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KCET 2021 Physics Solved Paper
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© examsnet.com
Question : 34
Total: 60
In figure E and
v
cm
represent the total energy and speed of centre of mass of an object of mass
1
kg
in pure rolling. The object is
sphere
ring
disc
hollow cylinder
Validate
Solution:
As we know, kinetic energy of an object in pure rolling motion is given as
E
=
1
2
m
v
2
cm
(
1
+
k
2
R
2
)
where,
k
is radius of gyration and
m
is the mass of the object.
⇒
E
v
2
c
m
=
1
2
[
1
+
k
2
R
2
]
...(i)
[
∵
Given,
m
=
1
kg
]
From the given graph, substituting the value of
E
v
2
in Eq. (i), we get
3
4
=
1
2
[
1
+
k
2
R
2
]
⇒
k
2
R
2
=
1
2
Since, for a
(a) Sphere,
k
2
R
2
=
2
5
(b) Ring,
k
2
R
2
=
1
(c) Hollow cylinder,
k
2
R
2
=
1
(d) Disc,
k
2
R
2
=
1
2
So, the given object is disc.
© examsnet.com
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