M3 June 2014 (R) Q5
5.
A uniform solid right circular cylinder has height \(h\) and radius \(r\). The centre of one plane face is \(O\) and the centre of the other plane face is \(Y\). A cylindrical hole is made by removing a solid cylinder of radius \(\dfrac{1}{4}r\) and height \(\dfrac{1}{4}h\) from the end with centre \(O\). The axis of the cylinder removed is parallel to \(OY\) and meets the end with centre \(O\) at \(X\), where \(OX = \dfrac{1}{4}r\). One plane face of the cylinder removed coincides with the plane face through \(O\) of the original cylinder. The resulting solid \(S\) is shown in Figure 3.
The solid \(S\) is freely suspended from \(O\). In equilibrium the line \(OY\) is inclined at an angle arctan(17) to the horizontal.
| Scheme | Marks |
|---|---|
| \(\pi r^2h \qquad \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right) \qquad \pi r^2h - \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right)\) | B2 |
| \(\tfrac{1}{2}h \qquad\qquad \tfrac{1}{8}h \qquad\qquad \bar{y}\) | B2 |
| \(\pi r^2h\,\tfrac{1}{2}h - \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right)\tfrac{1}{8}h = \left[\pi r^2h - \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right)\right]\bar{y}\) | M1 A1ft |
| \(\bar{y} = \dfrac{85h}{168}\) ** | A1 |
| (7) |
Notes
B2 masses or volumes B2 all correct; B1 two correct
B2 distances B2 all correct; B1 one of the known ones correct
M1 A1ft form a moments equation using their volumes and distances
A1 correct result with no errors in the working
(Corrected from the printed mark scheme: the mass of the remaining solid is printed as \(\pi r^2 - \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right)\), without the \(h\) in the first term.)
| Scheme | Marks |
|---|---|
| \(0 - \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right)\tfrac{1}{4}r = \left[\pi r^2h - \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right)\right]\bar{x}\) | M1 A1 |
| \(\bar{x} = -\dfrac{r}{252}\) | A1 |
| \(\tan\alpha = \dfrac{\;\dfrac{85h}{168}\;}{\dfrac{r}{252}} = 17\) | DM1 A1ft |
| \(r = 7.5h\) | A1 |
| (6) | |
| (13 marks) |
Notes
M1 A1 form an equation to find the distance of the centre of mass from the axis of the cylinder
A1 correct distance
DM1 using their two distances to find the tan of the required angle (may be inverted)
A1ft ratio is correct(inc correct way up) with their distances
A1 correct answer
(Corrected from the printed mark scheme: the mass of the remaining solid is printed as \(\pi r^2 - \pi\left(\tfrac{1}{4}r\right)^2\left(\tfrac{1}{4}h\right)\), without the \(h\) in the first term.)