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Question: Three point masses each of mass m are placed at the comes of an equilateral triangle of side 'a'. Th...

Three point masses each of mass m are placed at the comes of an equilateral triangle of side 'a'. Then the moment of inertia of this sytem about an axis passing along one side of the triangle is:

A

ma2

B

3 ma2

C

3/4 ma2

D

2/3 ma2.

Answer

3/4 ma2

Explanation

Solution

From the triangle BCD

CD2=BC2−BD2=a2−(a2)2C D ^ { 2 } = B C ^ { 2 } - B D ^ { 2 } = a ^ { 2 } - \left( \frac { a } { 2 } \right) ^ { 2 }

x2=3a24x ^ { 2 } = \frac { 3 a ^ { 2 } } { 4 }⇒ x=3a2x = \frac { \sqrt { 3 } a } { 2 }

Moment of inertia of system along the side AB.

=m×(0)2+m×(x)2+m×(0)2= m \times ( 0 ) ^ { 2 } + m \times ( x ) ^ { 2 } + m \times ( 0 ) ^ { 2 }

=mx2=m(3a2)2= m x ^ { 2 } = m \left( \frac { \sqrt { 3 } a } { 2 } \right) ^ { 2 } =3ma24= \frac { 3 m a ^ { 2 } } { 4 }