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GATE SOLVED PAPER
Mechanical Engineering
Theory of Machines
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GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2013
Q. 1
ONE MARK
A link OB is rotating with a constant angular velocity of 2 rad/s in counter
clockwise direction and a block is sliding radially outward on it with an uniform
velocity of 0.75 m/s with respect to the rod as shown in the figure. If OA = 1 m ,
the magnitude of the absolute acceleration of the block at location A in m/s2 is
.
a
(A) 3
(B) 4
(C) 5
Q. 2
Q. 3
n
i
.
o
c
i
d
o
(D) 6
A planar closed kinematic chain is formed with rigid links PQ = 2.0 m , QR = 3.0 m,
RS = 2.5 m and SP = 2.7 m with all revolute joints. The link to be fixed to obtain
a double rocker (rocker-rocker) mechanism is
(A) PQ
(B) QR
(C) RS
(D) SP
.n
w
w
If two nodes are observed at a frequency of 1800 rpm during whirling of a simply
supported long slender rotating shaft, the first critical speed of the shaft in rpm is
(A) 200
(B) 450
©
(C) 600
w
(D) 900
YEAR 2013
Q. 4
TWO MARKS
A compound gear train with gears P, Q, R and S has number of teeth 20, 40, 15
and 20, respectively. Gears Q and R are mounted on the same shaft as shown in
the figure below. The diameter of the gear Q is twice that of the gear R. If the
module of the gear R is 2 mm, the center distance in mm between gears P and
S is.
(A) 40
(C) 120
(B) 80
(D) 160
GATE SOLVED PAPER - ME
Q. 5
THEORY OF MACHINES
A flywheel connected to a punching machine has to supply energy of 400 Nm
while running at a mean angular speed of 20 rad/s. If the total fluctuation of
speed is not to exceed ! 2% , the mass moment of inertia of the flywheel in kgm2 is
(A) 25
(B) 50
(C) 100
Q. 6
(D) 125
A single degree of freedom system having mass 1 kg and stiffness 10 kN/m
initially at rest is subjected to an impulse force of magnitude 5 kN for 10-4 sec .
The amplitude in mm of the resulting free vibration is
(A) 0.5
(B) 1.0
(C) 5.0
(D) 10.0
n
i
.
o
c
YEAR 2012
Q. 7
The following are the data for two crossed helical gears used for speed reduction :
Gear I : Pitch circle diameter in the plane of rotation 80 mm and helix angle
30c.
Gear II : Pitch circle diameter in the plane of rotation 120 mm and helix
angle 22.5c.
If the input speed is 1440 rpm, the output speed in rpm is
(A) 1200
(B) 900
.n
w
Q. 10
(D) 720
A solid disc of radius r rolls without slipping on a horizontal floor with angular
velocity w and angular acceleration a. The magnitude of the acceleration of the
point of contact on the disc is
(A) zero
(B) ra
(C)
Q. 9
.
a
i
d
o
(C) 875
Q. 8
ONE MARK
©
w
w
(ra) 2 + (rw2) 2
(D) rw 2
In the mechanism given below, if the angular velocity of the eccentric circular disc
is 1 rad/s, the angular velocity (rad/s) of the follower link for the instant shown
in the figure is (Note. All dimensions are in mm).
(A) 0.05
(B) 0.1
(C) 5.0
(D) 10.0
A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg,
is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel,
in Joules is
(A) 395
(B) 790
(C) 1580
(D) 3160
GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2012
Q. 11
TWO MARKS
A concentrated mass m is attached at the centre of a rod of length 2L as shown
in the figure. The rod is kept in a horizontal equilibrium position by a spring of
stiffness k . For very small amplitude of vibration, neglecting the weights of the
rod and spring, the undamped natural frequency of the system is
(A)
k
m
(B)
2k
m
(C)
k
2m
(D)
4k
m
YEAR 2011
Q. 12
in
.
o
c
ONE MARK
A double-parallelogram mechanism is shown in the figure. Note that PQ is a
single link. The mobility of the mechanism is
.
a
i
(A) - 1
(C) 1
w
YEAR 2011
Q. 13
©
w
.
w
d
o
n
(B) 0
(D) 2
TWO MARKS
For the four-bar linkage shown in the figure, the angular velocity of link AB
is 1 rad/s. The length of link CD is 1.5 times the length of link AB. In the
configuration shown, the angular velocity of link CD in rad/s is
(A) 3
(C) 1
(B) 3
2
(D) 2
3
GATE SOLVED PAPER - ME
Q. 14
Q. 15
THEORY OF MACHINES
A mass of 1 kg is attached to two identical springs each with stiffness k = 20 kN/m
as shown in the figure. Under the frictionless conditions, the natural frequency of
the system in Hz is close to
(A) 32
(B) 23
(C) 16
(D) 11
A disc of mass m is attached to a spring of stiffness k as shown in the figure
The disc rolls without slipping on a horizontal surface. The natural frequency of
vibration of the system is
(A) 1
2p
k
m
(C) 1
2p
2k
3m
©
w
w
YEAR 2010
.
a
i
d
o
.n
w
n
i
.
o
c
(B) 1
2p
2k
m
(D) 1
2p
3k
2m
ONE MARK
Q. 16
Mobility of a statically indeterminate structure is
(A) #- 1
(B) 0
(C) 1
(D) $ 2
Q. 17
There are two points P and Q on a planar rigid body. The relative velocity
between the two points
(A) should always be along PQ
(B) can be oriented along any direction
(C) should always be perpendicular to PQ
(D) should be along QP when the body undergoes pure translation
Q. 18
Which of the following statements is INCORRECT ?
(A) Grashof’s rule states that for a planar crank-rocker four bar mechanism, the
sum of the shortest and longest link lengths cannot be less than the sum of
the remaining two link lengths
(B) Inversions of a mechanism are created by fixing different links one at a time
(C) Geneva mechanism is an intermittent motion device
(D) Gruebler’s criterion assumes mobility of a planar mechanism to be one
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Q. 19
The natural frequency of a spring-mass system on earth is wn . The natural
frequency of this system on the moon (g moon = g earth /6) is
(A) wn
(B) 0.408wn
(C) 0.204wn
(D) 0.167wn
Q. 20
Tooth interference in an external involute spur gear pair can be reduced by
(A) decreasing center distance between gear pair
(B) decreasing module
(C) decreasing pressure angle
(D) increasing number of gear teeth
YEAR 2010
Q. 21
TWO MARKS
A mass m attached to a spring is subjected to a harmonic force as shown in figure
The amplitude of the forced motion is observed to be 50 mm. The value of m (in
kg) is
.
a
i
(A) 0.1
(B) 1.0
(C) 0.3
(D) 0.5
Q. 22
in
.
o
c
.
w
w
w
d
o
n
For the epicyclic gear arrangement shown in the figure w 2 = 100 rad/s clockwise
(CW) and warm = 80 rad/s counter clockwise (CCW). The angular velocity w 5 (in
rad/s) is
©
(A) 0
(B) 70 CW
(B) 140 CCW
(D) 140 CW
GATE SOLVED PAPER - ME
Q. 23
THEORY OF MACHINES
For the configuration shown, the angular velocity of link AB is 10 rad/s
counterclockwise. The magnitude of the relative sliding velocity (in ms-1 ) of
slider B with respect to rigid link CD is
(A) 0
(B) 0.86
(C) 1.25
n
i
.
o
c
(D) 2.50
.
a
i
YEAR 2009
Q. 24
ONE MARK
A simple quick return mechanism is shown in the figure. The forward to return
ratio of the quick return mechanism is 2:1. If the radius of crank O1 P is 125 mm,
then the distance ‘d ’ (in mm) between the crank centre to lever pivot centre point
should be
d
o
©
.n
w
w
w
(A) 144.3
(B) 216.5
(C) 240.0
(D) 250.0
Q. 25
The rotor shaft of a large electric motor supported between short bearings at
both the ends shows a deflection of 1.8 mm in the middle of the rotor. Assuming
the rotor to be perfectly balanced and supported at knife edges at both the ends,
the likely critical speed (in rpm) of the shaft is
(A) 350
(B) 705
(C) 2810
(D) 4430
GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2009
Q. 26
TWO MARKS
An epicyclic gear train in shown schematically in the given figure. The run gear
2 on the input shaft is a 20 teeth external gear. The planet gear 3 is a 40 teeth
external gear. The ring gear 5 is a 100 teeth internal gear. The ring gear 5 is
fixed and the gear 2 is rotating at 60 rpm CCW (CCW=counter-clockwise and
CW=clockwise).
The arm 4 attached to the output shaft will rotate at
(A) 10 rpm CCW
(B) 10 rpm CW
(C) 12 rpm CW
(D) 12 rpm CCW
Q. 27
.
a
i
(C) 1424
Q. 28
d
o
n
.
w
(D) 955
A vehicle suspension system consists of a spring and a damper. The stiffness of
the spring is 3.6 kN/m and the damping constant of the damper is 400 Ns/m. If
the mass is 50 kg, then the damping factor (d ) and damped natural frequency
(fn), respectively, are
(A) 0.471 and 1.19 Hz
(B) 0.471 and 7.48 Hz
(C) 0.666 and 1.35 Hz
(D) 0.666 and 8.50 Hz
©
Q. 29
in
.
o
c
An automotive engine weighing 240 kg is supported on four springs with linear
characteristics. Each of the front two springs have a stiffness of 16 MN/m while
the stiffness of each rear spring is 32 MN/m. The engine speed (in rpm), at which
resonance is likely to occur, is
(A) 6040
(B) 3020
w
w
Match the approaches given below to perform stated kinematics/dynamics
analysis of machine.
Analysis
Approach
P.
Continuous relative rotation
1.
D’ Alembert’s principle
Q.
Velocity and acceleration
2.
Grubler’s criterion
R.
Mobility
3.
Grashoff’s law
S.
Dynamic-static analysis
4.
