This document contains, Dynamics of Machines: Question bank unitwise from vtu old question papers, Prepared by Hareesha N G, DSCE, Bengaluru

1. Dynamics of Machines:
Assignment Questions-Uniwise

2. DAYANANDA SAGAR COLLEGE OF ENGINEERING
DEPARTMENT OF AERONAUTICAL ENGINEERING
Assignment Questions Unit wise
Subject: Dynamics of Machinery Sub Code: 10AE53
Faculty In charge: Hareesha N G
Unit 1: Static Force Analysis:
Static force analysis: Introduction: Static equilibrium. Equilibrium of two and three force members, Members with two forces and
torque, free body diagrams, principle of virtual work. Static force analysis of four bar mechanism and slider-crank mechanism with
and without friction
1 a) Discuss the equilibrium of the following systems:
i) Two force members ii) three force members iii) Member with two forces and a torque
06 Jan 2009
Jul 2014
1 b) For the mechanism shown in fig. Q1(b), find the magnitude and direction of input torque T2 for the static
equilibrium. Take AB = 70mm, BC = 150mm, BD = 100mm and CD = 70mm, angle ABC = 90°. Also
determine the forces at pin joints A, B and C.
14 Jan 2010
Fig. Q.1 (b)
Fig. Q.2 (b)
2 a) With usual notations, explain the principle of virtual work, considering a slider crank mechanism 06
2 b) For the mechanism shown in Fig. Q2 (b), determine the torque on the link AB for static equilibrium of
the mechanism. Given, AB = 20 mm, BC = 60 mm, CD = 35 mm. AD = 50 mm, BE = 45 mm, CE = 20
mm and DG = 25 mm.
14 July 2014
Jun 2010
3 a) What is a free body diagram? List any two advantages of free body diagram. 04
3 b) In a four bar mechanism shown in fig Q 3(b) Torque T3 and T4 have magnitudes 30 N-m and 20 N-m
respectively. Link lengths are MN = 800mm NA = 300mm AB =700mm and MB 400mm. Determine
the Input torque T2 for the static equilibrium of the mechanism.
16 Jan 2009
Dec 2012
Fig. Q.3 (b) Fig. Q.4 (a)
4 a) A body shown in Q.4 (a) is subjected to three forces F1, F2 and F3. State the conditions for the static
equilibrium of the body. If force F1 is completely known, F2 known in direction only and F3 is
completely unknown; explain how the problem can be solved.
05 Jan 2010
4 b) For the static equilibrium of the mechanism shown in fig. Q 4 (b), find the required input torque. The
dimensions are AB = 150 mm, BC = AD = 500 mm; DC = 300 mm; CE = 100 mm and EF = 450 mm.
15 July 2013
Fig. Q.4 (b)
Fig. Q.5
5 A slider crank mechanism is shown in Q.5. The force applied to the piston is 2000 N. Determine the
torque T2 required for static equilibrium of the mechanism. Given AB = 300 mm, BC = 600 mm, BD =
200 mm, AE = 200 mm.
20 Jan 2014

3. 6 a) What are the free body diagrams of a mechanism? Write the free body diagram of 4 bar mechanism with
load acting on the rocker.
5
6 b) For a mechanism shown in Fig. Q.6 (b), find the required input torque for the static equilibrium. The
lengths OA and AB are 250 mm and 650 mm respectively. F = 500 N.
15 Jan 2014
Fig. Q.6(b)
Fig. Q.7 (b)
7 a) Explain the procedure for static force analysis of four bar mechanism. 8 Jun 2012
7 b) For the static equilibrium of the quick return mechanism shown in fig. Q.7 (b), determine the input
torque T2 to be applied on link AB for a force of 300N on the slider D. The dimensions of the various
links are OA=400mm, AB=200mm, OC=800mm, CD=300mm
12 Jul 2012

4. Unit 2: Dynamic Force Analysis:
D’Alembert’s principle, Inertia force, inertia torque, Dynamic force analysis of four-bar mechanism and slider crank mechanism.
Dynamically equivalent systems, Turning moment diagrams and flywheels, Fluctuation of Energy, Determination of size of flywheels
1 a) Explain D'Alembert’s principle and why it is used. 06 Jul 10, 13,
14
1 b) A punching press is required to punch 40 mm diameter holes in a plate of 15 mm thickness at the rate of
30 holes per minute. It requires 6 N-m of energy per mm2
of sheared area. If the punching takes 1/10 of
second and the speed of the flywheel varies from 160 to 140 rpm, determine the mass of the flywheel
having radius of gyration of 1 meter.
14 Jul 2014
Jul 2011
Dec 2012
2 a) Briefly discuss the following: (i) D'Alembert's principle (ii) Dynamically equivalent system. 6 Jul 2014
2 b) The turning moment diagram for a four stroke engine may be assumed for simplicity to be represented
by four isosceles triangles. The areas of the triangles are suction = - 0.5cm2
; Compression = - 2.1cm2
;
Expansion = +8.lcm2
and exhaust = - 0.8cm2
. 1 cm2
area represents 1400 N-m of work. Determine the
mass moment of inertia of the flywheel to keep the fluctuation of speed within 1% of mean speed, if the
mean speed is 400 rpm.
14 Jul 2014
Dec 2012
Jun 2010
3 a) Derive an equation for the maximum fluctuation of energy of a flywheel in terms of mean kinetic energy
and coefficient of fluctuation of Speed.
