Four point masses, each of mass m, are fixed at the corners of a square of side l. The square is rotating with angular frequency ω, about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in the figure. The angular momentum of the square about this axis is :

Solution

Physics sir · 2 steps

Measure each mass's perpendicular distance from the axis, not from the corner. Then add up mr².

1. Read the geometry. The axis passes through the top-left corner and is parallel to the dotted diagonal. That diagonal runs from the bottom-left corner to the top-right corner. Any corner sits at a perpendicular distance l/√2 from a diagonal it is not on, because that is half the other diagonal √2 l.

   Mass at	Distance from axis	mr²
   Top-left (on axis)	0	0
   Top-right	l/√2 (it lies on the dotted diagonal)	ml²/2
   Bottom-left	l/√2 (it lies on the dotted diagonal)	ml²/2
   Bottom-right	l/√2 + l/√2 = √2 l	2ml²

   I = 0 + ml²/2 + ml²/2 + 2ml² = 3ml²

2. So. L = Iω = 3ml²ω, which is option (a).

The idea to keep. For moment of inertia, r is always the perpendicular distance to the axis line. When the axis lies in the plane, picture parallel lines and measure straight across them.

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