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AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper
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© examsnet.com
Question : 119
Total: 160
A uniform solid sphere of radius
′
R
′
and radius of gyration
′
K
′
about an axis passing through the centre of mass, is rolling without slipping. Then the fraction of total energy associated with its rotation will be ___________
K
2
+
R
2
K
2
K
2
R
2
K
2
K
2
+
R
2
R
2
K
2
+
R
2
Validate
Solution:
For a body rolling without slipping.
v
C
M
=
R
ω
.....(i)
Translational kinetic energy,
K
E
T
=
1
2
m
v
C
M
2
Rotational kinetic energy,
K
E
R
=
1
2
I
ω
2
=
1
2
(
m
K
2
)
a
2
(
∵
I
=
m
K
2
, where
k
is radius of gyration.)
=
1
2
m
k
2
(
v
C
M
R
)
2
[from Eq. (i)]
=
1
2
m
v
C
M
2
(
k
2
R
2
)
∴
Total kinetic energy,
K
E
=
K
E
T
+
K
E
R
=
1
2
m
v
C
M
2
+
1
2
m
v
C
M
3
k
2
R
2
K
E
=
1
2
m
v
C
M
2
(
1
+
k
2
R
2
)
Fraction of total energy associated with its rotation, i.e.
K
E
R
K
E
=
1
2
m
v
C
M
2
(
k
2
R
2
)
1
2
m
v
C
M
2
(
1
+
k
2
R
2
)
=
k
2
k
2
+
R
2
© examsnet.com
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