06 Jan 2014
Dec 2012
3 b) A punching press is driven by a constant torque electric rotor. The press is provided with a flywheel that
rotates at a maximum speed of 225 rpm. The radius of gyration of the flywheel is 0.5 m. The press
punches 720 holes per hour, each punching operation takes two seconds and requires 15 kN-m energy.
Find the power of the motor and the minimum mass of the flywheel if speed of the same is not to fall
below 200 rpm.
14 Jan 2014
Jun 2010
4 During forward stroke of the piston of the double acting steam engine, the turning moment has the
maximum value of 2000 Km. when the crank makes an angle of 80 with the inner dead centre. During
the backward stroke the maximum turning moment is 1500 Nm when the crank makes 80° with the outer
dead center. The turning moment diagram for the engine may be assumed for simplicity to be
represented by two triangles. If the crank makes 100 rpm and the radius of gyration of the flywheel is
1.75 m, find the coefficient of fluctuation of energy and the mass of flywheel to keep the speed within
±0.75% of the mean speed. Also determine the crank angle at which the speed has its maximum and
minimum values.
20 Jan 2014
(2002
scheme)
5 a) What is the function of a flywheel? How does it differ from that of a governor? 5 Jul 2013
5 b) A single cylinder, four stroke I.C. engine develops 30 KW of power at 300 rpm. The turning moment
diagram for the expansion and compression strokes may be taken as isosceles triangles on bases 0 to π
and 3π to 4π radius respectively and the work done during compression is 25% of that of during
expansion. Work done during suction and exhaust is neglected. Find the mass moment of inertia of
flywheel to keep the speed fluctuations 1.5% on either side of the mean speed. Sketch the T.M. diagram
and mark the points of max and min speed on the diagram.
15 Jul 2013
6 a) Show that the max fluctuation of energy e = 2EKs where E = is the mean kinetic energy of the flywheel,
Ks = CO-efficient of fluctuation of speed.
6 Dec 2012
6 b) The turning moment diagram for a multi cylinder engine is drawn to a scale of 1mm = 500 N-m torque
and l mm = 6° of the crank displacement. The intercepted areas between output torque curve and mean
resistance line taken in order from one end are -30, +410, -280, +320. -330, +250, -360, +280, -260, sq-
mm, when flywheel rotating at 800 rpm. The engine has a stroke of 300 mm a fluctuation speed is not to
exceed ± 2% of mean speed. Determine the dia and c/s area of the flywheel rim for a limiting safe stress
of 7 x 106
N/m2
. The material density is 7200 kg/m2
. The width of the rim is to be 5 times the thickness.
14 Dec 2012
Dec 2011
7 a) Explain turning moment diagrams of 4-stroke IC engine and multi-cylinder engine. 6 Jun 2012
7 b) A three cylinder single acting engine has its cranks set equally at 120° and it runs at 600 rpm. The torque
crank angle diagram for each cycle is a triangle for the power stroke with a maximum torque of 90 N-m
at 60° from dead centre of corresponding crank. The torque on the return stroke is sensibly zero.
Determine : i) Power developed ; ii) Coefficient of fluctuation of speed ; iii) Coefficient of fluctuation of
energy and iv) Maximum angular acceleration of the flywheel. Take mass of flywheel is 12 kg and
radius of gyration of 80 mm.
14 Jun 2012
8 a) Derive an expression for the maximum fluctuation of energy of a flywheel in terms of mean kinetic
energy and coefficient of fluctuation of speed.
5 Jan 2010
8 b) The torque delivered by a two stroke engine is represented by T = (1000 + 300 sin 2θ - 500 cos 2θ) N-m,
where θ is the angle turned by the crank from inner dead center. The engine speed is 250 rpm. The mass
of the fly wheel is 400kg and radius of gyration is 400mm. Determine i) the power developed ii) the total
percentage fluctuation of speed iii) the angular acceleration of flywheel when the crank has turned
through an angle of 60° from IDC.
15 Jan 2010

5. Unit 3: Friction and Belt Drives:
Definitions: Types of friction: laws of friction, Friction in pivot and collar bearings. Belt drives: Flat belt drives, ratio of belt tensions,
centrifugal tension, and power transmitted.
1 a) Derive an expression for ratio of tensions in flat belt drive. 5 Jul 14
1 b) A belt 100 mm wide and 10 mm thick is to transmit power at speed of 1000 m/min. The net driving
tension is 1.8 times the tension on slack side. If the safe permissible stress is 2 MPa. Calculate the
maximum power that can be transmitted at this speed. Assume the density of leather as 1000 kg/m3
.
Also determine (i) the absolute maximum power. (ii) percentage increase in power.
15 Jul 14
2 a) State the laws of dynamic friction. 4 Jul 14,
Jan 14
2 b) Derive an expression for frictional torque in a flat collar bearing assuming uniform pressure. 6 Jul 14
2 c) A leather belt is required to transmit 7.5 kW from a pulley 1.2 m in diameter, running at 250 rpm, the
angle of contact is 165° and ji = 0.3. If the safe working stress for the leather belt is 1.5 MPa and density
of leather is 1000 kg/m3
and thickness of belt is 10 mm. determine the width of belt taking centrifugal
tension into account.