Kennedy’s theorem
(A) P-1, Q-2, R-3, S-4
(C) P-2, Q-3, R-4, S-1
(B) P-3, Q-4, R-2, S-1
(D) P-4, Q-2, R-1, S-3
GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2008
Q. 30
ONE MARK
A planar mechanism has 8 links and 10 rotary joints. The number of degrees of
freedom of the mechanism, using Gruebler’s criterion, is
(A) 0
(B) 1
(C) 2
(D) 3
YEAR 2008
Q. 31
Q. 32
TWO MARKS
The natural frequency of the spring mass system shown in the figure is closest to
(B) 10 Hz
(C) 12 Hz
(D) 14 Hz
.n
w
3
(D) - p 3 h sin c pq m
b
b 2
w
w
YEAR 2007
ONE MARK
For an under damped harmonic oscillator, resonance
(A) occurs when excitation frequency is greater than undamped natural
frequency
(B) occurs when excitation frequency is less than undamped natural frequency
(C) occurs when excitation frequency is equal to undamped natural frequency
(D) never occurs
YEAR 2007
Q. 35
d
o
A uniform rigid rod of mass m = 1 kg and length L = 1 m is hinged at its centre
and laterally supported at one end by a spring of spring constant k = 300 N/m .
The natural frequency wn in rad/s is
(A) 10
(B) 20
(C) 30
(D) 40
©
Q. 34
.
a
i
In a cam design, the rise motion is given by a simple harmonic motion
(SHM ) s = h/2 ^1 - cos (pq/b)h where h is total rise, q is camshaft angle, b is the
total angle of the rise interval . The jerk is given by
(A) h c1 - cos pq m
(B) p h sin c pq m
2
b
b2
b
2
(C) p 2 h cos c pq m
b
b 2
Q. 33
n
i
.
o
c
(A) 8 Hz
TWO MARKS
The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the
change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in
kg/m2 is
(A) 0.10
(B) 0.20
(C) 0.30
(D) 0.40
GATE SOLVED PAPER - ME
Q. 36
Q. 37
THEORY OF MACHINES
The natural frequency of the system shown below is
(A)
k
2m
(B)
k
m
(C)
2k
m
(D)
3k
m
The equation of motion of a harmonic oscillator is given by
d 2 x + 2xw dx + w2 x = 0
n
n
dt
dt2
and the initial conditions at t = 0 are x (0) = X, dx (0) = 0 . The amplitude of
dt
x (t) after n complete cycles is
(A) Xe-2npa k
.
a
i
x
1 - x2
(B) Xe2npa
(C) Xe-2np
(D) X
Q. 38
x
1 - x2
a
d
o
n
k
1 - x2
x
in
.
o
c
k
.
w
The input link O2 P of a four bar linkage is rotated at 2 rad/s in counter clockwise
direction as shown below. The angular velocity of the coupler PQ in rad/s, at an
instant when +O 4 O2 P = 180c, is
©
(A) 4
(B) 2 2
(C)1
(D) 1
2
w
w
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Common Data For Q. 39 and 40
A quick return mechanism is shown below. The crank OS is driven at 2 rev/s in
counter-clockwise direction.
Q. 39
If the quick return ratio is 1 : 2, then the length of the crank in mm is
(A) 250
(B) 250 3
a.
i
d
o
(C) 500
Q. 40
n
i
.
o
c
(D) 500 3
The angular speed of PQ in rev/s when the block R attains maximum speed
during forward stroke (stroke with slower speed) is
(A) 1
(B) 2
3
3
(C) 2
©
w
w
YEAR 2006
.n
w
(D) 3
ONE MARK
Q. 41
For a four-bar linkage in toggle position, the value of mechanical advantage is
(A) 0.0
(B) 0.5
(C) 1.0
(D) 3
Q. 42
The differential equation governing the vibrating system is
(A) mxp + cxo + k (x - y) = 0
(B) m (xp - yp) + c (xo - yo) + kx = 0
(C) mxp + c (xo - yo) + kx = 0
(D) m (xp - yp) + c (xo - yo) + k (x - y) = 0
Q. 43
The number of inversion for a slider crank mechanism is
(A) 6
(B) 5
(C) 4
(D) 3
GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2006
Q. 44
TWO MARKS
Match the item in columns I and II
Column I
Q. 45
Q. 46
Column II
P.
Addendum
1.
Cam
Q.
Instantaneous centre of velocity
2.
Beam
R.
Section modulus
3.
Linkage
S.
Prime circle
4.
Gear
(A) P-4, Q-2, R-3, S-1
(B)
P-4, Q-3, R-2, S-1
(C) P-3, Q-2, R-1, S-4
(D)
P-3, Q-4, R-1, S-2
If C f is the coefficient of speed fluctuation of a flywheel then the ratio of wmax /wmin
will be
2 - Cf
1 - 2C f
(B)
(A)
2 + Cf
1 + 2C f
1 + 2C f
2 + Cf
(C)
(D)
1 - 2C f
2 - Cf
Match the items in columns I and II
Column I
P.
Higher Kinematic Pair
Q.
Lower Kinemation Pair
R.
Quick Return Mechanism
S.
Mobility of a Linkage
©
.
w
w
w
(A) P-2, Q-6, R-4, S-3
(C) P-6, Q-2, R-5, S-3
Q. 47
ColumnII
1.
Grubler’s Equation
2.
Line contact
3.
Euler’s Equation
4.
Planar
5.
Shaper
6.
Surface contact
(B) P-6, Q-2, R-4, S-1
(D) P-2, Q-6, R-5, S-1
A machine of 250 kg mass is supported on springs of total stiffness 100 kN/m
. Machine has an unbalanced rotating force of 350 N at speed of 3600 rpm.
Assuming a damping factor of 0.15, the value of transmissibility ratio is
(A) 0.0531
(B) 0.9922
(C) 0.0162
Q. 48
a.
i
d
o
n
in
.
o
c
(D) 0.0028
In a four-bar linkage, S denotes the shortest link length, L is the longest link
length, P and Q are the lengths of other two links. At least one of the three
moving links will rotate by 360c if
(A) S + L # P + Q
(B) S + L > P + Q
(C) S + P # L + Q
(D) S + P > L + Q
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Common Data For Q. 49 and 50
A planetary gear train has four gears and one carrier. Angular velocities of the
gears are w1, w2, w3 and w4 , respectively. The carrier rotates with angular velocity
w5 .
Q. 49
Q. 50
n
i
.
o
c
What is the relation between the angular velocities of Gear 1 and Gear 4 ?
(B) w4 - w5 = 6
(A) w1 - w5 = 6
w4 - w5
w1 - w5
(C) w1 - w2 =-b 2 l
(D) w2 - w5 = 8
w4 - w5
3
w4 - w5 9
.
a
i
d
o
.n
w
For w1 = 60 rpm clockwise (CW) when looked from the left, what is the angular
velocity of the carrier and its direction so that Gear 4 rotates in counterclockwise
(CCW)direction at twice the angular velocity of Gear 1 when looked from the
left ?
(A) 130 rpm, CW
(B) 223 rpm, CCW
(C) 256 rpm, CW
(D) 156 rpm, CCW
©
w
w
Common Data For Q. 51 and 52
A vibratory system consists of a mass 12.5 kg, a spring of stiffness 1000 N/m, and
a dash-pot with damping coefficient of 15 Ns/m.
Q. 51
The value of critical damping of the system is
(A) 0.223 Ns/m
(B) 17.88 Ns/m
(C) 71.4 Ns/m
(D) 223.6 Ns/m
Q. 52
The value of logarithmic decrement is
(A) 1.35
(C) 0.68
(B) 1.32
(D) 0.66
YEAR 2005
Q. 53
ONE MARK
The number of degrees of freedom of a planar linkage with 8 links and 9 simple
revolute joints is
(A) 1
(B) 2
(C) 3
(D) 4
GATE SOLVED PAPER - ME
Q. 54
THEORY OF MACHINES
There are four samples P, Q, R and S, with natural frequencies 64, 96, 128
and 256 Hz, respectively. They are mounted on test setups for conducting
vibration experiments. If a loud pure note of frequency 144 Hz is produced by
some instrument, which of the samples will show the most perceptible induced
vibration?
(A) P
(B) Q
(C) R
(D) S
YEAR 2005
Q. 55
TWO MARKS
In a cam-follower mechanism, the follower needs to rise through 20 mm during
60c of cam rotation, the first 30c with a constant acceleration and then with a
deceleration of the same magnitude. The initial and final speeds of the follower
are zero. The cam rotates at a uniform speed of 300 rpm. The maximum speed
of the follower is
(A) 0.60 m/s
(B) 1.20 m/s
(C) 1.68 m/s
Q. 56
(D) 2.40 m/s
.
a
i
(C) 150 mm
Q. 57
in
.
o
c
A rotating disc of 1 m diameter has two eccentric masses of 0.5 kg each at radii of
50 mm and 60 mm at angular positions of 0c and 150c, respectively. A balancing
mass of 0.1 kg is to be used to balance the rotor. What is the radial position of
the balancing mass ?
(A) 50 mm
(B) 120 mm
d
o
n
(D) 280 mm
In a spring-mass system, the mass is 0.1 kg and the stiffness of the spring is
1 kN/m. By introducing a damper, the frequency of oscillation is found to be 90%
of the original value. What is the damping coefficient of the damper ?
(A) 1.2 Ns/m
(B) 3.4 Ns/m
(C) 8.7 Ns/m
.
w
w
w
(D) 12.0 Ns/m
©
Common Data For Q. 58 to 60
An instantaneous configuration of a four-bar mechanism, whose plane is horizontal
is shown in the figure below. At this instant, the angular velocity and angular
acceleration of link O 2 A are w = 8 rad/s and a = 0 , respectively, and the driving
torque (t ) is zero. The link O 2 A is balanced so that its centre of mass falls at O 2 .
GATE SOLVED PAPER - ME
Q. 58
THEORY OF MACHINES
Which kind of 4-bar mechanism is O 2 ABO 4 ?
(A) Double-crank mechanism
(B) Crank-rocker mechanism
(C) Double-rocker mechanism
(D) Parallelogram mechanism
Q. 59
At the instant considered, what is the magnitude of the angular velocity of O 4 B ?
