kg r1 = ______ m

M2 = ___ kg r2 = ______ m

M3 = ___ kg r3 = ______ m

M4 = ___ kg r4 = ______ m

Table:

Angular Position diagram:

Force & Couple polygon :

Result:

Inference:

Signature:

Date :

Exp. No: 08

Date :

MOTORIZED GYROSCOPE

Aim:

To study the gyroscopic effect and verify the relationship between the applied torque and

direction of rotation of precession.

Apparatus: Tachometer, Stopwatch and weights.

Theory: Consider a disc spinning with an angular velocity rad/s about the axis of spin OX,

in anti clockwise direction when seen from the front as shown in fig. Since the plane in which

the disc is rotating in parallel to the plane YOZ, it is called plane of spinning. The plane XOZ is

a horizontal plane and the axis of spin rotates in a plane parallel to the horizontal plane about an

axis OY (which is perpendicular to both the axes OX and OZ) at an angular velocity p rad/s.

This horizontal plane XOZ is called plane of precession and OY is the axis of precession.

Let I = Mass moment of inertia if the disc about OX, and

= Angular velocity of the disc

Angular momentum of the disc = I.

x x

1. The couple I. . p In the direction of the vector xx’ representing the change in angular

momentum is the active gyroscopic couple, which has to be applied over the disc when

the axis of spin is made to rotate with angular velocity p about the axis of procession.

The vector xx’ lies in the plane XOZ or the horizontal plane. In case of a very small

displacement , the vector xx’ will be perpendicular to the vertical plane XOY.

Therefore the couple causing this change in the angular momentum will lie in the plane

XOY. The vector xx’ as shown in figure represents an anticlockwise couple in the plane

XOY. Therefore, the plane XOY is called the plane of active gyroscopic couple and the

axis OZ perpendicular to the plane XOY, about which the couple acts, is called the axis

of active gyroscopic couple.

2. When the axis of spin itself moves with angular velocity p the disc is subjected to

reactive couple whose magnitude is same but opposite in direction to that of active

couple. This reactive couple to which the disc is subjected when the axis of spin rotates

about the axis of procession is known as reactive gyroscopic couple. The axis of the

reactive gyroscopic couple is represented by oz’ in figure.

3. The gyroscopic couple is usually applied through the bearings, which supported the

shaft. The bearings will resist equal and opposite couple.

4. The gyroscopic principle is used in an instrument or toy known as gyroscope. The

gyroscopes are installed in ships in order to minimize the rolling and pitching effects of

waves. They are also used in airplanes, monorail cars, gyrocompasses etc.

Procedure:

1. Balance the rotor position in the horizontal plane.

2. Start the motor and adjust the regulator by rotating in the clockwise direction.

3. Press the frame about the vertical axis by applying necessary force in clockwise

direction.

4. It will be observed that the rotor frame swings about the horizontal axis and also be

observed that the motor side is moving downwards.

5. Keep the weight block on the weighing pan and note down the time for 30 degrees of

precession and take three readings at particular speed.

6. Change the speed by using regulator and measure the speed by tachometer.

7. Again put one more weight block on weighing pan and note down time for 60 degrees of

precession and take three readings at particular speed.

8. Change the speed by using regulator and measure the speed by using tachometer.

9. Again put one more weight block 0.55Kg on weighing pan and note down time for 90

degrees of precession and take three readings at particular speed.

Specifications:

1. Mass of the rotor (m) = 6.3 kg

2. Weight of the rotor (W) = 6.3 x 9.81 N

3. Thickness of the rotor (t) = 10 mm

4. Distance between the rotor center and

center of weighing pan = __________mm

5. Diameter of the rotor (D) = 300mm

Formulae:

1. Angular speed = 60

2 N rad/s

2. Angular velocity of precession ( p ) = timedt

d

180

rad/s

3. Moment of Inertia of the disc (I) = mD2 / 8 kg.m2

4. Gyroscopic couple (C) = I. . p Nm

5. Applied torque (T) = lgm Nm

Tabular column:-

S. No Mass

(kg)

Angle of

precession

in rad.