10 Jul 14
Jan 14
Dec 12
Jan 10, 09
3 a) Derive an equation to calculate the centrifugal tension in a flat belt drive. 5 Jan 14
3 b) Determine the width of a 9.75 mm thick belt required to transmit 15 kW from a motor running at 900
rpm. The diameter of the driving pulley of the motor is 300 mm. The driven pulley runs at 300 rpm and
distance between centres of two pulleys is 3 m. The density of leather is 1000 kg/m3
. The maximum
allowable stress in leather is 2.5 MPa. The coefficient of friction between leather and pulley is 0.3.
Assume open belt drive and neglect slip in belt drive.
15 Jan 14
Jun 10
4 a) Explain: i) Slip, ii) Creep, iii) Initial tension and iv) Centrifugal tension in belt drive. 8 Jul 13
4 b) A flat belt is required to transmit 35 kW from a pulley of 1.5m effective diameter running at 300 rpm.
The angle of contact is spread over 11/24 of the circumference and the coefficient of friction between
belt and pulley surface is 0.3. Determine width of the belt required taking centrifugal tension into
account. It is given that the belt thickness is 9.5mm, density of its material is 1.1x103
kg/m3
and the
permissible working stress for belt is 2.5 N/mm2
.
12 Jul 13
Jun 12
5 a) Explain how centrifugal tension affects the maximum tension in the flat belt drive. Also derive the
equation for centrifugal tension in a flat belt drive.
06 Jul 13
Jun 10
5 b) Calculate the power lost in overcoming the friction and number of collars required for a thrust bearing
whose contact surfaces are 20cm external radius and 15cm in internal radius. The coefficient of friction
is 0.08. The total axial load is 30 kN. Intensity of pressure is not to exceed 3.5 x 105
N/m2
. Speed of the
shaft is 420 rpm.
14 Dec 12
Jul 09
6 a) Derive the expressions for frictional torque in a flat pivot bearing for uniform pressure and wear
conditions.
8 Jun 10, 12
6 b) An open belt drive is used to connect two parallel shafts, 4m apart. The diameter of the larger pulley is
1.5m and that of smaller pulley is 0.5m. The mass of the belt is Ikg/m length. The maximum tension is
not to exceed 1500N. The coefficient of friction is 0.25. The bigger pulley which is the driver runs at
250 rpm. Due to slip, the speed of driven pulley is 725 rpm. Calculate the power transmitted, power lost
is frictions, and the efficiency of the drive.
12 Dec 11
7 a) Derive the expressions for frictional torque in flat collar bearings for uniform pressure and wear
conditions.
8 Jul 11
7 b) An open belt drive connects two pulleys 1.5 m and 0.5m diameter on parallel shafts 3.5m apart. The belt
has a mass of 1 kg/m length and the maximum tension in the belt is not to exceed 2 kN. The 1.5m
pulley, which is the driver runs at 250 rpm. Due to belt slip, the velocity of the driven shaft is only 730
rpm. If the coefficient of friction between the belt and the pulley is 0.25 find, i) The torque on each shaft
ii) Power transmitted iii) The power lost in friction and iv) The efficiency of the drive.
12 Jul 11
8 a) Derive an expression for the ratio of tensions in a flat belt drive. 8 Dec 11, 12
8 b) 2.5 kW of power is transmitted by an open belt drive. The linear velocity of the belt is 2.5 m/s, The
angle of lap on the smaller pulley is 165°. The coefficient of friction is 0.3. Determine the effect on
power transmission in the following cases: 1) Initial tension in the belt is increased by 8%. 2) Initial
tension in the belt is decreased by 8%. 3) Angle of lap is increased by 8% by the use of an idler pulley,
for the same speed and tension on the tight side. 4) Coefficient of friction is increased by 8% by suitable
dressing to the friction sample of the belt.
12 Dec 10
9 a) Obtain condition for maximum power transmitted by a belt from one pulley to another. 6 Jan 10
9 b) A shaft rotating at 200 r.p.m. drives another shaft at 300 r.p.m. and transmits 6 kW through a belt. The
belt is 100 mm wide and 10 mm thick. The distance between the shafts is 4m. The smaller pulley is 0.5
m in diameter. Calculate the stress in the belt, if it is 1. an open belt drive, and 2. a cross belt drive. Take
μ= 0.3.
14 Jan 10

6. Unit 4: Balancing of Rotating Masses:
Static and dynamic balancing, Balancing of single rotating mass by balancing masses in same plane and in different planes,
Balancing of several rotating masses by balancing masses in same plane and in different planes.
1 a) What do you mean by static balancing and dynamic balancing?. Explain in detail taking a suitable
example.
5 Jun 10, 09
Jan 10
1 b) A rotating shaft carries four masses A, B, C and D, which are radially attached to it, along the shaft axis.
The mass centres are 40mm, 50mm, 60mm and 70mm respectively from the axis of rotation. The masses
B,C and D are 60kg, 50kg and 40kg respectively. The angles of the masses C and D with respect to mass
B are 90° and 210° in same sense, respectively. The planes containing B and C are 0.5m apart. For a
complete balanced system, determine ;i) The mass and angular position of mass A. ii) The position of
planes containing masses A and D.
