Question

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.​

Answer 1

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

Answer 2

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