M3 June 2012 Q4

EdexcelOld spec10 marksCentres of Mass

Figure 2: cross-section AVBC, a cone of height 2a with a cone of height a removed from its base, common axis VCO — Figure 2

Figure 2 shows the cross-section \(AVBC\) of the solid \(S\) formed when a uniform right circular cone of base radius \(a\) and height \(a\), is removed from a uniform right circular cone of base radius \(a\) and height \(2a\). Both cones have the same axis \(VCO\), where \(O\) is the centre of the base of each cone.

(a) Show that the distance of the centre of mass of \(S\) from the vertex \(V\) is \(\dfrac{5}{4}a\). (5)

The mass of \(S\) is \(M\). A particle of mass \(kM\) is attached to \(S\) at \(B\). The system is suspended by a string attached to the vertex \(V\), and hangs freely in equilibrium. Given that \(VA\) is at an angle 45° to the vertical through \(V\),

(b) find the value of \(k\). (5)

Mark scheme (a)

Scheme

Marks

	volume	Mass ratio	C of M from \(V\)
Large cone	\(\dfrac{1}{3}\pi a^2.2a = \dfrac{2}{3}\pi a^3\)	2	\(\dfrac{3}{4} \times 2a = \dfrac{3}{2}a\)
Small cone	\(\dfrac{1}{3}\pi a^2.a = \dfrac{1}{3}\pi a^3\)	1	\(a + \dfrac{3}{4}a = \dfrac{7}{4}a\)
S	\(\dfrac{1}{3}\pi a^2.a = \dfrac{1}{3}\pi a^3\)	1	\(D\)

B1, B1

\(1 \times D = 2 \times \dfrac{3}{2}a - 1 \times \dfrac{7}{4}a\)

M1A1

\(= \dfrac{12 - 7}{4}a = \dfrac{5}{4}a\) **

(5)

Diagram of S hanging from V: VA at 45 degrees, angle 26.6 degrees to VG, VB of length root 5 a at 81.87 degrees to the horizontal, weights Mg and kMg — Mark scheme (b)

Scheme	Marks
\(45^\circ + 26.6^\circ\left(= 71.6^\circ\right),\ \left(81.8698\ldots =\right) 81.9^\circ\)
Take moments about \(V\):	M1
\(Mg \times \dfrac{5}{4}a \times \cos 71.6 = kMg \times \sqrt{5}a \times \cos 81.9\)	A2
\(k = \dfrac{5\cos 71.6}{4\sqrt{5}\cos 81.9} = 1.25\)	M1A1
	(5)
	(10 marks)