15 July 14
Jun 12
Dec 11
Jun 10
Jan 10
2 a) Explain Graphical and analytical methods of balancing of several rotating masses in one plane 10 Jun 12, Jan 09
Jul 11, Dec 10
2 b) Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg respectively. The corresponding
radii of rotation are 0.2m, 0.15m, 0.25m and 0.3 m respectively. The angles between successive masses
are 45°, 75o
and 135°. Find the position and magnitude of the balance mass required, if its radius if
rotation is 0.2m.
10 Dec 12
3 a) Explain the method of balancing of single rotating mass by two masses in different planes. 06 July 13
3 b) A shaft carries four masses A, B. C and D. 200. 300, 240 and 360 kg respectively, revolving at radii 90.
70, 100 and 120 mm respectively. The distance from the plane A, other planes are at 270 mm. 420 mm
and 720 mm respectively. Angle between the crank A and B is 45°. B and C is 75°. C and D is 130°.
Balancing masses are replaced 120 mm and 100 mm from D and A respectively. The distance between
them being 500 mm. Find the balancing masses and their angular positions if they are placed at a radius
of 100 mm.
14 July 13
4 a) How do you balance single mass rotating in one plane by single mass on the same plane? Explain. 04 Dec 12
Jun 10
4 b) Four masses M1 = 100 kg, M2 = 175 kg, M3 = 200 kg and M4 = 125 kg are fixed to the crank of 200 mm
radius and revolve in planes 1, 2, 3 and 4 respectively. The angular position of the planes 2, 3 and 4 with
respect to first mass are 75°, 135° and 240° taken in the same sense. Distance of the planes 2, 3 and 4
from first are 600 mm, 1800 mm and 2400 mm. Determine the magnitude and position of the balancing
masses at radius 600 mm in planes 'L' and M' located in the middle of 1 and 2 and in the middle of 3 and
4 respectively.
16 Dec 12
5 A rotating shaft carries four masses A, B, C, D of 10 kg, 15 kg, 18 kg and 20 kg at radii 50 mm, 60 mm,
60 mm and 80 mm respectively. The masses B, C and D revolve in planes 400 mm, 600 mm and 800
mm respectively measured from plane of mass A and are angularly-located at 60°, 145° and 270°
respectively measured anticlockwise from mass A and are viewing from mass A. The shaft dynamically
balanced by two masses located at 50 mm radii and revolving in planes L and M placed midway
between the masses A and B and midway between those of masses C & D respectively. Determine the
magnitude of balance masses and their angular positions.
20 July 14
Jan 09
6 A 3.6 m long shaft carries three pulleys, two at its two ends and the third at the midpoint. The two end
pulleys have masses 79 kg and 40 kg with their radii 3 mm and 5 mm from axis of shaft respectively.
The middle pulley has a mass of 50 kg with radius 8 mm. The pulley are so keyed to the shaft that the
assembly is in static balance. The shaft rotates at 300 rpm in two bearings 2.4 m apart with equal
overhangs on either side. Determine (i) Relative angular positions of pulleys. (ii) Dynamic reaction on
the two bearings.
4 July 14
Jan 14
July 09
7 a) Explain why a single disturbing rotating mass in a plane can't be balanced by another single mass in a
different plane. And also show the position of the balancing masses with respect to the disturbing mass if
(i) Planes of both balancing masses arc on the same side. (ii) Planes of balancing masses are on either
side of disturbing mass plane.
06 July 13
7 b) A shaft carries four masses P. Q, R and S at the extremities of arms of radii 0.20m. 0.24m.0.3m and
0.24m respectively. The planes containing Q. R and S are at 0.3m. 0.48m and 0.72m respectively from
the plane containing P. Masses of P. Q. R & S are 20kg, 50kg. 30kg and 15 kg respectively. Determine
the necessary alternation to mass Q and angular position of all the masses so that the shaft is in complete
balance.
14 July 13
8 A shaft carries four masses A. B. C. and D of magnitude 200kg. 300kg. 400kg and 200kg respectively
and revolving at radii 80mm, 70mm. 60mm and 80mm in planes measured from A at 300mm. 400mm
and 700 mm. The angle between the crank measured anticlockwise are A to B 45°. B to C 70° and C to
D 120° the balancing masses arc to be placed in planes X and Y. The distance between the planes A and
X is 100mm between X and Y is 400mm and between Y and D is 200mm. If the balancing manes
revolved at a radius of 100mm find their magnitudes and angular position.
20 Dec 12
Jun 10
9 A rotating shaft carries four unbalanced masses 18kg, 14kg, 16kg and 12kg at radii 50mm. 60mm 70mm
and 60mm respectively. The 2nd
, 3rd
and 4lh
masses revolve in planes 80mm, 160mm and 280mm
20 Dec 2011

7. respectively, measured from the plane of the first mass and are angularly located at 60°, 135° and 270°
respectively, measured anti clockwise from the first mass, looking from this mass end of the shaft. The
shaft is dynamically balanced by two masses, both located at 50mm radii and revolving in planes
midway between those of lsl
and 2 masses and midway between those of 3 and 4 ' masses. Determine the
magnitude of the masses and their respective angular positions.
10 A shaft has three eccentrics, each 75 mm diameter and 25 mm thick, machined in one piece with the
shaft. The central planes of the eccentric are 60 mm apart. The distance of the centres from the axis of
rotation are 12 mm, 18 mm and 12 mm and their angular positions are 120° apart. The density of metal
is 7000 kg/m . Find the amount of out-of-balance force and couple at 600 r.p.m. If the shaft is balanced
by adding two masses at a radius 75 mm and at distances of 100 mm from the central plane of the
middle eccentric, find the amount of the masses and their angular positions.
