Question

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 \(\omega\), 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 :

• A. \(3ml^{2}\omega\)
• B. \(4ml^{2}\omega\)
• C. \(ml^{2}\omega\)
• D. \(2ml^{2}\omega\)

Hint:

Find the perpendicular distance of each of the four masses from the axis, then \(I = \sum m r^2\) and \(L = I\omega\).

Answer 1

The Correct Option is A

Take the square with corners (0,0), (l,0), (0,l), (l,l). The axis passes through the corner (0,l) parallel to the diagonal joining (0,0) and (l,l), i.e. the line \(y = x + l\).
Distance of a point (x, y) from the axis: \(\frac{|y - x - l|}{\sqrt2}\).
Corner (0,l): 0. Corner (l,l): \(\frac{l}{\sqrt2}\). Corner (0,0): \(\frac{l}{\sqrt2}\). Corner (l,0): \(\frac{2l}{\sqrt2} = \sqrt2\,l\).
\(I = m\left[0 + \frac{l^2}{2} + \frac{l^2}{2} + 2l^2\right] = 3ml^2\).
\(L = I\omega = 3ml^2\omega\).
Answer: (1) \(3ml^{2}\omega\)