M3 January 2005 Q2

Figure 1: solid cone of height 3r with vertex O joined at diameter AB to a hemisphere of base diameter 2r — Figure 1

A child’s toy consists of a uniform solid hemisphere, of mass \(M\) and base radius \(r\), joined to a uniform solid right circular cone of mass \(m\), where \(2m < M\). The cone has vertex \(O\), base radius \(r\) and height \(3r\). Its plane face, with diameter \(AB\), coincides with the plane face of the hemisphere, as shown in Figure 1.

(a) Show that the distance of the centre of mass of the toy from \(AB\) is \[\frac{3(M-2m)}{8(M+m)}r.\] (5)

The toy is placed with \(OA\) on a horizontal surface. The toy is released from rest and does not remain in equilibrium.

(b) Show that \(M > 26m\). (4)

Diagram: toy resting with OA on the surface, centre C of AB, D on the axis vertically above A, angle alpha at O — Mark scheme (b)

Scheme	Marks
\(\dfrac{3r}{4}\ ;\ \dfrac{3r}{8}\)	B1 ; B1
\(-m \cdot \dfrac{3r}{4} + M \cdot \dfrac{3r}{8} = (m+M)\bar{x}\)	M1 A1
\(\dfrac{3r(M-2m)}{8(M+m)} = \bar{x}\) *	A1
	(5)

Scheme	Marks
\(CD = r\tan\alpha = r \times \left(\dfrac{r}{3r}\right) = \dfrac{1}{3}r\)	M1 A1
No equil\(^\text{m}\) \(\Rightarrow \bar{x} > CD\)
\(\dfrac{3r(M-2m)}{8(M+m)} > \dfrac{r}{3}\)	M1
\(9(M-2m) > 8(M+m)\)
\(M > 26m\) *	A1
	(4)
	(9 marks)