Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 6 / b / ii

Table of contents
- Solution ii

Solution

Put

J_{0} = 4 σ D^{4} /3 = M D^{2} /3

. The normal to the lamina is already an eigenvector, with moment

3 J_{0}

. The planar block has characteristic polynomial

det (1 - λ - 1 - 1 2 - λ) = λ^{2} - 3 λ + 1,

so the principal moments of inertia are

I_{-} = \frac{J _{0}}{2} (3 - 5), I_{+} = \frac{J _{0}}{2} (3 + 5), I_{3} = 3 J_{0} .

A corresponding orthonormal set of principal axes is

n_{-} = \frac{( 2 , 5 - 1 , 0 )}{10 - 2 5}, n_{+} = \frac{( 2 , - 1 - 5 , 0 )}{10 + 2 5}, n_{3} = (0, 0, 1) .

The sign of each axis vector is immaterial. The smaller-moment axis makes angle

\frac{1}{2} arctan (2)

with the positive

x_{1}

direction, approximately

31. 7^{\circ}

; the other planar axis is perpendicular. As a check,

I_{-} + I_{+} = 3 J_{0} = I_{3}

, the perpendicular axis theorem for a planar lamina.

Figure 1.
The two planar principal axes of the sheared square lamina, with the third axis normal to the page
. Details.

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