M3 June 2013 (R) Q6

EdexcelOld spec15 marksCentres of Mass

(a) A uniform lamina is in the shape of a quadrant of a circle of radius \(a\). Show, by integration, that the centre of mass of the lamina is at a distance of \(\dfrac{4a}{3\pi}\) from each of its straight edges. (7)

Figure 3: lamina ABCDEFA with AC = AE = 2a, B the midpoint of AC, F the midpoint of AE, quadrant BCD centre B and quadrant DGE centre G — Figure 3

A second uniform lamina \(ABCDEFA\) is shown shaded in Figure 3. The straight sides \(AC\) and \(AE\) are perpendicular and \(AC = AE = 2a\). In the figure, the midpoint of \(AC\) is \(B\), the midpoint of \(AE\) is \(F\), and \(ABDF\) and \(DGEF\) are squares of side \(a\). \(BCD\) is a quadrant of a circle with centre \(B\). \(DGE\) is a quadrant of a circle with centre \(G\).

(b) Find the distance of the centre of mass of the lamina from the side \(AE\). (5)

The lamina is smoothly hinged to a horizontal axis which passes through \(E\) and is perpendicular to the plane of the lamina. The lamina has weight \(W\) newtons. The lamina is held in equilibrium in a vertical plane, with \(A\) vertically above \(E\), by a horizontal force of magnitude \(X\) newtons applied at \(C\).

Mark scheme (a)

Scheme	Marks
v1 Mass of quadrant \(= \rho\dfrac{\pi a^2}{4}\)	B1
\(\displaystyle\int_0^a \rho x\sqrt{a^2 - x^2}\,\mathrm{d}x = \rho\left[-\frac{1}{3}\left(a^2 - x^2\right)^{\frac{3}{2}}\right]_0^a\)	M1A1 A1
\(= \rho\left[0 + \dfrac{1}{3}a^3\right]\)	A1
\(\rho\dfrac{\pi a^2}{4}\bar{x} = \rho\dfrac{a^3}{3}\)	M1
\(\bar{x} = \dfrac{4a}{3\pi}\) , \(\bar{y} = \dfrac{4a}{3\pi}\) by symmetry AG	A1,A1
	(7)

Notes

(The marks printed for v1 add up to 8; the part is worth 7.)

6 (a) v2

Mass of quadrant \(= \rho\dfrac{\pi a^2}{4}\)	B1
\(\displaystyle\int_0^{\frac{\pi}{2}} \rho \cdot \tfrac{1}{2}a^2 \cdot \tfrac{2}{3}a\cos\theta\,\mathrm{d}\theta = \left[\frac{a^3}{3}\sin\theta\right]_0^{\frac{\pi}{2}} = \rho\frac{a^3}{3}\)	M1A1,=A1
\(\rho\dfrac{\pi a^2}{4}\bar{x} = \rho\dfrac{a^3}{3}\)	M1
\(\bar{x} = \dfrac{4a}{3\pi}\) , \(\bar{y} = \dfrac{4a}{3\pi}\) by symmetry AG	A1A1

Mark scheme (b)

Scheme

Marks

Area	\(2a^2\)	\(\dfrac{\pi a^2}{4}\)	\(-\dfrac{\pi a^2}{4}\)
Distance to \(AE\)	\(\dfrac{a}{2}\)	\(a + \dfrac{4a}{3\pi}\)	\(a - \dfrac{4a}{3\pi}\)

Moments about \(AE\): \(2a^2\bar{x} = 2a^2\dfrac{a}{2} + \dfrac{\pi a^2}{4}\left(a + \dfrac{4a}{3\pi}\right) - \dfrac{\pi a^2}{4}\left(a - \dfrac{4a}{3\pi}\right)\)

M1A2

\(= a^3 + \dfrac{2a^3}{3} = \dfrac{5a^3}{3}\)

\(\bar{x} = \dfrac{5a^3}{3} \times \dfrac{1}{2a^2} = \dfrac{5a}{6}\)

(5)

Scheme	Marks
Taking moments about \(E\): \(2aX = \dfrac{5a}{6}W\) their \(\bar{x}\)	M1A1ft
\(X = \dfrac{5}{12}W\)	A1
	(3)
	(15 marks)