On comparing equation (iv) with its general equation,
I
c
k
q
q
q
+
+
p
o
0
=
We get,
I
.
0 833
=
,
c
80
=
,
k
1500
=
So, undamped natural frequency of oscillations is given by
n
w
I
k
=
.
.
0 833
1500
1800 72
=
=
42.43
/sec
rad
=
Sol. 79
Option (C) is correct.
From the previous part of the question
Damping coefficient,
c
80
/
Nms rad
=
Sol. 80
Option (C) is correct.
From the Kutzbach criterion the degree of freedom,
n
(
)
l
j
h
3
1
2
=



For single degree of Freedom
(
)
n
1
=
,
1
(
)
l
j
h
3
1
2
=



l
j
h
3
2
4



0
=
...(i)
The simplest possible mechanisms of single degree of freedom is fourbar
mechanism. For this mechanism
j
4
=
,
h
0
=
From equation (i), we have
l
3
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,
d
EI
MgL
3
3
=
And natural frequency,
n
w
g
d
=
ML
EI
3
3
=
...(i)
If the length of the cantilever beam is halved, then
n
w
l
M
L
EI
ML
EI
2
3
8
3
3
3
#
=
=
b
b
l
l
From equation (i)
n
w
l
8
n
w
=
So, natural frequency is increased by a factor
8
.
Sol. 83
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
30
c
to
35
c
are about the largest that can be used without causing
difficulties.
Sol. 84
Option (B) is correct.
Let initial length of the spring
L
=
Potential energy at
A
,
PE
A
(
)
mg L
d
=

and at
B
,
PE
B
(
)
mg L
x
kx
2
1
2
d
=

+
+
6
@
So, change in potential energy from position
A
to position
B
is
PE
AB
T
PE
PE
B
A
=

mgL
mg
mgx
kx
mgL
mg
2
1
2
d
d
=


+

+
PE
AB
T
kx
mgx
2
1
2
=

Sol. 85
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 viceversa.
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
N
N
N
2
1
2
=
+
b
l
of the engine.
Sol. 86
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
N
S
Planet
Gear
N
P
Ring Gear
N
G
(i)
Arm is fixed & sun
gear rotates
1rpm
+
(CW)
0
1
+
Z
Z
P
S

Z
Z
R
S

(ii)
Sun Gear rotates
through
rpm
x
+
(CW)
0
x
+
x
Z
Z
P
S

x
Z
Z
R
S

(iii)
Add
y
+
revolution to
all elements
y
+
y
+
y
+
y
+
(iv)
Total Motion
y
+
x
y
+
y
x
Z
Z
P
S

y
x
Z
Z
R
S

Let Teethes and speed of the sum gear, planet gear and ring gear is represented
by
Z
G
,
Z
P
,
Z
R
and
N
G
,
N
P
,
N
R
respectively.
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 Natural Frequency, Sol., Q., Epicyclic gearing, Flywheel