20

8. Unit 5: Balancing of Reciprocating Masses
Inertia effect of crank and connecting rod, single cylinder engine, balancing in multi cylinder-inline engine primary & Secondary
forces, V-type engine; Radial engine - Direct and reverse crank method.
1 a) A single cylinder reciprocating engine has the following data: Speed the engine = 120rpm; Stroke -
320mm; Mass of reciprocating parts = 45kg; Mass of revolving parts 35kg at crank radius of 200mm. If
60% of the reciprocating parts and all revolving parts are to balanced, find i) Balance mass required at
radius of 300mm and ii) unbalanced force when the crank has turned 60° from the TDC
10 Jan 14
Dec 11
1 b) A 90° V engine has 2 cylinders, placed symmetrically. The two connecting rods operate a common
crank. The length of connecting rods is 320mm each and crank radium is 80mm. The reciprocating mass
per cylinder is 12kg. If the engine runs at 600rpm, determine the resultant primary and secondary forces.
Also, find the maximum resultant secondary force
10 Dec 11
2 a) With usual notations, explain the primary and secondary unbalanced forces of reciprocating masses. 5
2 b) A five cylinder in-line engine running at 750 rpm has successive cranks 144° apart; the distance between
the cylinder centre lines being 375m. Piston stroke is 225 mm and the ratio of the connecting rod to the
crank is 4. Examine the engine for balance of primary and secondary forces and couples. Find the
maximum values of these and the position of the central crank, at which, these maximum values occur.
The reciprocating mass for each Cylinder is 15 kg
15 Jul 14
Dec 10
3 a) Explain how balancing of rotary engines are carried out using direct and reverse crank method. 05 Jan 14
Dec 11, 10
3 b) A four cylinder vertical engine has cranks 300mm long. The planes of rotation of the first, third and
fourth cranks are 750mm, 1050mm and 1650mm respectively from that of the second crank and their
reciprocating masses are 150kg, 400kg and 250kg respectively. Find the mass of the reciprocating parts
for the second cylinder and the relative angular positions of the crank in order that the engine may be in
complete primary balance. If each connecting rod of all four cylinders is 1.35m long and the speed is
300rpm, find, the maximum unbalanced secondary force and couple.
15 Jul 14
Jul 12
Jun 10
Jan 10
Jul 09
4 a) Explain how balancing of Inline Engine is done. 04
4 b) The pistons of a four cylinder vertical inline engine reach their uppermost position at 90o
interval in
order of their axial position. The cylinder centre lines are spaced at 0.35 m. Length of crank = 0.12 m.
Length of connecting rod = 0.42 m. The reciprocating mass per cylinder is 2.5 kg and the engine runs at
600 rpm. Determine the out of balance primary and secondary forces and couples on this engine taking
the central plane of engine as reference plane.
16 Jul 13
5) The firing order in a 6 cylinder vertical four stroke in-line engine is 1-4-2-6-3-5. The piston stroke is 100
mm and the length of each connecting rod is 200 mm. The pitch distances between the cylinder centre
lines are WO mm, 100 mm, 150 mm, 100 mm, and 100 mm respectively. The reciprocating mass per
cylinder is I kg and the engine runs at 3000 r.p.m. Determine the out-of-balance primary and secondary
forces and couples on this engine, taking a plane midway between the cylinder 3 and 4 as the reference
plane.
20 Jan 14
Jul 13
Jan 12
Jun 12
Jan 09
Jul 08
6 a) Derive the equations for primary and secondary resultant forces of V engine when cylinders are placed
at 1200
apart.
10 Jul 13
Jul 11
6 b) A twin cylinder V-engine has the cylinders set at an angle of 45°, with both pistons connected to the
single crank. The crank radius is 62.5 mm and the connecting rods are 275 mm long. The reciprocating
mass per line is 1.5 kg and the total rotating mass is equivalent to 2 kg at the crank radius. A balance
mass fitted opposite to the crank, is equivalent to 2.25 kg at a radius of 87.5 mm. Determine for an
engine speed of 1800 r.p.m. ; the maximum and minimum values of the primary and secondary forces
due to the inertia of reciprocating and rotating masses.
10 Jul 11
7 a) With usual notations, explain primary and secondary unbalanced forces of reciprocating masses. 05 Jul & Jan 14
7 b) An air compressor has four vertical cylinders 1, 2. 3 and 4 in line and driving cranks at 90° intervals
reach their uppermost positions in this order. The cranks are of 150 mm radius, the connecting rods 500
mm long and their cylinder centre line 400 mm apart. The mass of the reciprocating parts for each
cylinder is 22.5 kg and speed of rotation is 400 rpm. Show that there are no out of balance primary and
secondary forces and determine the corresponding couples, indicating the positions of No.l crank for
maximum values. The central plane of the machine may be taken as reference plane.
15 Dec 12
8 a) Check the conditions of primary and secondary balancing of 2-cylinder and 3-cylinder in line engine. 06 Jun 12
8 b) The cranks and connecting rods of a 4-cylinder in-line engine running at 1800 r.p.m are 60 mm and 240
mm each respectively and the cylinders are spaced 150 mm apart. If the cylinders are numbered 1 to 4 in
sequence from one end, the cranks appear at intervals of 90° in an end view in the order 1-4-2-3. The
reciprocating mass corresponding to each cylinder is 1.5 kg. Det : 1. Unbalanced primary and secondary
forces, if any, and 2. Unbalanced primary and secondary couples wrt central plane of the engine.
