M3 June 2017 Q4
4.
A thin uniform right hollow cylinder, of radius \(a\) and height \(4a\), has a base but no top. A thin uniform hemispherical shell, also of radius \(a\), is made of the same material as the cylinder. The hemispherical shell is attached to the open end of the cylinder forming a container \(C\). The open circular rim of the cylinder coincides with the rim of the hemispherical shell. The centre of the base of \(C\) is \(O\), as shown in Figure 3.
The container is placed with its circular base on a plane which is inclined at \(\theta^\circ\) to the horizontal. The plane is sufficiently rough to prevent \(C\) from sliding. The container is on the point of toppling.
| Scheme | Marks |
|---|---|
| Ratio of masses: \(\pi a^2 \quad 2\pi a \times 4a \quad 2\pi a^2 \qquad 11\pi a^2\) | M1A1 |
| Distances: \(0 \qquad 2a \qquad 4\tfrac{1}{2}a \qquad \bar{y}\) | B1 |
| \((0 +)\,2a \times 8 + 4\tfrac{1}{2}a \times 2 = 11\bar{y}\) | M1A1ft |
| \(\bar{y} = \dfrac{25}{11}a\ \ (= 2.272\ldots a)\) | A1 |
| (6) |
Notes
M1 Attempt the ratio of the masses of the separate parts. Formulae used must be correct; allow if cylinder has a top as well as a bottom - ignore top - or neither. Similarly if hemisphere has a base. Allow if the base of the cylinder and the curved surface are combined.
A1 Correct ratio seen eg 1:8:2:11 (or any equivalent) (Top of cylinder/base of hemisphere not ignored now; combined curved surface and base for cylinder not allowed.)
B1 Correct distances from \(O\) or any other point for the curved surface of the cylinder and the hemispherical shell.
M1 Form a moments equation about their chosen point. Must be dimensionally correct (ie no \(a^3\) seen). Extra terms score M0.
A1ft Correct equation, follow through their mass ratio and distances. 1 or 2 signs may be negative if a point other than \(O\) has been used.
A1cao Correct distance from \(O\) exact or min 2sf. (Must be obtained from a correct equation.)
ALT Find c of m of cylinder (inc base) first and then combine with hemisphere. All marks available. Award second M when combining with the hemisphere.
NB If c of m of cylinder with base is given as \(2a\) then only M1A0B0M0A0A0 available.
| Scheme | Marks |
|---|---|
| \(\tan\theta^\circ = \dfrac{a}{\dfrac{25}{11}a}\ \ \left(= \dfrac{11}{25}\right)\) | M1A1ft |
| \(\theta = 23.749\ldots\) Accept 24 or better | A1 |
| (3) | |
| (9 marks) |
Notes
M1 Use \(\tan\theta = \dfrac{a}{\bar{y}}\) or \(\dfrac{\bar{y}}{a}\) with their answer from (a)
A1ft Correct expression for \(\tan\theta\) follow through their \(\bar{y}\)
A1cao Correct value for \(\theta\) Min 2 sf. (NB Equivalent in radians scores A0)