1. Derive an expression for K.E, of a rotating body. Show that M. I of body rotating
about a given axis is twice the rotational K.E. If it has unit angular velocity
Answers
Explanation:
Consider a rigid body rotating with a constant angular velocity
ω
about an axis passing through the point O.
As the body rotates, all the particles perform uniform circular motion.
The linear speed of the particle with mass m
1
is V
1
=r
1
ω. Therefore, its kinetic energy is
E
1
=
2
1
m
1
V
1
2
=
2
1
m
1
r
1
2
ω
2
Similarly, the kinetic energy of the particle with mass m
2
is E
2
=
2
1
m
2
V
2
2
=
2
1
m
2
r
2
2
ω
2
and so on. The rotational kinetic energy of the body is
E
rot
=E
1
+E
2
+....+E
N
=
2
1
m
1
r
2
2
ω
2
+
2
1
m
2
r
2
2
ω
2
+...+
2
1
m
N
r
N
2
ω
2
=
2
1
[m
1
r
1
2
+m
2
r
2
2
+....+m
N
r
N
2
]ω
2
=
2
1
(
i=1
∑
N
m
i
r
i
2
)ω
2
∴E
rot
=
2
1
Iω
2
KE=
2
1
Iω
2
L=Iω,ω=
I
L
KE=
2
1
I×(
I
L
)
2
=
2
1
I
I
2
L
2
(∵I=MK
2
)
=
2
1
I
L
2
=
2
1
MK
2
L
2
=
2M
1
[
K
L
]
2