M2 June 2018 Q3

3. [The centre of mass of a semicircular lamina of radius \(r\) is \(\dfrac{4r}{3\pi}\) from the centre.]

Figure 2: square ABDE of side 2a with semicircle BCD on diameter BD — Figure 2

Figure 2 shows the uniform lamina \(ABCDE\), such that \(ABDE\) is a square with sides of length \(2a\) and \(BCD\) is a semicircle with diameter \(BD\).

(a) Show that the distance of the centre of mass of the lamina from \(BD\) is \(\dfrac{20a}{3(8 + \pi)}\). (5)

The lamina is freely suspended from \(D\) and hangs in equilibrium.

(b) Find, to the nearest degree, the angle that \(DE\) makes with the downward vertical. (3)

Mark scheme (a)

Scheme	Marks
Mass ratios : \(\ 4a^2 : \dfrac{\pi a^2}{2} : a^2\left(4 + \dfrac{\pi}{2}\right)\)	B1
Distances relative to \(BD\): \(\ a, -\dfrac{4a}{3\pi}, (d)\)	B1
Moments about \(BD\) (or a parallel axis)	M1
\(4a^2 \times a + \dfrac{\pi a^2}{2} \times \dfrac{-4a}{3\pi} = \left(4 + \dfrac{\pi}{2}\right)a^2 \times d\)	A1
\(a\left(4 - \dfrac{4}{6}\right) = \left(4 + \dfrac{\pi}{2}\right)d\)
\(d = a \times \dfrac{10}{3} \times \dfrac{2}{(8 + \pi)} = \dfrac{20a}{3(8 + \pi)}\)	A1
	(5)

Notes

B1 Or equivalent

B1 Or equivalent . Condone sign errors

M1 Dimensionally correct. All terms required. Condone sign errors. Accept in a vector equation.

A1 Correct unsimplified equation

(Distance from \(AE = \dfrac{(28 + 6\pi)a}{3(8 + \pi)}\))

A1 Obtain given answer from correct working. Condone -ve becoming positive with no explanation at the end

Diagram: D with horizontal distance a and vertical distance d to the centre of mass, angle theta at D between DE and the line to the centre of mass — Mark scheme (b)

Scheme	Marks
Use trig to find relevant angle	M1
\(\tan\theta = \dfrac{a}{d} = \dfrac{3(8 + \pi)}{20}\left(= \dfrac{1}{0.598}\right)\)	A1
\(\theta\ (= 59.12....) = 59^\circ\)	A1
	(3)
	(8 marks)