A-LevelMathematicsMechanics9709/52 · Feb/Mar 2019 · Q68 Marks
Fig. 1 shows the cross-section of a solid cylinder through which a cylindrical hole has been drilled to make a uniform prism. The radius of the cylinder is 5r and the radius of the hole is r. The centre of the hole is a distance 2r from the centre of the cylinder.
- (i)Find, in terms of r, the distance of the centre of mass of the prism from the centre of the cylinder.[4]
- (ii)The prism has weight WN and is placed with its curved surface on a rough horizontal plane. The axis of symmetry of the cross-section makes an angle of 30° with the vertical. A horizontal force of magnitude PN acting in the plane of the cross-section through the centre of mass is applied to the cylinder at the highest point of this cross-section (see Fig. 2). The prism rests in limiting equilibrium. Find the coefficient of friction between the prism and the plane.[4]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2019). The mark or mark type is shown in brackets.
- [M1]
- Area of hole = πr² and Area of original circle = 25n r2
- [A1]
- Area of cross-section = 24πr²
- [M1] Take moments about the centre of the cylinder
- πr² (2r) = 24πr² (d)
- [A1]
- d = r/12 ( = 0.083333....r)
- [M1] Take moments about the point of contact with the plane
- P(2 × 5r) = W(r/12)cos60
- [A1]
- P = Wcos60/120 = W/240 = 0.00417W ( = F )
- [M1] Use F = µR Note R = W by resolving vertically
- μ = (Wcos60/120)/W
- [A1]
- μ = 1/240 = 0.00417
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