14 Jun 09
9 a) Derive an expression for resultant unbalanced force in a partially balanced single cylinder engine. 05 Jun 10
9 b) The three cylinders of an air compressor have their axes 120° to one another, and their connecting rods
are coupled to a single crank. The stroke is 100 mm and the length of each connecting rod is 150 mm.
The mass of the reciprocating parts per cylinder is 1.5 kg. Find the maximum primary and secondary
forces acting on the frame of the compressor when running at 3000 r.p.m. Describe clearly a method by
which such forces may be balanced.
15 Dec 08

9. Unit 6: Governors:
Types of governors, force analysis of Porter and Hartnell governors, Controlling force, stability, sensitiveness, isochronism, effort and
power
1 a) Define the following, with respect to the working of governors: i) Sensitiveness ii) Isochronism
iii)Hunting of governors iv) Effort of a governor v) Stability of a governor
06 Jul , Jan 14
Jul 13
1 b) The arms of a porter governor are each 300mm long and are hinged on the axis of rotation. The mass of
each ball is 5kg and mass of the sleeve is 15kg. The radius of rotation of the ball is 200mm, when the
governor begins to lift and 250mm, when the governor is at the maximum speed. Determine: i) Range
of speed, neglecting the sleeve friction. ii) Range of speed, if the frictional force at the sleeve is 30N
14 Jul , Jan 14
Jul 13
Jan 09
2 a) Explain the function and types of governors. 04 Jun 12
2 b) In a Hartnell governor, the lengths of ball and sleeve arms of a bell crank lever are 120 mm and 100 mm
respectively. The distance of the fulcrum of the bell crank lever from the governor axis is 140 mm. Each
governor ball has a mass of 4 kg. The governor runs at a mean speed of 300 r.p.m. with the ball arms
vertical and sleeve arms horizontal. For an increase of speed of 4 per cent, the sleeve moves 10 mm
upwards. Neglecting friction, find: 1, the minimum equilibrium speed if the total sleeve movement is
limited to 20 mm, 2. the spring stiffness, 3. the sensitiveness of the governor, and 4. the spring stiffness
if the governor is to be isochronous at 300 r.p.m.
16 Jun 12
3 a) Obtain an expression for height of a porter governor. 05 Dec 12
3 b) The mass of each ball of a Hartnell type governor is 1.4kg. The length of ball arm of the bell - crank
lever is 100mm whereas the length of arm towards sleeve is 50mm. The distance of the fulcrum of bell -
crank lever from the axis of rotation is 80mm. The extreme radii of rotation of the balls are 75mrn and
112.5mm. The maximum equilibrium speed is 6% greater than the minimum equilibrium speed which is
300rpm. Determine i) stiffness of the spring and ii) equilibrium speed when radius of rotation of the ball
is 90mm. Neglect the obliquity of the arms.
15 Jul 14
4 a) Obtain an expression for power of the porter governor without obliquity effect. 06 Dec 11
4 b) The arms of a porter governor are 300 mm long. The upper arms are pivoted on the axis of rotation and
the lower arms are attached to the sleeve at a distance of 35 mm from the axis of rotation. The mass of
sleeve is 54 kg and the mass of each ball is 7 kg. Determine the equilibrium speed when the radius of
rotation of the ball is 225 mm. What will be the range of speed for this position, if the frictional
resistance to the motion of the sleeve is equivalent to a force of 30 N at the sleeve.
14 Jan 14
5 a) Derive an expression for power an porter governor considering obliquity effect. 06
5 b) In an engine governor of the Porter type, the upper and lower arms are 200 mm and 250 mm respectively
and pivoted on the axis of rotation. The mass of the central load is 15 kg, the mass of each ball is 2 kg
and friction of the sleeve together with the resistance of the operating gear is equal to a load of 25 N at
the sleeve. If the limiting inclinations of the upper arms to the vertical are 30° and 40°, find, taking
friction into account, range of speed of the governor.
14 Jun 13
6 a) Explain the principle of operation of a centrifugal governor with a neat sketch. 06 Jun 10
6 b) The upper arms of a Porter governor has lengths 350 mm and are pivoted on the axis of rotation. The
lower arms has lengths 300 mm and are attached to the sleeve at a distance of 40 mm from the axis.
Each hall has a mass of 4 kg and mass on the sleeve is 45 kg. Determine the equilibrium speed for a
radius of rotation of 200 mm and find also the effort and power of the governor for 1% speed change.
14 Dec 2012
7 a) Obtain an expression for stiffness of a spring loaded type Hartnell governor. 08
7 b) All the arms of a Porter governor are 178 mm long and are hinged at a distance of 38 mm from the axis
of rotation. The mass of each ball is 1.15 kg and mass of the sleeve is 20 kg. The governor sleeve begins
to rise at 280 r.p.m. when the links are at an angle of 30° to the vertical. Assuming the friction force to
be constant, determine the minimum and maximum speed of rotation when the inclination of the arms to
the vertical is 45°.
