M3 June 2008 Q4

EdexcelOld spec13 marksCentres of Mass

Figure 3: bowl formed from a hemisphere of radius 6a with a hemisphere of radius 2a, both centre O, removed — Figure 3

A uniform solid hemisphere, of radius \(6a\) and centre \(O\), has a solid hemisphere of radius \(2a\), and centre \(O\), removed to form a bowl \(B\) as shown in Figure 3.

(a) Show that the centre of mass of \(B\) is \(\dfrac{30}{13}a\) from \(O\). (5)

Figure 4: the bowl fixed to the plane face of a cylinder of radius 2a and height 6a, on a common axis through O — Figure 4

The bowl \(B\) is fixed to a plane face of a uniform solid cylinder made from the same material as \(B\). The cylinder has radius \(2a\) and height \(6a\) and the combined solid \(S\) has an axis of symmetry which passes through \(O\), as shown in Figure 4.

(b) Show that the centre of mass of \(S\) is \(\dfrac{201}{61}a\) from \(O\). (4)

The plane surface of the cylindrical base of \(S\) is placed on a rough plane inclined at 12\(^\circ\) to the horizontal. The plane is sufficiently rough to prevent slipping.

Sketch: large hemisphere equals small hemisphere plus bowl — Mark scheme (a)

Sketch: bowl plus cylinder equals S — Mark scheme (b)

Sketch: S on the inclined plane with angle theta at the centre of mass, distance 12a minus y-bar to the plane and 2a along it — Mark scheme (c)

Scheme	Marks
\(12a - \dfrac{201}{61}a\)	M1
\(\tan\theta = \dfrac{2a}{12a - \frac{201}{61}a}\) \(\tan\theta = \dfrac{2a}{....}\)	M1
\(\theta = 12.93.....\)	A1
so critical angle \(= 12.93\ldots\) \(\therefore\) if \(\theta = 12^\circ\) it will NOT topple.	A1ft
	(4)
	(13 marks)