M5 June 2016 Q5

Figure 1: wire ABC bent at a right angle at B, with AB = 2a vertical and BC = 2a horizontal — Figure 1

A uniform piece of wire \(ABC\), of mass \(2m\) and length \(4a\), is bent into two straight equal portions, \(AB\) and \(BC\), which are at right angles to each other, as shown in Figure 1. The wire rotates freely in a vertical plane about a fixed smooth horizontal axis \(L\) which passes through \(A\) and is perpendicular to the plane of the wire.

(a) Show that the moment of inertia of the wire about \(L\) is \(\dfrac{20ma^2}{3}\) (3)

(b) By writing down an equation of rotational motion for the wire as it rotates about \(L\), find the period of small oscillations of the wire about its position of stable equilibrium. (8)

Mark scheme (a)

Scheme	Marks
\(I_L = \dfrac{4}{3}ma^2 + \dfrac{1}{3}ma^2 + m(a^2 + (2a)^2)\) PRINTED ANSWER	M1 A1
\(= \dfrac{20}{3}ma^2\)	A1
	(3)

Notes

First M1 for a complete method to find MI with correct no. of terms

First A1 for a correct expression

Second A1 for a correct PRINTED ANSWER.

M0 for \(4/3ma^2 + 4/3ma^2 + m(2a)^2\)

Mark scheme (b)

Scheme	Marks
\(m\begin{pmatrix}0\\a\end{pmatrix} + m\begin{pmatrix}a\\0\end{pmatrix} = 2m\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} \Rightarrow \bar{x} = \bar{y} = \dfrac{a}{2}\)	M1 A1
\(AG = \dfrac{a}{2}\sqrt{1^2 + 3^2} = \dfrac{a}{2}\sqrt{10}\)
\(M(A),\quad 2mg\dfrac{a}{2}\sqrt{10}\sin\theta = -\dfrac{20}{3}ma^2\ddot{\theta}\)	M1 A1 A1
\(-\dfrac{3g\sqrt{10}}{20a}\theta = \ddot{\theta}\), for small \(\theta\)	M1
\(T = 2\pi\sqrt{\dfrac{2a\sqrt{10}}{3g}}\)	DM1 A1
	(8)
	(11 marks)

Notes

First M1 for attempt to find both coordinates of the CM

First A1 for correct position (They could find this by inspection)

Second M1 for moments about \(A\), with correct no. of terms, and usual rules, in particular the RHS must be dimensionally correct.

Second and third A marks for a correct general equation with \(\theta\) the angle between \(AG\) and the vertical, A1 for each side.

Third M1 for use of small angle approximation and putting into SHM form. N.B. Not available if \(\theta\) is not the angle between \(AG\) and the vertical

Fourth DM1, dependent on third M1, for \(\dfrac{2\pi}{\omega}\)

Fourth A1 for a correct answer in any form