(A) 1 rad/s
(B) 3 rad/s
(C) 8 rad/s
(D) 64 rad/s
3
Q. 60
At the same instant, if the component of the force at joint A along AB is 30 N,
then the magnitude of the joint reaction at O 2
(A) is zero
(B) is 30 N
(C) is 78 N
n
i
.
o
c
(D) cannot be determined from the given data
YEAR 2004
Q. 61
For a mechanism shown below, the mechanical advantage for the given configuration is
d
o
.n
w
(A) 0
(C) 1.0
Q. 62
.
a
i
w
(B) 0.5
(D) 3
A vibrating machine is isolated from the floor using springs. If the ratio of
excitation frequency of vibration of machine to the natural frequency of the
isolation system is equal to 0.5, then transmissibility ratio of isolation is
(A) 1/2
(B) 3/4
©
w
(C) 4/3
(D) 2
YEAR 2004
Q. 63
ONE MARK
TWO MARKS
The figure below shows a planar mechanism with single degree of freedom. The
instant centre 24 for the given configuration is located at a position
(A) L
(C) N
(B) M
(D) 3
GATE SOLVED PAPER - ME
Q. 64
Q. 65
THEORY OF MACHINES
A uniform stiff rod of length 300 mm and having a weight of 300 N is pivoted at
one end and connected to a spring at the other end. For keeping the rod vertical
in a stable position the minimum value of spring constant k needed is
(A) 300 N/m
(B) 400 N/m
(C) 500 N/m
(D) 1000 N/m
in
.
o
c
Match the following
Type of Mechanism
P.
Motion achieved
1.
Intermittent Motion
2.
Quick return Motion
R. Off-set slider-crank Mechanism
3.
Simple Harmonic Motion
S.
4.
Straight Line Motion
Scott-Russel Mechanism
.
a
i
Q. Geneva Mechanism
(A)
(C)
Q. 66
P-2
P-4
Q-3
Q-1
R-1
R-2
S-4
S-3
.n
w
(B)
(D)
P-3
P-4
Q-2
Q-3
R-4
R-1
S-1
S-2
In the figure shown, the relative velocity of link 1 with respect to link 2 is
12 m/sec . Link 2 rotates at a constant speed of 120 rpm. The magnitude of
Coriolis component of acceleration of link 1 is
©
w
w
(A) 302 m/s2
(C) 906 m/s2
Q. 67
od
Scotch Yoke Mechanism
(B) 604 m/s2
(D) 1208 m/s2
Match the following with respect to spatial mechanisms.
Types of Joint
P.
Revolute
Degree of constraints
1.
Three
Q. Cylindrical
2.
Five
R. Spherical
3.
Four
4.
Two
5.
Zero
GATE SOLVED PAPER - ME
(A)
(C)
P-1
P-2
Q-3
Q-3
THEORY OF MACHINES
R-3
R-1
(B)
(D)
P-5
P-4
Q-4
Q-5
R-3
R-3
Common Data For Q. 68 and 69
A compacting machine shown in the figure below is used to create a desired
thrust force by using a rack and pinion arrangement. The input gear is mounted
on the motor shaft. The gears have involute teeth of 2 mm module.
.
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Q. 68
If the drive efficiency is 80%, the torque required on the input shaft to create 1000
N output thrust is
(A) 20 Nm
(B) 25 Nm
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(C) 32 Nm
Q. 69
(D) 50 Nm
If the pressure angle of the rack is 20c, then force acting along the line of action
between the rack and the gear teeth is
(A) 250 N
(B) 342 N
(C) 532 N
Q. 70
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(D) 600 N
A mass M , of 20 kg is attached to the free end of a steel cantilever beam of
length 1000 mm having a cross-section of 25 # 25 mm. Assume the mass of the
cantilever to be negligible and E steel = 200 GPa. If the lateral vibration of this
system is critically damped using a viscous damper, then damping constant of
the damper is
(A) 1250 Ns/m
(C) 312.50 Ns/m
(B) 625 Ns/m
(D) 156.25 Ns/m
GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2003
Q. 71
ONE MARK
The mechanism used in a shaping machine is
(A) a closed 4-bar chain having 4 revolute pairs
(B) a closed 6-bar chain having 6 revolute pairs
(C) a closed 4-bar chain having 2 revolute and 2 sliding pairs
(D) an inversion of the single slider-crank chain
Q. 72
The lengths of the links of a 4-bar linkage with revolute pairs are p, q, r, and s
units. given that p < q < r < s . Which of these links should be the fixed one, for
obtaining a “double crank” mechanism ?
(A) link of length p
(B) link of length q
(C) link of length r
(D) link of length s
Q. 73
When a cylinder is located in a Vee-block, the number of degrees of freedom
which are arrested is
(A) 2
(B) 4
(C) 7
(D) 8
in
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YEAR 2003
Q. 74
Q. 75
TWO MARKS
For a certain engine having an average speed of 1200 rpm, a flywheel approximated
as a solid disc, is required for keeping the fluctuation of speed within 2% about
the average speed. The fluctuation of kinetic energy per cycle is found to be 2 kJ.
What is the least possible mass of the flywheel if its diameter is not to exceed 1
m?
(A) 40 kg
(B) 51 kg
(C) 62 kg
(D) 73 kg
.
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A flexible rotor-shaft system comprises of a 10 kg rotor disc placed in the middle
of a mass-less shaft of diameter 30 mm and length 500 mm between bearings
(shaft is being taken mass-less as the equivalent mass of the shaft is included
in the rotor mass) mounted at the ends. The bearings are assumed to simulate
simply supported boundary conditions. The shaft is made of steel for which the
value of E 2.1 # 1011 Pa. What is the critical speed of rotation of the shaft ?
(A) 60 Hz
(B) 90 Hz
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(C) 135 Hz
(D) 180 Hz
Common Data For Q. 76 and 77
The circular disc shown in its plan view in the figure rotates in a plane parallel
to the horizontal plane about the point O at a uniform angular velocity w. Two
other points A and B are located on the line OZ at distances rA and rB from O
respectively.
GATE SOLVED PAPER - ME
Q. 76
THEORY OF MACHINES
The velocity of Point B with respect to point A is a vector of magnitude
(A) 0
(B) w (rB - rA) and direction opposite to the direction of motion of point B
(C) w (rB - rA) and direction same as the direction of motion of point B
(D) w (rB - rA) and direction being from O to Z
Q. 77
The acceleration of point B with respect to point A is a vector of magnitude
(A) 0
(B) w (rB - rA) and direction same as the direction of motion of point B
(C) w2 (rB - rA) and direction opposite to be direction of motion of point B
(D) w2 (rB - rA) and direction being from Z to O
Q. 78
Q. 79
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The undamped natural frequency of oscillations of the bar about the hinge point
is
(A) 42.43 rad/s
(B) 30 rad/s
(C) 17.32 rad/s
(D) 14.14 rad/s
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The damping coefficient in the vibration equation is given by
(A) 500 Nms/rad
(B) 500 N/(m/s)
(C) 80 Nms/rad
w
(D) 80 N/(m/s)
w
YEAR 2002
©
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ONE MARK
Q. 80
The minimum number of links in a single degree-of-freedom planar mechanism
with both higher and lower kinematic pairs is
(A) 2
(B) 3
(C) 4
(D) 5
Q. 81
The Coriolis component of acceleration is present in
(A) 4 bar mechanisms with 4 turning pairs
(B) shape mechanism
(C) slider-crank mechanism
(D) scotch yoke mechanism
YEAR 2002
Q. 82
TWO MARKS
If the length of the cantilever beam is halved, the natural frequency of the mass
M at the end of this cantilever beam of negligible mass is increased by a factor of
(A) 2
(B) 4
(C)
8
(D) 8
GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2001
Q. 83
ONE MARK
For a spring-loaded roller follower driven with a disc cam,
(A) the pressure angle should be larger during rise than that during return for
ease of transmitting motion.
(B) the pressure angle should be smaller during rise than that during return for
ease of transmitting motion.
(C) the pressure angle should be large during rise as well as during return for
ease of transmitting motion.
(D) the pressure angle does not affect the ease of transmitting motion.
Q. 84
In the figure shown, the spring deflects by d to position A (the equilibrium
position) when a mass m is kept on it. During free vibration, the mass is at
position B at some instant. The charge in potential energy of the spring mass
system from position A to position B is
in
.
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(A) 1 kx2
2
(B) 1 kx2 - mgx
2
(C) 1 k (x + d) 2
2
(D) 1 kx2 + mgx
2
Q. 85
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Which of the following statements is correct ?
(A) Flywheel reduces speed fluctuations during a cycle for a constant load, but
flywheel does not control the mean speed of the engine, if the load changes.
(B) Flywheel does not reduce speed fluctuation during a cycle for a constant
load, but flywheel does not control the mean speed of the engine, if the load
changes.
(C) Governor controls speed fluctuations during a cycle for a constant load, but
governor does not control the mean speed of the engine, if the load changes.
(D) Governor controls speed fluctuations during a cycle for a constant load, and
governor also controls the mean speed of the engine, if the load changes.
GATE SOLVED PAPER - ME
THEORY OF MACHINES
YEAR 2001
Q. 86
TWO MARKS
The sun gear in the figure is driven clockwise at 100 rpm. The ring gear is held
stationary. For the number of teeth shown on the gears, the arm rotates at
(A) zero
(C) 33.33 rpm
Q. 87
(B) 20 rpm
(D) 66.67 rpm
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The assembly shown in the figure is composed of two massless rods of length L
with two particles, each of mass m . The natural frequency of this assembly for
small oscillations is
.
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(A)
(C)
g
L
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g
(L cos a)
(B)
(D)
*********
2g
(L cos a)
(g cos a)
L
GATE SOLVED PAPER - ME
THEORY OF MACHINES
SOLUTION
Sol. 1
Sol. 2
Option (C) is correct.
Absolute acceleration of given link at location A is
2
2
a = a radial
+ a tangential
Where aradial = w2 r = w2 # OA
= 2 2 # 1 = 4 m /s 2
atangential = 2vw = 2 # 0.75 # 2 = 3 m/s2
So that from equation (i), we get
a = 42 + 32 = 5 m/s
...(i)
Option (B) is correct.
For helical gears, speed ratio is given by
cos b2
N1 = D 2
D1 # cos b1
N2
in
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...(i)
N1 = 1440 rpm, D1 = 80 mm, D2 = 120 mm , b1 = 30c, b 2 = 22.5c
Hence from Eq. (i)
cos b1
cos 30c
N = 80
N 2 = D1 #
1440
D2
cos b 2 # 1 120 # cos 22.5c #
.
a
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= 899.88 - 900 rpm
Sol. 3
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Option (C) is correct.
©
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As given in problem, a planer closed kinematic chain is shown in above figure.
In given figure, if the link opposite to the shortest link, i.e. link RS = 2.5 m is
fixed and the shortest link PQ is made a coupler, the other two links QR and
SP would oscillate. The mechanism is known as rocker-rocker or double-rocker or
double lever or oscillating-oscillating converter.
Sol. 4
Option (B) is correct.
