A2 June 2025 Q6

EdexcelCurrent spec12 marksCentres of Mass

6. A uniform semicircular lamina has radius \(r\) and the midpoint of the diameter of the lamina is at \(O\). The distance of the centre of mass of the lamina from \(O\) is \(d\).

(a) Use algebraic integration to show that \(d = \dfrac{4r}{3\pi}\) (4)

The uniform plane template \(T\), shown shaded in Figure 7, is formed from a rectangle and a semicircular ring.
The rectangle has \(AD = BC = 2a\) and \(AB = DC = 8a\).
The midpoint of \(CD\) is \(O\).
The semicircular ring has centre \(O\), outer radius \(OD = 4a\) and inner radius \(OE = 2a\).
The template is modelled as a uniform lamina.

Figure 7: template T: rectangle ABCD with AB = 8a and AD = BC = 2a, joined along DC to a semicircular ring with centre O, outer radius 4a and inner radius 2a; E and F are the ends of the inner semicircle on DC — Figure 7

(b) Show that the distance of the centre of mass of \(T\) from \(DC\) is \(\dfrac{32a}{3(8 + 3\pi)}\) (5)

The template is freely suspended from \(A\). The weight of \(T\) is \(W\) newtons.
A horizontal force is applied to \(T\) at \(B\) so that \(T\) is held in equilibrium with \(AB\) vertical.
The force acts in the same vertical plane as \(T\) and has magnitude \(\lambda W\) newtons.

Diagrams: the semicircle y = sqrt(r^2 − x^2) with a thin vertical strip, or the semicircle x^2 + y^2 = r^2 above the x-axis with a thin vertical strip — Mark scheme (a)

Scheme	Marks	AO

\(\left(\dfrac{1}{2}\pi r^2 d =\right) \displaystyle\int_0^r 2yx\,\mathrm{d}x\) OR \(\left(\dfrac{1}{2}\pi r^2 d =\right) \displaystyle\int_{-r}^r \dfrac{1}{2}y^2\,\mathrm{d}x\)	M1	2.1
\(\displaystyle\int_0^r 2xy\,\mathrm{d}x = \int_0^r 2x\sqrt{r^2 - x^2}\,\mathrm{d}x\) OR \(\displaystyle\int_{-r}^r \dfrac{1}{2}y^2\,\mathrm{d}x = \int_{-r}^r \dfrac{1}{2}\left(r^2 - x^2\right)\mathrm{d}x\)	M1	3.1a
\(= -\left[\dfrac{2}{3}\left(r^2 - x^2\right)^{\frac{3}{2}}\right]_0^r = \dfrac{2}{3}r^3\) OR \(= \left[\dfrac{1}{2}\left(r^2x - \dfrac{1}{3}x^3\right)\right]_{-r}^r = \dfrac{2}{3}r^3\)	A1ft	1.1b
\(\Rightarrow d = \dfrac{\frac{2}{3}r^3}{\frac{1}{2}\pi r^2} = \dfrac{4r}{3\pi}\) *	A1*	2.2a
	(4)

Scheme	Marks	AO

CoM of “triangle” is \(\dfrac{2}{3}r\cos\theta\) from \(O\) OR CoM of “arc” is \(\dfrac{2x}{\pi}\) from \(O\)	M1	2.1
\(\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{2}r^2 \times \dfrac{2}{3}r\cos\theta\,\mathrm{d}\theta\) OR \(\displaystyle\int_0^r \pi x \times \dfrac{2x}{\pi}\,\mathrm{d}x\)	M1	3.1a
\(= \dfrac{1}{3}r^3\Big[\sin\theta\Big]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \dfrac{2}{3}r^3\) OR \(= \left[\dfrac{2}{3}x^3\right]_0^r = \dfrac{2}{3}r^3\)	A1ft	1.1b
\(\Rightarrow d = \dfrac{\frac{2}{3}r^3}{\frac{1}{2}\pi r^2} = \dfrac{4r}{3\pi}\) *	A1*	2.2a
	(4)

Mark scheme (b)

Scheme

Marks

	Rectangle	Large sc	Small sc	Total
Mass ratio	16	\(8\pi\)	\(2\pi\)	\(16 + 6\pi\)
From \(DC\)	\(-a\)	\(\dfrac{16a}{3\pi}\)	\(\dfrac{8a}{3\pi}\)	\(\bar{x}\)

B1
B1

1.1b
1.1b

Moments about \(DC\)

2.1

\(8\pi \times \dfrac{16a}{3\pi} - 2\pi \times \dfrac{8a}{3\pi} - 16a = (16 + 6\pi)\bar{x}\)

1.1b

\(\bar{x} = \dfrac{a\left(\dfrac{7 \times 16}{3} - 16\right)}{16 + 6\pi} = \dfrac{64a}{3(16 + 6\pi)} = \dfrac{32a}{3(8 + 3\pi)}\) *

A1*

2.2a

(5)

Notes

B1: Correct mass ratios

B1: Distances from \(DC\) or a parallel axis

M1: Moments about \(DC\) or a parallel axis. All terms required. Dimensionally correct.

A1: Correct unsimplified equation for their axis

A1*: Obtain given answer from correct working.

Mark scheme (c)

Scheme	Marks	AO
Moments about \(A\):	M1	3.1a
\(W \times \left(2a + \dfrac{32a}{3(8 + 3\pi)}\right) = 8a\lambda W\)	A1	1.1b
\(\lambda = \left(\dfrac{(9\pi + 40)}{12(8 + 3\pi)}\right) = 0.33\) or better (0.3265…)	A1	1.1b
	(3)
	(12 marks)

Notes

M1: Moments about \(A\). Dimensionally correct. Allow if \(a\) missing throughout.

A1: Correct unsimplified equation

A1: Correct only. Accept 0.33 or better (0.3265…).
Condone an exact answer in terms of \(\pi\) if given as a single fraction.

Notes

Alternative (a)

Notes

Notes