AS June 2025 Q3

EdexcelCurrent spec9 marksCentres of Mass

3. [In this question you may quote, without proof, the formula for the distance of the centre of mass of a circular arc from its centre.]

Figure 2: framework: A vertically above B with AB = 1.5a, C vertically below B with BC = 2a, BD horizontal with a right angle at B, straight wire AD, and a quarter-circle arc from D down to C — Figure 2

Uniform wire is used to form the rigid framework shown in Figure 2.

The framework consists of four straight pieces of wire, \(AB\), \(BC\), \(BD\) and \(AD\) and one piece in the shape of an arc of a circle, \(CD\).

In the framework

\(AB = 1.5a\) and \(BC = BD = 2a\)
\(CD\) is an arc of a circle of radius \(2a\) and centre \(B\)
\(ABC\) is a straight line and \(ABD\) is a right angle
\(A, B, C\) and \(D\) all lie in the same plane

(a) Show that the distance of the centre of mass of arc \(CD\) from \(AC\) is \(\dfrac{4a}{\pi}\) (2)

(b) Show that the distance of the centre of mass of the framework from \(AC\) is \(\dfrac{17a}{2(8+\pi)}\) (4)

The framework is freely pivoted at \(A\).

The framework is held in equilibrium, with \(BD\) horizontal, by an upward vertical force that is applied to the framework at \(D\).

The mass of the framework is \(M\).

The magnitude of the force exerted on the framework by the pivot at \(A\) is \(kMg\).

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{2a\sin\frac{\pi}{4}}{\frac{\pi}{4}}\cos\dfrac{\pi}{4}\)	M1	2.1
\(\dfrac{4a}{\pi}\) *	A1*	1.1b
	(2)

Notes

M1: Correct expression using either trigonometry or Pythagoras’:
\(\sqrt{2 \times \text{“distance”}^2} = \dfrac{2a\sin\frac{\pi}{4}}{\frac{\pi}{4}}\) leading to “distance” = …

A1*: Correct given answer correctly obtained. Condone \(\dfrac{4}{\pi}a\).

Mark scheme (b)

Scheme	Marks	AO
\(AD\) \(BD\) Arc \(CD\) framework
\(2.5a\) \(2a\) \(\pi a\) \(8a + \pi a\)	B1	1.2
Moments about \(AC\)	M1	3.1a
\(2.5a \times a + 2a \times a + \pi a \times \dfrac{4a}{\pi} = (8a + \pi a)\bar{x}\)	A1	1.1b
\(\bar{x} = \dfrac{17a}{2(8+\pi)}\) *	A1*	2.2a
	(4)

Notes

B1: Any equivalent ratios

M1: Or moments about a parallel axis (distances consistent with framework not lamina). Allow consistently cancelled \(a\). Must have the correct number of terms.

A1: Correct unsimplified equation for their axis

A1*: Correct given answer correctly obtained. Condone \(\dfrac{17a}{2(\pi+8)}\)

Mark scheme (c)

Scheme	Marks	AO
Moments about \(D\) or any other complete method	M1	3.1a
\(kMg \times 2a = Mg\left(2a - \dfrac{17a}{2(8+\pi)}\right)\)	A1	1.1b
\(k = \dfrac{15 + 4\pi}{4(8+\pi)}\)	A1	1.1b
	(3)
	(9 marks)

Notes

M1: Correct no. of terms, dim correct or e.g. M(\(A\)) and vertical resolution.

A1: Correct unsimplified equation in \(k\), \(Mg\), \(a\) and \(\pi\) only. Condone correctly cancelled terms.

A1: cao (denominator does not need to be factorised)