M3 January 2005 Q3

Figure 2: region R under y = sin x between O and pi, maximum height 1 — Figure 2

A uniform lamina occupies the region \(R\) bounded by the \(x\)-axis and the curve

\[y = \sin x, \qquad 0 \leqslant x \leqslant \pi,\]

as shown in Figure 2.

(a) Show, by integration, that the \(y\)-coordinate of the centre of mass of the lamina is \(\dfrac{\pi}{8}\). (6)

Figure 3: cross-section R with its flat face on a plane inclined at theta degrees to the horizontal — Figure 3

A uniform prism \(S\) has cross-section \(R\). The prism is placed with its rectangular face on a table which is inclined at an angle \(\theta^\circ\) to the horizontal. The cross-section \(R\) lies in a vertical plane as shown in Figure 3. The table is sufficiently rough to prevent \(S\) sliding. Given that \(S\) does not topple,

(b) find the largest possible value of \(\theta\). (3)

Mark scheme (a)

Scheme	Marks
\(\displaystyle\int_0^\pi \tfrac{1}{2}y^2\,\mathrm{d}x = \int_0^\pi \tfrac{1}{2}\sin^2 x\,\mathrm{d}x\)	M1
\(\displaystyle = \tfrac{1}{4}\int_0^\pi (1-\cos 2x)\,\mathrm{d}x\)	M1
\(= \tfrac{1}{4}\left[x - \tfrac{1}{2}\sin 2x\right]_0^\pi\)	A1
\(= \dfrac{\pi}{4}\)	A1
\(\bar{y} = \dfrac{\pi/4}{\int_0^\pi \sin x\,\mathrm{d}x} = \dfrac{\frac{\pi}{4}}{2}\)	M1
\(= \dfrac{\pi}{8}\)	A1
	(6)

Scheme	Marks
\(\tan\theta = \dfrac{\pi/2}{\bar{y}}\)	M1
\(= 4\)	A1ft
\(\theta = 75.96^\circ\)	A1
	(3)
	(9 marks)