M5 June 2013 Q2

2. A uniform square lamina \(S\) has side \(2a\). The radius of gyration of \(S\) about an axis through a vertex, perpendicular to \(S\), is \(k\).

(a) Show that \(k^2 = \dfrac{8a^2}{3}\). (4)

The lamina \(S\) is free to rotate in a vertical plane about a fixed smooth horizontal axis which is perpendicular to \(S\) and passes through a vertex.

(b) By writing down an equation of rotational motion for \(S\), find the period of small oscillations of \(S\) about its position of stable equilibrium. (5)

Mark scheme (a)

Scheme	Marks
\(I_G = \tfrac{1}{3}ma^2 + \tfrac{1}{3}ma^2 = \tfrac{2}{3}ma^2\) (perp axes)	M1 A1
\(I_B = \tfrac{2}{3}ma^2 + m(a\sqrt{2})^2 = \tfrac{8}{3}ma^2\) i.e. \(k^2 = \tfrac{8}{3}a^2\) **	M1 A1
	(4)

Notes

First M1 for use of perpendicular axes rule with appropriate no. of terms

First A1 for correct expression for \(I_G\) (or from formulae sheet)

Second M1 for use of parallel axes rule to obtain \(I_B\)

Second A1 for PRINTED ANSWER.

Alternative, using result(s) on formulae sheet:

First M1A1 \(I_{AB} = I_{BC} = 4/3ma^2\) (from formulae sheet)

Second M1 for use of perpendicular axes rule with appropriate no. of terms

\(I_B = I_{AB} + I_{BC} = 8/3ma^2\)

(Corrected from the printed mark scheme: the second line is printed as \(I_G = \tfrac{2}{3}ma^2 + m(a\sqrt{2})^2\); this is the moment of inertia about the vertex, \(I_B\), as the notes say.)

Mark scheme (b)

Scheme	Marks
\(-mga\sqrt{2}\sin\theta = \tfrac{8}{3}ma^2\ddot{\theta}\)	M1 A1
\(\ddot{\theta} = -\dfrac{3g\sqrt{2}}{8a}\theta\), for small \(\theta\)	DM1
\(T = 2\pi\sqrt{\dfrac{8a}{3g\sqrt{2}}}\)	M1 A1
	(5)
	(9 marks)

Notes

First M1 for moments equation (dim correct and \(mg\) resolved)

First A1 for correct equation

Second M1 dependent for use of \(\sin\theta = \theta\) and SHM equation

Third M1 for use of \(T = 2\pi/\omega\) (only if proper SHM equn with -)

Third A1 for answer.