We have
ZP = 20 , ZQ = 40 , ZR = 15 , ZS = 20 , DQ = 2DR, mR = 2 mm
module
mR = DR
ZR
or
DR = 2 # 15 = 30 mm
and
DQ = 2 # 30 = 60 mm
Also
DP = ZP
DQ
ZQ
GATE SOLVED PAPER - ME
THEORY OF MACHINES
DP = 20 # 60 = 30 mm
40
Z
Again
DS = S # D R
ZR
= 20 # 30 = 40 mm
15
Thus the centre distance between P and S is
D
dPS = DP + Q + DR + DS
2
2
2
2
= 15 + 30 + 15 + 20 = 80 mm
or
Sol. 5
Option (A) is correct.
We have E = 400 N-m , w = 20 rad/ sec , Cs = 0.04
The energy of flywheel is given by
E = Iw 2 C s
Sol. 6
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400
I = E
=
= 25 kg-m2
2
w 2 Cs
20
0
.
04
^ h#
or
Option (C) is correct.
Given
m = 1 kg , k = 10 kN/m , u = 0 , F = 5 kN , t = 10-4 sec
Here the time duration is very less.
The acceleration is calculated by
F = ma
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a = F = 5000 = 5000 m/ sec2
m
1
Also from law of motion
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v = u + at
w
v = 0 + 5000 # 10-4
= 0.5 m/ sec
Now equating the energies of the system and spring by using conservation law.
1 mv2 = 1 kx2
2
2
2
1 # ^0.5h2
or
x2 = mv =
10000
k
-3
or
x = 5 # 10 m = 5 mm
©
Sol. 7
w
Option (A) is correct.
Since two nodes are observed at frequency of 1800 rpm, therefore third critical
speed of shaft
f3 = 1800 rpm
because two nodes can be observed only in 3 rd mode.
The whirling frequency of shaft
g
f = p # n2
2
d
where
n = Numbers of mode
g
For n = 1,
f1 = p
2 d
Therefore
fn = n2 # f1
or
f3 = 32 # f1
f
or
f1 = 3 2 = 1800 = 200 rpm
9
]3g
GATE SOLVED PAPER - ME
Sol. 8
THEORY OF MACHINES
Option (D) is correct.
For A solid disc of radius (r) as given in figure, rolls without slipping on a
horizontal floor with angular velocity w and angular acceleration a.
The magnitude of the acceleration of the point of contact (A) on the disc is only
by centripetal acceleration because of no slip condition.
...(i)
v = wr
Differentiating Eq. (1) w.r.t. (t)
dw
dv
dv = r dw = r : a
b dt = a, dt = a l
dt
dt
or,
a = r:a
Instantaneous velocity of point A is zero
So at point A, Instantaneous tangential acceleration = zero
Therefore only centripetal acceleration is there at point A.
.
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Centripetal acceleration = rw 2
Sol. 9
Option (B) is correct.
From similar DPQO and DSRO
PQ
= PO
SR
SO
od
©
Sol. 10
PQ =
(50) 2 - (25) 2 = 43.3 mm
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From Eq. (i)
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43.3 = 50
5
SR
SR = 43.5 # 5 = 4.33 mm
50
Velocity of Q = Velocity of R (situated at the same link)
VQ = VR = SR # w = 4.33 # 1 = 4.33 m/s
V
Angular velocity of PQ .
wPQ = Q = 4.33 = 0.1 rad/s
43.3
PQ
Option (B) is correct.
For flywheel
K.E = 1 Iw 2
2
w = 2pN = 2 # p # 600 = 62.83 rad/s
60
60
...(i)
GATE SOLVED PAPER - ME
Sol. 11
THEORY OF MACHINES
I (for solid circular disk) = 1 mR2 = 1 # 20 # (0.2) 2 = 0.4 kg - m2
2
2
Hence,
K.E = 1 # ^0.4h # ^62.83h2 = 789.6 - 790 Joules .
2
Option (D) is correct.
For a very small amplitude of vibration.
From above figure change in length of spring
x = 2L sin q = 2Lq
Mass moment of inertia of mass (m) about O is
(is very small so sin q - q)
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I = mL2
As no internal force acting on the system. So governing equation of motion from
Newton’s law of motion is,
Iqp + kx # 2L = 0
.
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mL2 qp + k 2Lq # 2L = 0
2
qp + 4kL 2q = 0
mL
or
qp + 4kq = 0
m
p
Comparing general equation q + wn2 q = 0 we have
w n2 = 4k & wn =
m
Option (C) is correct.
or,
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Sol. 12
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w
w
4k
m
Given that PQ is a single link.
Hence : l = 5 , j = 5 , h = 1
It has been assumed that slipping is possible between the link l5 & l1 . From the
kutzbach criterion for a plane mechanism,
Numbers of degree of freedom or movability.
n = 3 (l - 1) - 2j - h = 3 (5 - 1) - 2 # 5 - 1 = 1
Sol. 13
Option (D) is correct.
Given wAB = 1 rad/ sec , lCD = 1.5lAB
Let angular velocity of link CD is wCD
From angular velocity ratio theorem,
wAB = lCD
wCD
lAB
& lCD = 1.5
lAB
GATE SOLVED PAPER - ME
THEORY OF MACHINES
wCD = wAB # lAB = 1 # 1 = 2 rad/ sec
3
1.5
lCD
Sol. 14
Option (A) is correct.
Given k = 20 kN/m , m = 1 kg
From the Given spring mass system, springs are in parallel combination. So,
keq = k + k = 2k
Natural Frequency of spring mass system is,
keq
wn =
m
2pfn =
keq
m
fn = Natural Frequency in Hz.
keq
2k =
1
fn = 1
= 1
2p m
2p m
2 # 3.14
= 200 = 31.84 Hz - 32 Hz
6.28
Sol. 15
2 # 20 # 1000
1
Option (C) is correct.
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q = x & x = rq
r
Total energy of the system remains constant.
So,
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T.E. = K.E. due to translatory motion
+ K.E. due to rotary motion + P.E. of spring
T.E. = 1 mxo2 + 1 Iqo2 + 1 kx2
2
2
2
From equation (i) xo = rqo
= 1 mr2 qo2 + 1 Iqo2 + 1 kr2 q 2
2
2
2
w
w
...(i)
= 1 mr2 qo2 + 1 # 1 mr2 qo2 + 1 kr2 q 2
2
2
2
2
= 3 mr2 qo2 + 1 kr2 q2 = Constant
2
4
For a disc I = mr
2
On differentiating above equation w.r.t. t , we get
3 mr2 (2qq
1 2
# op) + 2 kr (2qqo) = 0
4
3 mr2 qp + kr2 q = 0
2
qp + 2k q = 0
3m
wn2 = 2k &
3m
wn =
Therefore, natural frequency of vibration of the system is,
2k
fn = wn = 1
2p 3m
2p
2k
3m
2
GATE SOLVED PAPER - ME
Sol. 16
THEORY OF MACHINES
Option (A) is correct.
Given figure shows the six bar mechanism.
We know movability or degree of freedom is n = 3 (l - 1) - 2j - h
The mechanism shown in figure has six links and eight binary joints (because
there are four ternary joints A, B, C & D , i.e. l = 6, j = 8 h = 0
So,
n = 3 (6 - 1) - 2 # 8 =- 1
Therefore, when n =- 1 or less, then there are redundant constraints in the chain,
and it forms a statically indeterminate structure. So, From the Given options (A)
satisfy the statically indeterminate structure n #- 1
Sol. 17
Option (C) is correct.
.
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Velocity of any point on a link with respect to another point (relative velocity)
on the same link is always perpendicular to the line joining these points on the
configuration (or space) diagram.
vQP = Relative velocity between P & Q
= vP - vQ always perpendicular to PQ.
Sol. 18
Option (A) is correct.
According to Grashof’s law “For a four bar mechanism, the sum of the shortest
and longest link lengths should not be greater than the sum of remaining two link
lengths if there is to be continuous relative motion between the two links.
l 4 + l 2 $ l1 + l 3
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Sol. 19
Option (A) is correct.
We know natural frequency of a spring mass system is,
k
...(i)
wn =
m
This equation (i) does not depend on the g and weight (W = mg)
So, the natural frequency of a spring mass system is unchanged on the moon.
Hence, it will remain wn , i.e. wmoon = wn
Sol. 20
Option ( D) is correct.
When gear teeth are produced by a generating process, interference is automatically
eliminated because the cutting tool removes the interfering portion of the flank.
This effect is called undercutting. By undercutting the undercut tooth can be
considerably weakened.
So, interference can be reduced by using more teeth on the gear. However, if the
gears are to transmit a given amount of power, more teeth can be used only by
increasing the pitch diameter.
Sol. 21
Option (A) is correct.
in
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Given k = 3000 N/m , c = 0 , A = 50 mm , F (t) = 100 cos (100t) N
wt = 100t
It is a forced vibratory system.
w = 100
From the Newton’s law,
...(i)
mxp + kx = F
And its general solution will be,
©
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w
.
w
x = A cos wt
dx = xo =- Aw sin wt
dt
where w =
k
m
d2 x = xp =- Aw2 cos wt
dt2
Substitute these values in equation (i), we get
- mAw2 cos wt + kA cos wt = 100 cos (wt)
- mAw2 + kA = 100
Now substitute k = 3000 N/m , A = 0.05 m , in above equation, we get
- m # 0.05 # (100) 2 + 3000 # 0.05 = 100
- 5m + 1.5 = 1
m = 0.1 kg
Alternate Method:
We know that, in forced vibration amplitude is given by :
FO
A=
2
2
^k - mwh + (cw)
Here, F (t) = 100 cos (100t), FO = 100 N , A = 50 mm = 50 # 10-3 m
...(i)
GATE SOLVED PAPER - ME
THEORY OF MACHINES
w = 100 rad/ sec , k = 3000 Nm-1 , c = 0
So, from equation (i), we get
A = FO 2
k - mw
k - mw2 = FO
A
100
3000 - m # (100) 2 =
50 # 10-3
10000m = 1000
Sol. 22
& m = 0.1 kg
Option (C) is correct.
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Given Ni = No. of teeth for gear i ,
N2 = 20 , N 3 = 24 , N 4 = 32 , N5 = 80 , w2 = 100 rad/ sec (CW)
warm = 80 rad/ sec (CCW) =- 80 rad/ sec
The table of the motion given below :
Take CCW =- ve and CW =+ ve
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S. Condition of Motion
No.
