Two particles of masses m1, and m2, respectively are connected by a rigid massless rod of length a and move freely in a plane. Show that the moment of inertia of the system about an axis perpendicular to the plane and passing through the center of mass is ua²?, where u is the reduced mass.
Step-by-step explanation:
The position of COM from m
1
mass is
(m
1
+m
2
)
m
2
r
So MI of m
1
mass about COM is m
1
×(
m
1
+m
2
m
2
r
)
2
=
(m
1
+m
2
)
2
m
1
m
2
2
r
2
So MI of m
2
mass about COM is m
2
×(
m
1
+m
2
m
1
r
)
2
=
(m
1
+m
2
)
2
m
2
m
1
2
r
2
So total MI is
(m
1
+m
2
)
2
m
1
m
2
2
r
2
+
(m
1
+m
2
)
2
m
2
m
1
2
r
2
=
m
1
+m
2
m
1
m
2
r
2
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