12 Jan 12
8 a) Prove that Hartnell governor can be isochronous, whereas porter governor is not. 04
8 b) A spring loaded governor of the Hartnell type has arms of equal length. The masses rotate in a circle of
130 mm diameter when the sleeve is in the mid position and the ball arms are vertical. The equilibrium
speed for this position is 450 r.p.m., neglecting friction. The maximum sleeve movement is to be 25 mm
and the maximum variation of speed taking in account the friction to be 5 per cent of the mid position
speed. The mass of the sleeve is 4 kg and the friction may be considered equivalent to 30 N at the sleeve.
The power of the governor must be sufficient to overcome the friction by one per cent change of speed
either way at mid-position. Determine, neglecting obliquity effect of arms ; i. The value of each rotating
mass : 2. The spring stiffness in N/mm ; and 3. The initial compression of spring.
16 Jan 14
9 a) What is controlling force? Explain. Sketch the variation of controlling force vs radius rotation for porter
governor.
04 Jan 14
9 b) The radius of rotation of the balls of a Hartnell governor is 80 mm at the minimum speed of 300 r.p.m.
Neglecting gravity effect, determine the speed after the sleeve has lifted by 60 mm. Also determine the
initial compression of the spring, the governor effort and the power. The particulars of the governor are
given below: Length of ball arm = 150 mm ; length of sleeve arm = 100 mm ; mass of each ball = 4kg
and stiffness of the spring = 25 N/mm.
16 Jun 09

10. Unit 7: Gyroscope:
Vectorial representation of angular motion, gyroscopic couple, Effect of gyroscopic couple on ship, plane disc, aeroplane, stability of
two wheelers and four wheelers.
1 a) With usual notations and diagram, derive an expression for the gyroscopic couple, produced by a
rotating disc.
06 Jul 14
Dec 12
1 b) The turbine rotor of a ship has a mass of 3500 kg. It has a radius of gyration of 0.45 m and a speed of
3000 r.p.m. clockwise when looking from stern. Determine the gyroscopic couple and its effect upon the
ship: When the ship is steering to the left on a curve of 100 m radius at a speed of 36 km/h, When the
ship is pitching in a simple harmonic motion, the bow falling with its maximum velocity The period of
pitching is 40 seconds and the total angular displacement between the two extreme positions of pitching
is 12 degrees.
14 Dec 12
2 a) With neat sketches, explain the effect of gyroscopic couple on a ship during pitching, steering and
rolling.
08 Jan 14, 10
Jul 13, 10
2 b) The turbine rotor of a ship has a mass of 2000 kg and rotates at a speed of 3000 r.p.m. clockwise when
looking from a stern. The radius of gyration of the rotor is 0.5 m. Determine the gyroscopic couple and
its effects upon the ship when the ship is steering to the right in a curve of 100 m radius at a speed of
16.1 knots (1 knot = 1855 m/hr). Calculate also the torque and its effects when the ship is pitching in
simple harmonic motion, the bow falling with its maximum velocity. The period of pitching is 50
seconds and the total angular displacement between the two extreme positions of pitching is 12°. Find
the maximum acceleration during pitching motion.
12 Jun 12
3 a) With neat sketches, explain the effect of gyroscopic couple on an airplane. 07 Jun 12
3 b) A rear engine automobile is traveling along a track of 100 m mean radius. Each of the four wheels has a
moment of inertia of 2 kg.m2
and an effective diameter of 0.6 m. The rotating parts of the engine have a
moment of inertia of 1.25 kg m2. The engine axis is parallel to rear axle and the crank shaft rotates in the
same direction as the wheels. The gear ratio of engine to back axle is 3:1. The automobile mass is 1500
kg and the centre of gravity is 0.5m above the road level. The width of track of the vehicle is 1.5m.
Determine the limiting speed of the vehicle around the curve for all four wheels to maintain contact with
the road surface if it is not banked.
13 Jan 14
Jun 12
Jun 10
4 a) Derive an expression for heel angle of a motor cycle to avoid skidding. 06 Jul 13, 10
Dec 12
4 b) Each wheel of a motorcycle is 600mm diameter and has a moment of inertia of 1.2 kg.m . The total mass
of motorcycle and the rider is 180 kg and the combined centre of mass is 580mm above the ground level.
When the motorcycle is upright, the moment of inertia of the rotating part of the engine is 0.2 kg.m2, the
engine speed is five times the speed of the wheel and is in the same sense. Determine the angle of heel
necessary, when the motorcycle takes a turn of 35m radius at a speed of 54 km/hr.
14 Jul 14
5 a) Obtain the expressions for the reaction forces on the outer and inner wheels of a four wheel automobile
is subjected to gyroscopic effect and centrifugal force.
08
5 b) A four - wheeled trolley car has a total mass of 3000 kg. Each axle with its two wheels and gears has a
total M.I of 32 kg-m2
. Each wheel is of 450mm radius. The centre distance between two wheels is 1.4m.
Each axle is driven by a motor with speed ratio of 1:3. Each motor along with its gear has a moment of
inertia of 16 kg-m2 and rotates in the opposite direction to that of axle. The center of mass of the car is
1m above the rails. Calculate the limiting speed of the car when it has to travel around a curve of 250m
radius without the wheels leaving the rails.