©
Revolution of elements
Arm Gear Compound
2 w2 Gear 3 - 4,
w3 = w4
Gear 5
w5
1.
Arm ‘a ’ is fixed & Gear 2 0
rotates through + 1 revolution
(CW)
+1
- N2
N3
- N2 # N 4
N3
N5
2.
Gear 2 rotates through + x 0
revolution (CW)
+x
- x N2
N3
- x N2 # N 4
N3
N5
3.
Add + y revolutions to all + y
elements
+y
+y
+y
4.
Total motion.
Note.
i.e.
+y
x + y y - x N2
N3
y - x N2 # N 4
N3
N5
Speed of driver
= No.of teeth on driven
Speed of driven No. of teeth on driver
w1 = N2
w2
N1
Speed ratio =
Gear 3 & 4 mounted on same shaft, So w3 = w4
And
warm = y
y =- 80 rad/ sec (CCW)
From the table
GATE SOLVED PAPER - ME
THEORY OF MACHINES
From the table
x + y = w2 = 100
x = 100 - (- 80) = 180 rad/ sec (CW)
And
From the table
w5 = y - x # N2 # N 4
N3
N5
=- 80 - 180 # 20 # 32 =- 140 rad/ sec
80
24
Negative sign shows the counter clockwise direction.
Sol. 23
Option (D) is correct.
Let, vB is the velocity of slider B relative to link CD
The crank length AB = 250 mm and velocity of slider B with respect to rigid link
CD is simply velocity of B (because C is a fixed point).
Hence,
vB = (AB) # wAB = 250 # 10-3 # 10 = 2.5 m/ sec
Alternate Method :
From the given figure, direction of velocity of CD is perpendicular to link AB &
direction of velocity of AB is parallel to link CD.
So, direction of relative velocity of slider B with respect to C is in line with link
BC.
= 2.5 m/ sec
Sol. 24
.
a
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Option (D) is correct.
©
in
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vC = 0
vBC = vB - vC = AB # wAB - 0 = 0.025 # 10
Hence
Or
.
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Given O1 P = r = 125 mm
Forward to return ratio = 2 : 1
Time of cutting (forward) stroke
b
We know that,
= = 360 - a
a
a
Time of return stroke
Substitute the value of Forward to return ratio, we have
2 = 360 - a
a
1
2a = 360 - a
& a = 120c
a
120
c
And angle
RO1 O2 = =
= 60c
2
2
Now we are to find the distance ‘d ’ between the crank centre to lever pivot centre
point (O1 O2). From the DRO2 O1
sin a90c - a k = O1 R = r
2
O1 O 2 O1 O 2
sin (90c - 60c) = r
O1 O 2
GATE SOLVED PAPER - ME
THEORY OF MACHINES
O1 O 2 =
Sol. 25
r
= 125 = 250 mm
sin 30c 1/2
Option (B) is correct.
Given d = 1.8 mm = 0.0018 m
The critical or whirling speed is given by,
g
wc =
d
2pNc = g
60
d
g
60
=
Nc = 60
2p d
2 # 3.14
NC = Critical speed in rpm
9.81
0.0018
= 9.55 5450 = 704.981 - 705 rpm
Sol. 26
Option (A) is correct.
Given Z2 = 20 Teeth , Z 3 = 40 Teeth , Z5 = 100 Teeth , N5 = 0 ,
N2 = 60 rpm (CCW)
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If gear 2 rotates in the CCW direction, then gear 3 rotates in the clockwise
direction. Let, Arm 4 will rotate at N 4 rpm. The table of motions is given below.
Take CCW =+ ve , CW =- ve
©
w
S.
Condition of Motion
No.
Revolution of elements
Sun
Planet
Gear 2 Gear 3
Arm 4 Ring Gear 5
N2
N3
N4
N5
1.
Arm fixed and sun gear 2 + 1
rotates + 1 rpm (CCW)
- Z2
Z3
0
- Z2 # Z 3
Z3
Z5
2.
Give + x rpm to gear 2 + x
(CCW)
- Z2 x
Z3
0
- x Z2
Z5
3.
Add + y revolutions to + y
all elements
+y
+y
+y
4.
Total motion.
y - x Z2
Z3
+y
y - x Z2
Z5
Note :
y+x
Speed ratio =
Speed of driver
= No. of teeth on dirven
Speed of driven
No. of teeth on driver
Ring gear 5 is fixed. So,
N5 = 0
GATE SOLVED PAPER - ME
THEORY OF MACHINES
y - x Z2 = 0
Z5
y = Z2 x = 20 x = x
5
100
Z5
From the table
...(i)
N2 = 60 rpm (CCW)
y + x = 60
x + x = 60
5
Given,
From table
x = 10 # 5 = 50 rpm
And from equation (i),
y = 50 = 10 rpm (CCW)
5
From the table the arm will rotate at
N 4 = y = 10 rpm (CCW)
Sol. 27
Option (A) is correct.
in
.
o
c
.
a
i
d
o
n
.
w
Given k1 = k2 = 16 MN/m , k 3 = k 4 = 32 MN/m , m = 240 kg
Here, k1 & k2 are the front two springs or k 3 and k 4 are the rear two springs.
These 4 springs are parallel, So equivalent stiffness
w
w
keq = k1 + k2 + k 3 + k 4 = 16 + 16 + 32 + 32 = 96 MN/m2
k
We know at resonance
w = wn =
m
©
2pN =
60
keq
m
N =Engine speed in rpm
6
keq
= 60 96 # 10
N = 60
2p m
2p
240
= 60 # 102 # 40 = 6042.03 - 6040 rpm
2p
Sol. 28
Option (A) is correct.
Given k = 3.6 kN/m , c = 400 Ns/m , m = 50 kg
We know that, Natural Frequency
k = 3.6 # 1000 = 8.485 rad/ sec
wn =
50
m
And damping factor is given by,
400
d or e = c = c
=
cc 2 km
2 # 3.6 # 1000 # 50
400
= 0.471
=
2 # 424.26
...(i)
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Damping Natural frequency,
wd =
fd = wd
2p
fd = wn # 1 - e2 = 8.485 # 1 - (0.471) 2 = 1.19 Hz
2p
2 # 3.14
Option (B) is correct.
2pfd =
Sol. 29
1 - e2 wn
1 - e2 wn
Analysis
Approach
P.
Continuous relative rotation
3.
Grashoff law
Q.
Velocity and Acceleration
4.
Kennedy’s Theorem
R.
Mobility
2.
Grubler’s Criterion
S.
Dynamic-static Analysis
1.
D’Alembert’s Principle
n
i
.
o
c
So, correct pairs are P-3, Q-4, R-2, S-1
Sol. 30
Option (B) is correct.
From Gruebler’s criterion, the equation for degree of freedom is given by,
Given l = 8 and
.
a
i
n = 3 (l - 1) - 2j - h
j = 10 , h = 0
d
o
n = 3 (8 - 1) - 2 # 10 = 1
Sol. 31
Sol. 32
.n
w
...(i)
from equation(i)
Option (B) is correct.
Given m = 1.4 kg , k1 = 4000 N/m , k2 = 1600 N/m
In the given system k1 & k2 are in parallel combination
So,
keq = k1 + k2 = 4000 + 1600 = 5600 N/m
Natural frequency of spring mass system is given by,
keq
5600 = 1
= 1
63.245 = 10.07 - 10 Hz
fn = 1
2p m
2p
2p #
1.4
Option (D) is correct.
Jerk is given by triple differentiation of s w.r.t. t ,
3
Jerk = d s3
dt
p (wt)
Given
s = h c1 - cos pq m = h ;1 - cos
q = wt
2
2
b
b E
Differentiating above equation w.r.t. t , we get
ds = h - pw - sin p (wt)
2= b '
dt
b 1G
Again Differentiating w.r.t. t ,
d 2 s = h p2 w2 cos p (wt)
2 b2 ;
b E
dt2
Again Differentiating w.r.t. t ,
d 3 s =- h p3 w3 sin pq
2 b3
b
dt3
Let w = 1 rad/ sec
d 3 s =- h p3 sin pq
cb m
2 b3
dt3
©
w
w
GATE SOLVED PAPER - ME
Sol. 33
THEORY OF MACHINES
Option (C) is correct.
Given m = 1 kg , L = 1 m , k = 300 N/m
We have to turn the rigid rod at an angle q about its hinged point, then rod
moves upward at a distance x and also deflect in the opposite direction with the
same amount. Let q is very very small and take tan q b q
q = x & x= Lq
2
L /2
From DAOB ,
in
.
o
c
:
and
q = wt & q = w
By using the principal of energy conservation,
1 Iw2 + 1 kx2 = Constant
2
2
1 Iqo2 + 1 k L q 2 = c
2
2 b2 l
.
a
i
d
o
n
...(i)
...(ii)
From equation (i) and (ii)
1 Iqo2 + 1 L2 kq 2 = c
2
8
On differentiating w.r.t. t , we get
2
1 I 2qq
op + kL # 2qqo = 0
...(iii)
2 #
8
For a rigid rod of length L & mass m , hinged at its centre, the moment of inertia,
2
I = mL
12
Substitute I in equation (iii), we get
2
1 mL2 2qq
op + kL qqo = 0
#
#
2
12
4
©
w
w
.
w
qp + 3k q = 0
m
Compare equation (iv) with the general equation,
qp + wn2 q = 0
So, we have
wn2 = 3k
m
wn =
Sol. 34
...(iv)
3k =
m
3 # 300 = 30 rad/ sec
1
Option (C) is correct.
For an under damped harmonic oscillator resonance occurs when excitation
frequency is equal to the undamped natural frequency
wd = wn
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Sol. 35
Option (A) is correct.
Given w1 = 210 rad/ sec , w2 = 190 rad/ sec , DE = 400 Nm
As the speed of flywheel changes from w1 to w2 , the maximum fluctuation of
energy,
DE = 1 I 6(w1) 2 - (w2) 2@
2
2DE
2 # 400
I =
2
2
2
2 =
6(w1) - (w2) @ 6(210) - (190) @
800
=
= 0.10 kgm2
400 # 20
Sol. 36
Option (A) is correct.
n
i
.
o
c
d
o
.
a
i
The springs, with stiffness k & k are in parallel combination. So their resultant
2
2
stiffness will be,
k1 = k + k = k
2 2
As k1 & k are in series, so the resultant stiffness will be,
2
keq = k # k = k = k
2
k+k
2k
The general equation of motion for undamped free vibration is given as,
mxp + keq x = 0
mxp + k x = 0
2
xp + k x = 0
2m
Compare above equation with general equation xp + wn2 x = 0 , we get
Natural frequency of the system is,
k
wn2 = k & wn =
2m
2m
©
.n
w
w
w
Alternative :
Sol. 37
keq = k
2
We know, for a spring mass system,
k/2
keq
k
wn =
=
=
m
m
2m
Option (A) is correct.