Time taken

(in s)

Speed

(rpm)

Ang.

Velocity

’ p ’rad/s

Gyroscopic

Couple in Nm

Applied

torque in

Nm T Tavg

1. 0.50 kg

2. 1.0 kg

3. 1.5 kg

Result:

Inference:

Signature:

Date :

Exp. No: 9

Date :

Undamped natural vibrations an equivalent of a spring mass system Aim:

To study the undamped natural vibrations of an equivalent spring mass system. Procedure:

1. Attach a thicker beam between the trunion bracket and the spring mount the

exciter assembly over it.

2. Adjust the position of the beam in such a way that the beam remains in the

horizontal position (approximately) using the spring supporting screw.

3. Weigh the exciter assembly along with discs and bearing and weight platform.

4. Clamp the assembly at any convenient position.

5. Measure the distance L1 of the assembly from pivot.

6. Allow the system to vibrate freely. Record the time required for 10 oscillations

and find the periodic time and natural frequency of vibrations.

7. Repeat the experiment by adding different weights on the platform.

Observations:

m = mass of the exciter assembly along with the weight platform = 15 kg.

mt = Weight attached to the exciter assembly.

L1 = Distance of weight from pivot.

L = Distance of spring from pivot.

We = Equivalent mass at the spring.

g = gravitational acceleration

K = Stiffness of the spring = 0.4kg / cm = ____________ N/m.

Calculations:

Weight of exiter assembly along with mass (W) = (m + me) /g (in N)

Equivalent weight at the spring (We) = W (L12 / L2) (in N)

S. No Load Spring

stiffness (N/m)

Time for 10

oscillations,

t seconds.

Experimental

Frequency

Theoretical

Frequency

1.

2.

3.

Calculation:

Texpt. = Time for 10 oscillations /10; Ttheo =

Experimental frequency =

Theoretical frequency =

Result:

Inference:

Signature:

Date :

Exp. No : 10

Date :

PROELL GOVERNOR

Aim:

To conduct an experiment on the Proell governor set up and to plot the following graphs.

1. Speed Vs Sleeve displacement

2. Radius of rotation Vs Force

Apparatus Required: Tachometer, Scale

Description: The drive unit consists of a small DC electric motor connected through belt and

pulley arrangement. Motor and Test set up are mounted on a MS fabricated stand. The

governor spindle is driven by motor through V belt and is supported in a ball bearing.

The optional governor mechanism can be mounted on spindle. The electronic control

unit controls speed, an extension to the spindle shaft allows the use of a hand tachometer to

determine the speed. A graduated scale is fixed to the sleeve and guided in vertical direction.

1. Height of governor: The axes of the arms intersect the spindle axis. It is usually

denoted by h.

2. Equilibrium speed: It is the speed at which the governor balls, arms etc., are in complete

equilibrium and the sleeve does not tend to move upwards or downwards.

3. Mean equilibrium speed: It is the speed at the mean position of the balls or the sleeve.

4. Maximum and minimum equilibrium speeds: The speeds at the maximum and minimum

radius of rotation of the balls, without tending to move either way are known as

maximum and minimum equilibrium speeds respectively.

5. Sleeve lift: It is the vertical distance, which the sleeve travels due to change in

equilibrium speed.

Proell governor:

Fig. 1 (a) Fig. 1 (b)

The Proell governor has the balls fixed at B and C to the extension of the links DF and

EG, as shown in Fig. 1 (a). The arms FP and GQ are pivoted at P and Q respectively. Consider

the equilibrium of the forces on one-half of the governor as shown in Fig. 1 (b). The

instantaneous centre (I) lies on the intersection of the line PF produced and the line from D

drawn perpendicular to the spindle axis. The perpendicular BM is drawn on ID. The arms of the

governor may be connected to the spindle in the following three ways:

1. The pivot P may be on the spindle axis as shown in figure.

2. The pivot P may be offset form the spindle axis and the arms when produced intersect at

another point on the spindle axis.

