An massless equilateral triangle EFG of side ‘a’ (As shown in figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the line EX perpendicular to EG in the plane of EFG is N/20 ma² where N is an integer. The value of N is _____.

Solution

Physics sir · 3 steps

For each mass, measure the perpendicular distance to the line EX, not the distance along the triangle's sides.

1. Set up distances. The axis EX is a vertical line through E, so a particle's distance from it is its horizontal distance from E. E is on the axis, so its distance is 0. G is at the other end of the base, so its distance is a.
2. Distance of F. F sits directly above the midpoint of EG, so its horizontal distance from E is a/2. Its height √3/2a runs parallel to EX and does not count.

   I = m(0)² + m(a/2)² + m(a)² = 5/4ma² = 25/20ma²

3. So. Comparing with N/20ma² gives N = 25.

The common slip: 35. Taking F's height √3/2a as its distance gives 7/4ma² = 35/20ma², so N = 35. That height is parallel to EX, so it does not count.

The idea to keep. Moment of inertia uses each mass's perpendicular distance from the axis. Any displacement parallel to the axis adds nothing, so drop the height when the axis is vertical.

Answer from the NTA final answer key. Solution by PrepWiser.