Given The equation of motion of a harmonic oscillator is
d 2 x + 2xw dx + w2 x = 0
n
n
dt
dt2
...(i)
GATE SOLVED PAPER - ME
THEORY OF MACHINES
xp + 2xwn xo + wn2 x = 0
Compare equation (i) with the general equation,
mxp + cxo + kx = 0
xp + c xo + k x = 0
m
m
c = 2xw
We get,
n
m
k = w2 ,
n
m
x =
From equation (ii) & (iii),
...(ii)
& wn =
c
2m #
k
m
=
k
m
...(iii)
c
2 km
Logarithmic decrement,
2pc
c c2 - c2
2p # 2x km
= ln a x1 k =
=
x2
(2 km ) 2 - (2x km ) 2
2px
=
1 - x2
d = ln a x1 k =
x2
...(iv)
4px km
4km - 4x 2 km
in
.
o
c
x1 = e 2px
1-x
x2
If system executes n cycles, the logarithmic decrement d can be written as
d = 1 loge x1
n
xn + 1
end = x1
xn + 1
2
.
a
i
Where
d
o
n
x1 = amplitude at the starting position.
xn + 1 = Amplitude after n cycles
The amplitude of x (t) after n complete cycles is,
end = X
x (t)
Sol. 38
©
w
w
.
w
x (t) = e-nd # X = Xe-
Option (C) is correct.
Given, O 4 O2 P = 180c, wO P = 2 rad/sec
The instantaneous centre diagram is given below,
2
Let, velocity of point P on link O2 P is VP ,
n2px
1 - x2
From equation (iv)
GATE SOLVED PAPER - ME
THEORY OF MACHINES
VP = wO P # O2 P = wO P # (I12 I23) = 2a
And P is also a point on link QP ,
So,
VP = wPQ # O 4 P = wPQ # (I13 I23)
...(i)
= wPQ # 2a
Both the links O2 P and QP are runs at the same speed
From equation (i) and (ii), we get
...(ii)
2
2a = wPQ # 2a
wPQ = 1 rad/sec
or,
Sol. 39
2
Option (A) is correct.
.
a
i
d
o
©
.n
w
n
i
.
o
c
w
w
Given Quick return ratio = 1: 2 , OP = 500 mm
Here OT = Length of the crank. We see that the angle b made by the forward
stroke is greater than the angle a described by the return stroke.
Since the crank has uniform angular speed, therefore
Quick return ratio = Time of return stroke
Time of cutting stroke
a
1=a =
360 - a
2 b
and Angle
From the DTOP ,
360 - a = 2a
3a = 360
a = 120c
TOP = a = 120 = 60c
2
2
cos a = OT = r
500
2
OP
cos 60c = r
500
r = 500 # 1 = 250 mm
2
OT = r
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Sol. 40
Option (B) is correct.
We know that maximum speed during forward stroke occur when QR & QP are
perpendicular.
So,
V = OS # wOS = PQ # wPQ
V = rw
250 # 2 = 750 # wPQ
wPQ = 500 = 2 rad/ sec
750 3
Sol. 41
Option (D) is correct.
velocity ratio theorem
M.A = T4 = w 2 = RPD
w4
T2
RPA
in
.
o
c
from angular
Construct BlA and C lD perpendicular to the line PBC . Also, assign lables b
and g to the acute angles made by the coupler.
RPD = RClD = RCD sin g
RBlA RBA sin b
RPA
R sin g
So,
M.A. = T4 = w2 = CD
T2 w4 RBA sin b
When the mechanism is toggle,then b = 0c and 180c.
So
M.A = 3
.
a
i
Sol. 42
d
o
n
w
.
w
Option (C) is correct.
Assume any arbitrary relationship between the coordinates and their first
derivatives, say x > y and xo > yo. Also assume x > 0 and xo > 0 .
A small displacement gives to the system towards the left direction. Mass m is
fixed, so only damper moves for both the variable x and y .
Note that these forces are acting in the negative direction.
©
w
Differential equation governing the above system is,
2
dy
- kx = 0
/ F =- m ddtx2 - cb dx
dt dt l
mxp + c (xo - yo) + kx = 0
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Sol. 43
Option (C) is correct.
For a 4 bar slider crank mechanism, there are the number of links or inversions
are 4. These different inversions are obtained by fixing different links once at
a time for one inversion. Hence, the number of inversions for a slider crank
mechanism is 4.
Sol. 44
Option (B) is correct.
Column I
Column II
P.
Addendum
4.
Gear
Q.
Instantaneous centre of velocity
3.
Linkage
R.
Section modulus
2.
Beam
S.
Prime circle
1.
Cam
So correct pairs are, P-4, Q-3, R-2, S-1
Sol. 45
.
a
i
d
o
©
.n
w
w
w
wmax (C f - 2) = wmin (- 2 - C f )
Hence,
Sol. 46
n
i
.
o
c
Option (D) is correct.
The ratio of the maximum fluctuation of speed to the mean speed is called the
coefficient of fluctuation of speed (C f ).
Let,
N1 & N2 = Maximum & Minimum speeds in r.p.m. during
the cycle
...(i)
N = Mean speed in r.p.m. = N1 + N2
2
2 (N1 - N2)
from equation (i)
Therefore,
C f = N1 - N 2 =
N
N1 + N 2
2 (w1 - w2)
= w1 - w2 =
w
w1 + w2
2 (wmax - wmin)
Cf =
w1 = wmax ,
wmax + wmin
w2 = wmin
C f wmax + C f wmin = 2wmax - 2wmin
wmax =- (2 + C f ) = 2 + C f
wmin
Cf - 2
2 - Cf
Option (D) is correct.
In this question pair or mechanism is related to contact & machine related to it.
Column I
P.
Column II
Higher Kinematic Pair
2.
Line Contact
Q. Lower Kinematic Pair
6.
Surface Contact
R. Quick Return Mechanism
5.
Shaper
S.
1.
Grubler’s Equation
Mobility of a Linkage
So correct pairs are, P-2, Q-6, R-5, S-1
Sol. 47
Option (C) is correct.
Given m = 250 kg , k = 100 kN/m , N = 3600 rpm , e = c = 0.15
cc
2p
N
2
3
.
14
3600
#
#
w =
=
= 376.8 rad/ sec
60
60
Natural frequency of spring mass system,
k = 100 # 1000 = 20 rad/ sec
wn =
250
m
GATE SOLVED PAPER - ME
THEORY OF MACHINES
w = 376.8 = 18.84
wn
20
So,
T.R. = FT =
F
=
=
Sol. 48
2
1 + a 2e w k
wn
w 22
w 2
:1 - a wn k D + 92e wn C
1 + (2 # 0.15 # 18.84) 2
2 2
61 - (18.84) @ + 62 # 0.15 # 18.84@2
32.945 = 0.0162
1 + 31.945
=
2
125309
+
1
354
.
945
31
.
945
6
@
Option (A) is correct.
Here P, Q, R, & S are the lengths of the links.
According to Grashof’s law : “For a four bar mechanism, the sum of the shortest
and longest link lengths should not be greater than the sum of remaining two link
lengths, if there is to be continuous relative motion between the two links
S+L G P+Q
Sol. 49
©
.
a
i
d
o
n
Option (A) is correct.
.
w
w
w
in
.
o
c
The table of motions is given below. Take CW =+ ve , CCW =- ve
S. Condition of Motion
No.
Revolution of elements
Gear 1 Compound
Gear 2-3,
N1
N2 = N 3
Gear 4
N4
Carrier
N5
1.
Carrier 5 is fixed + 1
& Gear 1 rotates
+ 1 rpm (CW)
- Z1
Z2
Z3
Z1
Z2 # Z 4
0
2.
Gear 1 rotates through + x
+ x rpm (CW)
- x Z1
Z2
x Z1 Z 3
Z2 Z 4
0
GATE SOLVED PAPER - ME
THEORY OF MACHINES
3.
Add + y revolutions + y
to all elements
4.
Total motion.
x+y
+y
+y
+y
y - x Z1
Z2
y + x # Z1 Z 3 + y
Z2 Z 4
Note
Speed of driver
= No. of teeth on driven
Speed of driven
No.of teeth on driver
N1 = Z 2
N2
Z1
Speed ratio =
(i)
i.e.
CCW = Counter clock wise direction (- ve)
CW = Clock wise direction (+ ve)
(ii) Gear 2 & Gear 3 mounted on the same shaft (Compound Gears)
n
i
.
o
c
N2 = N3
We know,
w = 2pN , & w\ N
60
(x + y) - y
N1 - N5 = w1 - w5 =
Hence,
w4 - w5 y + x Z1 Z 3 - y
N 4 - N5
#Z Z
2 4
w1 - w5 =
Z
x
2 Z4
=
w4 - w5
Z1 Z 3
x # Z1 Z 3
Z2 Z 4
w1 - w5 = 45 # 40 = 3 2 = 6
#
w4 - w5
15 # 20
Option (D) is correct.
Given w1 = 60 rpm (CW), w4 =- 2 # 60 (CCW) =- 120 rpm
From the previous part,
w1 - w5 = 6
w4 - w5
60 - w5 = 6
- 120 - w5
So,
Sol. 50
d
o
.
a
i
©
.n
w
w
w
60 - w5 =- 720 - 6w5
w5 =- 780 =- 156 rpm
5
Negative sign show the counter clock wise direction.
So,
Sol. 51
w5 = 156 rpm , CCW
Option (D) is correct.
Given m = 12.5 kg , k = 1000 N/m , c = 15 Ns/m
Critical Damping,
cc = 2m k = 2 km
m
On substituting the values, we get
cc = 2 1000 # 12.5 = 223.6 Ns/m
Sol. 52
None of these
We know logarithmic decrement,
d = 2pe 2
1-e
c
And
e=
= 15 = 0.0671
223.6
cc
Now, from equation (i), we get
...(i)
cc = 223.6 Ns/m
GATE SOLVED PAPER - ME
Sol. 53
THEORY OF MACHINES
= 0.422
d = 2 # 3.14 # 0.0671
1 - (0.0671) 2
Option (C) is correct.