Procedure:

1. Make the proper connections of the motor to the governor.

2. Increase the motor speed gradually by using regulator.

3. Take sleeve displacement reading when the pointer remains at steady state.

4. Note down the speed of the governor for different sleeve displacements by using

Tachometer.

5. Determine the force and radius of the rotations for each and every reading. Repeat the

experiment and take minimum six readings and tabulate the readings.

Specifications:

Initial height of the Governor h0 = _________mm

Length of each link (l) = 125mm

Mass of the each ball (m) = 0.7 kg

Mass of the sleeve Ms = 1.5 kg

Extension of length FB = 75 mm.

Ratio (q) = 1

Formulae:

Height of the Governor (h) = (h0 – x) /2

Radius of rotation (r) = ahl 22

Where a = Eccentric position = ____________mm

Angular speed ( ) = 60

2 N

Force (F)

N

Tabulation:

S. No Displacement

‘x’ in mm

Speed

(rpm)

Height

(h) in cm

Angular Speed

( )

Radius of

Rotation

(r) (mm)

Controlling

Force

(N)

1.

2.

3.

4.

5.

Graphs:

Radius of rotation (X axis) Vs Controlling force ( Y axis)

Result:

Inference:

Signature:

Date :

Exp. No: 11 (a)

Date :

Determination of Torsional Vibration Frequency of a Single Rotor System

Aim:

To study the torsional vibration of system and find the natural frequency of single rotor system

and compare it with the theoretical values

Apparatus Required: Single rotor system, Stop watch, Vernier Caliper, Steel rule.

Torsional Vibrations:

When the particles of the shaft or disc move in a circle about the axis of the shaft, then

this type of vibration is known as torsional vibrations. In this case, the shaft is twisted and

untwisted alternately and the torsional shear stresses are induced in the shaft.

Free Torsional Vibrations of a Single Rotor System:

For a shaft fixed at one end and carrying a rotor at the free end, the natural frequency of

torsional vibration,

fn = 2

1

I

q=

2

1

LI

GJ q=

L

GJ = Torsional stiffness of shaft

where G = Modulus of rigidity for the shaft material = 0.8 x 106 kg / cm2

J = Polar Moment of Inertia of shaft

d = Diameter of Shaft = 3 mm

L = Length of shaft

m = Mass of rotor

k = Radius of gyration of rotor, and

I = Mass moment of inertia of rotor = m.k2

It may be noted that the point or the section of the shaft whose amplitude of torsional vibration is

zero, is known as node. In other words, at the node, the shaft remains unaffected by the

vibration.

A torsionally equivalent system is one, which twists through exactly the same angle as

the actual shaft when equal and opposite torques are applied to the rotor. Based on this

principle we can correlate this system with torsional system in which there is heavy rotating

mass is converted to the shaft at one end and the other end is fixed.

Description:

One end of the shaft is gripped in the chuck and heavy disc free to rotate in ball bearing

is fixed at the other end of the shaft.

The bracket with fixed end of shaft can be clamped at any convenient position along the

beam. Thus length of the shaft can be varied during the experiments. Specially designed

chuck is used for clamping the end of the shaft. The ball bearing support to the flywheel

provides negligible damping during experiment. The bearing housing is fixed to side member of

the main frame.

Procedure:

1. Fix the bracket at any convenient position in the beam.

2. Grip the shaft at the bracket by means of chuck.

3. Twist the rotor through some angle and set into oscillation.

4. Note down the time taken for ‘n’ oscillations.

5. The experiment is repeated for different positions of bracket, i.e., for different lengths of

shaft.

Calculation:

G – torsional modulus (Modulus of rigidity) = 0.8 x 106 kg / cm2 = ____________ N / m2.

J - Polar moment of inertia of shaft (m4) = 32

4d

= ____________ m4

L - Length of the shaft “m” = ____________ m.