12 Jan 14
Jun 12
Jul 11
Jan 10
6 a) Explain the following; i) Velocity of precession ii) Gyroscopic couple iii) Axis of spin iv)
Plane of precession v) plane of gyroscopic couple iv) Angular momentum
12 Jul 11
Jan 08
6 b) A ship is propelled by a turbine rotor of mass 2000kg and has a speed of 1800 rpm. The rotor has a
radius of gyration of 0,35m and rotates in the clockwise direction, when viewed from the bow.
Determine the gyroscopic couple and its effect when the ship; i) Turns right at a radius of 200m with a
speed of 15 knots (1 knot = 1.853km/hr) ii) Pitches with bow rising, with an angular velocity of 0.08
rad/sec. iii) Rolls at angular velocity of 0.1 rad/sec.
08 Jul 14
Dec 12
Dec 11
7 a) Derive an expression to determine gyroscopic couple. 06 Jul 14, Dc11
Jul 13, 12
7 b) A uniform disc of 150 mm diameter has a mass of 5 kg. It is mounted centrally in bearings which
maintain its axle in a horizontal plane. The disc spins about it axle with a constant speed of 1000 r.p.m.
while the axle precesses uniformly about the vertical at 60 r.p.m. The directions of rotation are as shown
in Fig. 14.3. If the distance between the bearings is 100 mm, find the resultant reaction at each bearing
due to the mass and gyroscopic effects.
14
8 A four wheeled motor car of mass 2000 kg has a wheel base 2.5 m, (rack width 1.5 m and height of
centre of gravity 500 mm above the ground level and lies at 1 metre from the front axle. Each wheel has
an effective diameter of 0.8 m and a moment of inertia of 0.8 kg-m2
. The drive shaft, engine flywheel
and transmission are rotating at 4 times the speed of road wheel, in a clockwise direction when viewed
from the front, and is equivalent to a mass of 75 kg having a radius of gyration of 100 mm. If the car is
taking a right turn of 60 m radius at 60 km/h,find the load on each wheel.
20

11. Unit 8: Analysis of CAMS:
Analysis of Tangent cam with roller follower and Circular arc cam, operating flat faced and roller followers, Undercutting in Cams.
1 A symmetrical tangent cam has the following data: Roller radius = 17.5 mm, Minimum radius of the
cam = 30 mm, Maximum lift = 20 mm, Cam displacement during the outstroke = 75°, Cam shaft speed -
600 rpm, Find the acceleration of the follower at the beginning of the ,outstroke. Kind also the nose,
radius, the distance between the cam and nose centre and the angle subtended by the straight flank at the
centre.
20 Jul 14
Jan 14
Jan 10
2 A straight sided cam has both sides tangential to the base circle, with a radius of 25 mm. The total angle
of action is 120°. A lilt of 10 mm is given to the roller 20 mm diameter, the centre of which moves along
a straight line, passing through the axis of the cam. The crank shaft has a speed of 240 rpm. Determine
(i) The radius of the nose arc. (ii) The velocity and acceleration of the roller centre when the roller in
contact with the cam at the end of one of the straight flanks adjacent to the nose and (iii) The
acceleration of roller centre at peak.
20 Jul 14
Dec 11
3 Derive the expression for displacement, velocity and acceleration, when roller is tn contact with straight
Hank of the tangent cam..
20 Jan 14
4 For a symmetrical tangent cam operating a roller follower, the least radius of the cam is 30 mm and the
roller radius is 15 mm. the angle of ascent is 60°. the total lift is 15 mm and the speed of the cam shaft is
300 rpm. Calculate: i) Principal dimensions of the cam (i.e., the distance between the cam centre and the
nose centre, nose radius and the angle of contact of cam with straight front) ii) Acceleration of the
follower at the beginning of the lift, where the roller just touches the nose (i.e., flank merges into the
nose) and the apex of the circular nose. Assume that there is no dwell between ascent and descent.
20 Jul 13
5 For a tangent cam with roller follower, derive the expressions for the velocity and acceleration of the
follower when i) roller is on the Hank ii) roller is on the nose
20 Jul 13
6 In a four stroke petrol engine, the crank angle is 4° after-IDC when the suction valve open and 50o
after
BDC when the suction valve closes. The lift is 10mm, the hole radius is 2.5mm and the least radius of
the cam 20mm, The shaft rotates at 600 rpm the cam is of the circular type with a circular nose and Hank
while the follower is flat faced. Determine the maximum velocity, maximum acceleration and
retardation of the valve. What is the minimum force exerted by the springs to overcome the inertia of
moving parts weighting 250 gram.
20 Dec 12
7 The following data relate to a circular cam operating a flat faced follower. Least radius = 40mm. lift =
12mm. angle of action - 160°. speed = 500 rpm. If the period of acceleration of the follower is 60° of the
retardation during the lift, determine i) Principle dimensions of CAM. ii) Acceleration at the main
points. What is the max acceleration and deceleration during the lift?
20 Dec 12
Jun 12
Dec 11, 09
Jun 10
8 a) Find the velocity and acceleration of a roller follower operated by a tangent cam when the roller is
making contact: i) On the flank, ii) With the nose.
12 Jun 10
8 b) A tangent cam with straight working faces tangential to a base circle of 120 mm diameter has a roller
follower of 48 mm diameter. The nose circle radius of the cam is 12 mm and the angle between the
tangential faces of the cam is 90°. If the speed of the cam is 180 rpm, determine the acceleration of the
follower when : i) During lift, the roller just leaves the straight flank ii) The roller is at the top of the
nose.
08 Jun 10