Given l = 8 , j = 9
We know that, Degree of freedom,
n = 3 (l - 1) - 2j = 3 (8 - 1) - 2 # 9 = 3
Sol. 54
Sol. 55
Option (C) is correct.
The speed of sound in air = 332 m/s
For frequency of instrument of 144 Hz, length of sound wave
LI = 332 = 2.30 m
144
For sample P of 64 Hz,
LP = 332 = 5.1875 m
64
Q of 96 Hz
LQ = 332 = 3.458 m
96
R of 128 Hz
LR = 332 = 2.593 m
128
S of 250 Hz
LS = 332 = 1.2968 m
256
Here, the length of sound wave of sample R (LR = 2.593 m) is most close to the
length of sound wave of Instrument (LI = 2.30 m). Hence, sample R produce most
perceptible induced vibration.
in
.
o
c
.
a
i
d
o
n
Option (B) is correct.
Given N = 300 r.p.m
w = 2pN = 10p rad/ sec
60
p
1
# 30
180
Time taken to move 30c is,
= 6 = 1 sec
t =
10p
10 60
Now, Cam moves 30c with a constant acceleration & then with a deceleration,
so maximum speed of the follower is at the end of first 30c rotation of the cam
and during this 30c rotation the distance covered is 10 mm, with initial velocity
u = 0.
From Newton’s second law of motion,
S = ut + 1 at2
2
2
0.01 = 0 + 1 # a # b 1 l
2
60
.
w
Angular velocity of cam,
©
w
w
Maximum velocity,
Sol. 56
Option (C) is correct.
a = 0.01 # 2 # (60) 2 = 72 m/ sec2
v max = u + at = 72 # 1 = 1.2 m/ sec
60
GATE SOLVED PAPER - ME
THEORY OF MACHINES
Given m1 = m2 = 0.5 kg , r1 = 0.05 m , r2 = 0.06 m
Balancing mass m = 0.1 kg
Let disc rotates with uniform angular velocity w and x & y is the position of
balancing mass along X & Y axis.
Resolving the forces in the x -direction, we get
n
i
.
o
c
.
a
i
SFx = 0
0.5 6- 0.06 cos 30c + 0.05 cos 0c@ w2
0.5 # (- 0.00196)
x
Similarly in y -direction,
SFy
d
o
.n
w
= 0.1 # x # w2
= 0.1x
Fc = mrw2
=- 0.0098 m =- 9.8 mm
=0
0.5 (0.06 # sin 30c + 0.05 # sin 0) w = 0.1 # y # w2
0.5 # 0.03 = 0.1 # y
y = 0.15 m = 150 mm
Position of balancing mass is given by,
Sol. 57
©
w
w
r =
x2 + y2 =
2
(- 9.8) 2 + (150) 2 = 150.31 mm b 150 mm
Option (C) is correct.
Given m = 0.1 kg , k = 1 kN/m
Let, wd be the frequency of damped vibration & wn be the natural frequency of
spring mass system.
...(i)
Hence,
wd = 90% of wn = 0.9wn (Given)
Frequency of damped vibration
wd = (1 - e2) wn
From equation (i) and equation (ii), we get
...(ii)
(1 - e2) wn = 0.9wn
On squaring both the sides, we get
1 - e2 = (0.9) 2 = 0.81
e2 = 1 - 0.81 = 0.19
e = 0.19 = 0.436
And Damping ratio is given by,
e=c = c
cc
2 km
c = 2 km # e = 2 1000 # 0.1 # 0.436 = 8.72 Ns/m b 8.7 Ns/m
GATE SOLVED PAPER - ME
Sol. 58
THEORY OF MACHINES
Option (B) is correct.
From Triangle ABC ,
AB =
(100) 2 + (240) 2 =
67600 = 260 mm
in
.
o
c
Length of shortest link l1 = 60 mm
Length of longest link l 3 = 260 mm
From the Grashof’s law,
.
a
i
l1 + l 3 $ l 2 + l 4
60 + 260 $ 160 + 240
d
o
n
320 $ 400
.
w
l1 + l 3 < l 2 + l 4
So,
Also, when the shortest link O2 A will make a complete revolution relative to
other three links, if it satisfies the Grashof’s law. Such a link is known as crank.
The link O 4 B which makes a partial rotation or oscillates is known as rocker. So,
crank rocker mechanism is obtained.
Here,
O2 A = l1 = 60 mm is crank (fixed link)
©
w
w
Adjacent link,
O2 O 4 = 240 mm is fixed
So, crank rocker mechanism will be obtained.
Sol. 59
Option (B) is correct.
Let, w4 is the angular velocity of link O 4 B
From the triangle ABC ,
tan q = 100 = 5
240 12
q = tan-1 b 5 l = 22.62c
12
Also from the triangle O1 O2 A ,
tan q = O2 A
O1 O 2
...(i)
GATE SOLVED PAPER - ME
THEORY OF MACHINES
O1 O2 = O2 A = 60 = 144 mm
5
tan q
12
From the angular velocity ratio theorem.
V24 = w4 # I24 I14 = w # I24 I12
n
i
.
o
c
144
w4 = I24 I12 # w =
8 = 144 # 8
384
I24 I14
(240 + 144) #
= 3 rad/ sec
Sol. 60
Sol. 61
.
a
i
Option (D) is correct.
From the given data the component of force at joint A along AO2 is necessary to
find the joint reaction at O2 . So, it is not possible to find the magnitude of the
joint reaction at O2 .
d
o
.n
w
Option (D) is correct.
Mechanical advantage in the form of torque is given by,
w
T
M.A. = output = input
woutput
Tinput
w
w
Here output link is a slider, So, woutput = 0
©
Therefore,
Sol. 62
M.A. = 3
Option (C) is correct.
Given w = r = 0.5
wn
And due to isolation damping ratio,
e = c =0
cc
For isolation c = 0
We know the transmissibility ratio of isolation is given by,
2
1 + a 2e w k
wn
1+0
= 1 =4
T . R. =
=
2 2
2 2
2
0.75 3
w
w
+
1
(
0
.
5
)
0
6
@
:1 - a wn k D + 92e wn C
Sol. 63
Option ( D) is correct.
Given planar mechanism has degree of freedom, N = 1 and two infinite parallel
lines meet at infinity. So, the instantaneous centre I24 will be at N , but for single
degree of freedom, system moves only in one direction.
Hence, I24 is located at infinity(3).
GATE SOLVED PAPER - ME
Sol. 64
THEORY OF MACHINES
Option (C) is correct.
Given l = 300 mm = 0.3 m , W = 300 N
Let, rod is twisted to the left, through an angle q.
From the similar triangle OCD & OAB ,
y
= x
tan q =
0.15 0.30
y
If q is very very small, then tan q - q =
= x
0.15 0.30
in
.
o
c
x = 0.30q and y = 0.15q ..(i)
On taking moment about the hinged point O
.
a
i
kx # 300 + W # y = 0
Wy
y
k ==- 300 # a k =- 1 =- 0.5 N/mm
2
300x
300
x
y
From equation (i) = 0.15q =- 500 N/m
x 0.30q
d
o
n
.
w
Negative sign shows that the spring tends to move to the point B.
In magnitude,
k = 500 N/m
Sol. 65
w
Option (C) is correct.
w
Types of Mechanisms
P.
Q.
R.
S.
©
Motion Achieved
Scott-Russel Mechanism
4.
Straight Line Motion
Geneva Mechanism
1.
Intermittent Motion
Off-set slider-crank Mechanism
2.
Quick Return Mechanism
Scotch Yoke Mechanism
3.
Simple Harmonic Motion
So, correct pairs are, P-4, Q-1, R-2, S-3
Sol. 66
Sol. 67
Option (A) is correct.
Given N2 = 120 rpm , v1 = 12 m/ sec
So, coriolis component of the acceleration of link 1 is,
c
= 2w2 v1 = 2 # 2p # 120 # 12 = 301.44 m/s2 - 302 m/s2
a 12
60
Option (C) is correct.
Types of Joint
Degree of constraints
P.
Revolute
2.
Five
Q.
Cylindrical
3.
Four
R.
Spherical
1.
Three
GATE SOLVED PAPER - ME
THEORY OF MACHINES
So, correct pairs are P-2, Q-3, R-1
Sol. 68
Option (B) is correct.
.
a
i
d
o
.n
w
n
i
.
o
c
w
Let, Z is the number of teeth and motor rotates with an angular velocity w1 in
clockwise direction & develops a torque T1 .
Due to the rotation of motor, the gear 2 rotates in anti-clockwise direction & gear
3 rotates in clock wise direction with the same angular speed.
Let, T2 is the torque developed by gear.
Now, for two equal size big gears,
(Pitch circle diameter)
Module
m =D=
Z
(No.of teeths)
©
w
D = mZ = 2 # 80 = 160 mm
(Due to rotation of gear 2 & gear 3 an equal force (F ) is generated in the
downward direction because teeth are same for both the gears)
For equilibrium condition, we have
Downward force = upward force
And
F + F = 1000
F = 500 N
Power Output
h =
= 2 # T2 w2
T1 w1
Power Input
Output power is generated by the two gears
2 # bF # D l w2
2
...(i)
=
T1 w1
We know velocity ratio is given by
GATE SOLVED PAPER - ME
THEORY OF MACHINES
N1 = w1 = Z2
w2 Z1
N2
2 # bF # D l
2
Z
h =
# Z1
T1
2
Z
F
D
500
# 0.160 # 20 = 25 N-m
T1 = # # b 1 l =
h
0.8
80
Z2
From equation (i),
Sol. 69
w = 2pN
60
Option (C) is correct.
in
.
o
c
Given pressure angle f = 20c, FT = 500 N from previous question.
From the given figure we easily see that force action along the line of action is F .
From the triangle ABC ,
cos f = FT
F
F = FT = 500 = 532 N
cos f
cos 20c
Sol. 70
.
a
i
d
o
n
Option (A) is correct.
Given M = 20 kg , l = 1000 mm = 1 m , A = 25 # 25 mm2
Esteel = 200 GPa = 200 # 109 Pa
Mass moment of inertia of a square section is given by,
4
(25 # 10-3) 4
I =b =
= 3.25 # 10-8 m 4
12
12
Deflection of a cantilever, Loaded with a point load placed at the free end is,
3
20 # 9.81 # (1) 3
mgl3
= 196.2 = 0.01 m
d = Wl =
=
19500
3EI
3EI
3 # 200 # 109 # 3.25 # 10-8
g
= 9.81 = 31.32 rad/ sec
wn =
0.01
d
Therefore, critical damping constant
©
w
w
.
w
cc = 2 Mwn = 2 # 20 # 31.32
= 1252.8 Ns/m - 1250 Ns/m
Sol. 71
Option (D) is correct.
