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AMU Engineering 2018 solved paper Solved Paper
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© examsnet.com
Question : 13
Total: 150
The time-period of a physical pendulum is
2
π
√
I
∕
m
g
d
, where
I
is the moment of inertia of the pendulum about the axis of rotation and
d
is perpendicular distance between the axis of rotation and the centre of mass of the pendulum. A circular ring hangs from a nail on a wall. The mass of the ring is
3
k
g
and its radius is
20
c
m
. If the ring is slightly displaced, the time of resulting oscillations will be
1.0
s
1.3
s
1.8
s
2.1
s
Validate
Solution:
Given,
T
=
2
π
√
I
m
g
.
d
.......(i)
m
=
3
k
g
,
d
or
r
=
20
c
m
=
0.2
m
Moment of inertia of the ring about
X
X
′
I
=
I
0
+
m
r
2
=
m
r
2
+
m
r
2
=
2
m
r
2
=
2
×
3
×
(
0.2
)
2
=
0.24
k
g
−
m
2
Putting values in Eq (i)
T
=
2
π
√
0.24
3
×
10
×
0.2
=
2
π
√
0.04
=
2
π
×
0.2
=
1.2566
s
T
≈
1.3
s
© examsnet.com
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