I – Mass moment of inertia = mk2 = M

2

2

D= ____________ kg-m2

k – Radius of gyration – D/2

Period of oscillation, Texp = ‘ ’

=

Frequency of the system fexp = 1 / Texp

Frequency of the system, fthe = 1 / Ttheo

Observations:

Diameter of the shaft, d = 0.3 m

Diameter of the rotor, D = 225 mm

Mass of the rotor, M = 3.12 kg

Modulus of Rigidity, G = _____________ N/m2

Table:

S. No Length, L

(m)

Time ‘t’ for 10

oscillations (s)

Torsional stiffness

q= L

GJ N-m

Tthe Texp fthe fexp

1.

2.

3.

Result:

Inference:

Signature:

Date :

Exp. No: 11 (b)

Date :

Determination of Torsional Vibration Frequency of a two Rotor System

Aim:

To study the torsional vibration of system and find the natural frequency of a two rotor system

and compare it with the theoretical values

Apparatus Required: Two rotor system set up, stop watch, vernier Caliper, steel rule.

Free Torsional vibrations of a two rotor system:

For a shaft fixed at one end and carrying a rotor at the free end, the natural frequency of

torsional vibration,

fn = 2

1

q=

L

GJ Torsional stiffness of shaft

where G = Modulus of rigidity for the shaft material = 0.8 x 106 kg / cm2

J = Polar Moment of Inertia of shaft

d = Diameter of Shaft = 0.3 m

L = Length of shaft

m = Mass of rotor

k = Radius of gyration of rotor, and

= Mass moment of inertia of rotor A= mA.kA2

= Mass moment of inertia of rotor B= mB.kB2

Procedure:

1. Fix the bracket at any convenient position in the beam.

2. Grip the shaft at the bracket by means of chuck.

3. Twist both the rotor through some angle in opposite direction and set into oscillation.

4. Note down the time taken for ‘n’ oscillations.

Calculation:

G – torsional modulus (Modulus of rigidity) = 0.8 x 106 kg / cm2 = ____________ N / m2.


4d

= ____________ m4


= Mass moment of inertia of rotor A = mA.kA2 = mA

2

2

AD= ____________ kg-m2

= Mass moment of inertia of rotor A = mB.kB2 = mB

2

2

BD= ____________ kg-m2


=



Observations:

Diameter of the shaft, d = 3 mm

Diameter of the rotor, DA = 225 mm

Diameter of the rotor, DB = 190 mm

Mass of the rotor, WA = 3.12 kg

Mass of the rotor, WB = 2.22 kg

Modulus of Rigidity, G = _____________ N/m2

Length of the shat detween rotors = _____________ m.

Table:

S. No Length, L

(m)

Time ‘t’ for 10

oscillations (s)

Torsional stiffness

q= L

GJ N-m

Tthe Texp fthe fexp

1.

2.

3.

Result:

Inference:

Signature:

Date :

Exp. No: 12

Date :

Damped torsional vibrations of a shaft Aim:

To study the damped torsional vibrations of a shaft and determine the damping

coefficient Ct.

Procedure:

2. Put thin mineral oil in the drum and note the depth of immersion.

3. Put the sketching pen in its sprocket.

4. Allow the flywheel to vibrate.

5. Allow the pen to decent.

6. Measure the time for some oscillations.

7. Determine Xn (i.e.) amplitude at any position and Xntr amplitude after ‘r’ cycles.

8. Repeat the experiment by adding different weights on the platform.

Calculation:

G – torsional modulus (Modulus of rigidity) = 0.8 x 106 kg / cm2 = ________N / m2.


4d

= ____________ m4


I – Mass moment of inertia = mk2 = M

2

2

D= ____________ kg-m2

k – Radius of gyration – D/2


=



Logarithmic decrement δ =

Damping factor

Table:

S.No Depth of immersion Xn (m) Xntr (m)

Result:

Inference:

Signature:

Date :

dynamics of mahinery lab manual_cycle 2

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