A single slider crank chain is a modification of the basic four bar chain. It is find,
that four inversions of a single slider crank chain are possible. From these four
inversions, crank and slotted lever quick return motion mechanism is used in
shaping machines, slotting machines and in rotary internal combustion engines.
Sol. 72
Option (A) is correct.
Given p < q < r < s
“Double crank” mechanism occurs, when the shortest link is fixed. From the
given pairs p is the shortest link. So, link of length p should be fixed.
GATE SOLVED PAPER - ME
Sol. 73
THEORY OF MACHINES
Option (B) is correct.
We clearly see from the figure that cylinder can either revolve about x -axis or
slide along x -axis & all the motions are restricted.
Hence, Number of degrees of freedom = 2 & movability includes the six degrees of
freedom of the device as a whole, as the ground link were not fixed. So, 4 degrees
of freedom are constrained or arrested.
Sol. 74
.
a
i
d
o
Sol. 75
n
i
.
o
c
Option (B) is correct.
Given N = 1200 rpm , DE = 2 kJ = 2000 J , D = 1 m , Cs = 0.02
Mean angular speed of engine,
w = 2pN = 2 # 3.14 # 1200 = 125.66 rad/ sec
60
60
Fluctuation of energy of the flywheel is given by,
2
For solid disc I = mR
DE = Iw2 Cs = 1 mR2 w2 Cs
2
2
2 # 2000
m = 22D2E = 1 2
2
R w Cs
(
^ 2 h # 125.66) # 0.02
2000 = 50.66 kg - 51 kg
= 4#2#
(125.66) 2 # 0.02
Option (B) is correct.
Given m = 10 kg , d = 30 mm = 0.03 m , l = 500 mm = 0.5 m ,
Eshaft = 2.1 # 1011 Pa
©
.n
w
w
w
We know that, static deflection due to 10 kg of Mass at the centre is given by,
3
mgl 3
...(i)
d = Wl =
48EI
48EI
The moment of inertia of the shaft,
...(ii)
I = p d 4 = p (0.03) 4 = 3.974 # 10-8 m 4
64
64
Substitute values in equation (i), we get
10 # 9.81 # (0.5) 3
d =
48 # 2.1 # 1011 # 3.974 # 10-8
= 12.2625 3 = 3.06 # 10-5 m
400.58 # 10
If wc is the critical or whirling speed in r.p.s. then,
g
g
& 2pfc =
wc =
d
d
GATE SOLVED PAPER - ME
Sol. 76
THEORY OF MACHINES
g
1
9.81
fc = 1
=
2p d
2 # 3.14 3.06 # 10-5
9.81
= 90.16 Hz - 90 Hz
= 1
6.28 30.6 # 10-6
Option (C) is correct.
Given, the circular disc rotates about the point O at a uniform angular velocity
w.
in
.
o
c
Let vA is the linear velocity of point A & vB is the linear velocity of point B.
vA = wrA and vB = wrB .
Velocity of point B with respect to point A is given by,
vBA = vB - vA = wrB - wrA = w (rB - rA)
.
a
i
From the given figure,
rB > rA
wrB > wrA
So,
d
o
n
vB > vA
Therefore, relative velocity w (rB - rA) in the direction of point B.
Sol. 77
Sol. 78
.
w
Option (D) is correct.
Acceleration of point B with respect to point A is given by,
...(i)
aBA = wvBA = w # w (rB - rA) = w2 (rB - rA)
This equation (i) gives the value of centripetal acceleration which acts always
towards the centre of rotation.
So, aBA acts towards to O i.e. its direction from Z to O
©
w
w
Option (A) is correct.
Given m = 10 kg , k = 2 kN/m , c = 500 Ns/m , k q = 1 kN/m/rad
GATE SOLVED PAPER - ME
THEORY OF MACHINES
l1 = 0.5 m , l2 = 0.4 m
Let, the rigid slender bar twist downward at the angle q. Now spring & damper
exert a force kx1 & cx 2 on the rigid bar in the upward direction.
From similar triangle OAB & OCD ,
tan q = x2 = x1
0.4 0.5
Let q be very very small, then tan q - q,
q = x 2 = x1
0.4 0.5
x2 = 0.4q or x1 = 0.5q
On differentiating the above equation, we get
xo2 = 0.4qo or xo1 = 0.5qo
...(i)
...(ii)
We know, the moment of inertia of the bar hinged at the one end is,
2
10 (0.5)2
I = ml 1 = #
= 0.833 kg - m2
3
3
As no external force acting on the system. So, governing equation of motion from
the Newton’s law of motion is,
Iqp + cxo2 l2 + kx1 l1 + k q q = 0
0.833qp + 500 # 0.4xo2 + 2000 # (0.5) x1 + 1000q = 0
n
i
.
o
c
.
a
i
d
o
0.833qp + 200xo2 + 1000x1 + 1000q = 0
0.833qp + 200 # 0.4qo + 1000 # 0.5q + 1000q = 0
0.833qp + 80qo + 1500q = 0
Sol. 79
Sol. 80
.n
w
...(iii)
...(iv)
On comparing equation (iv) with its general equation,
Iqp + cqo + kq = 0
We get, I = 0.833 , c = 80 , k = 1500
So, undamped natural frequency of oscillations is given by
1500 = 1800.72 = 42.43 rad/ sec
wn = k =
0.833
I
Option (C) is correct.
From the previous part of the question
Damping coefficient,
c = 80 Nms/rad
©
w
w
Option (C) is correct.
From the Kutzbach criterion the degree of freedom,
n = 3 (l - 1) - 2j - h
For single degree of Freedom (n = 1),
1 = 3 (l - 1) - 2j - h
...(i)
3l - 2j - 4 - h = 0
The simplest possible mechanisms of single degree of freedom is four-bar
mechanism. For this mechanism j = 4 , h = 0
From equation (i), we have
3l - 2 # 4 - 4 - 0 = 0 & l = 4
Sol. 81
Option (B) is correct.
When a point on one link is sliding along another rotating link, such as in quick
return motion mechanism, then the coriolis component of the acceleration must be
calculated. Quick return motion mechanism is used in shaping machines, slotting
GATE SOLVED PAPER - ME
THEORY OF MACHINES
machines and in rotary internal combustion engines.
Sol. 82
Option (C) is correct.
The deflection of a cantilever beam loaded at the free end is given by,
MgL3
d =
3EI
And natural frequency,
g
= 3EI3
wn =
d
ML
If the length of the cantilever beam is halved, then
3EI
wnl =
8 b 3EI3 l
3 =
L
ML
M #b l
2
...(i)
From equation (i)
wnl =
8 wn
So, natural frequency is increased by a factor
Sol. 83
Sol. 84
8.
Option (C) is correct.
For a spring loaded roller follower driven with a disc cam, the pressure angle
should be large during rise as well as during return for ease of transmitting
motion.
If pressure angle is large, then side thrust will be minimum. Pressure angles of
up to about 30c to 35c are about the largest that can be used without causing
difficulties.
.
a
i
d
o
n
Option (B) is correct.
Let initial length of the spring = L
Potential energy at A,
in
.
o
c
.
w
PEA = mg (L - d)
PEB = mg 6L - (d + x)@ + 1 kx2
2
So, change in potential energy from position A to position B is
and at B ,
©
Sol. 85
Sol. 86
w
w
TPEAB = PEB - PEA
TPEAB
= mgL - mgd - mgx + 1 kx2 - mgL + mgd
2
= 1 kx2 - mgx
2
Option (A) is correct.
The mean speed of the engine is controlled by the governor. If load increases then
fluid supply increases by the governor and vice-versa.
Flywheel stores the extra energy and delivers it when needed. So, Flywheel
reduces speed fluctuations.
Flywheel reduce speed fluctuations during a cycle for a constant load, but Flywheel
does not control the mean speed bN = N1 + N2 l of the engine.
2
Option (B) is correct.
First make the table for the motion of the gears.
Take CW =+ ve , CCW =- ve
GATE SOLVED PAPER - ME
THEORY OF MACHINES
S.
No.
Condition of Motion
Arm Sun Gear
NS
(i)
Arm is fixed & sun
gear rotates + 1 rpm
(CW)
0
+1
(ii)
Sun Gear rotates
through + x rpm
(CW)
0
+x
(iii)
Add + y revolution to
all elements
+y
(iv)
Total Motion
.n
w
Ring Gear
NG
- ZS
ZP
- ZS
ZR
- x ZS
ZP
- x ZS
ZR
+y
+y
+y
x+y
y - x ZS
ZP
y - x ZS
ZR
n
i
.
o
c
.
a
i
d
o
+y
Planet
Gear NP
Let Teethes and speed of the sum gear, planet gear and ring gear is represented
by ZG , ZP , ZR and NG , NP , NR respectively.
Given sun gear is driven clockwise at 100 rpm. So, From the table
w
w
x + y = 100
Ring gear is held stationary. From the table
y - x ZS = 0
ZP
y = x # 20
80
y =x
& x = 4y
4
From equation on (i) and (ii)
©
...(i)
...(ii)
4y + y = 100
y = 20 rpm
Sol. 87
Option (D) is correct.
Give a small displacement q to the assembly. So assembly oscillates about its
mean position.
GATE SOLVED PAPER - ME
THEORY OF MACHINES
From this a restoring torque is acts along the line of oscillation.
Net restoring torque,
T = mg sin (a + q) # L - mg sin (a - q) # L
T
= mgL 6sin a cos q + cos a sin q - sin a cos q + cos a sin q@
T = 2mgL cos a sin q
For very small deflection q,
sin q , q
T = 2mgLq cos a
Now from newton’s law,
Iqp + T = 0
in
.
o
c
.
a
i
Iqp + 2mgLq cos a = 0
d
o
n
2
2mL2 d q2 + (2mgL cos a) q = 0
dt
d 2 q + g cos a q = 0
L
dt2
2
p
On comparing with q + wn q = 0 , we get
g cos a
wn2 =
L
g cos a
wn =
L
©
w
w
.
w
*********
I = mL2 + mL2