| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Centre of mass with variable parameter |
| Difficulty | Standard +0.8 This is a multi-part centre of mass problem requiring systematic application of standard techniques: finding COM with variable density materials, applying tipping conditions with moments, and solving resulting equations. While it involves multiple steps and careful algebraic manipulation (particularly the 'show that' proofs), the methods are standard M2 fare with no novel conceptual insights required. The variable parameter h adds complexity but follows predictable patterns for this topic. |
| Spec | 6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let centre of mass be at G; G is on CO by symmetry | B1 | |
| Let \(CG = Y\) and curved surface density be \(\sigma\) | Accept \(\sigma\) taken to be 1 without comment | |
| \(\left(\pi(0.1)^2 \times 4\sigma + 2\pi \times 0.1 \times h \times \sigma\right)Y = \pi(0.1)^2 \times 4\sigma \times h + 2\pi \times 0.1 \times h \times \sigma \times \frac{h}{2}\) | M1 | Complete method |
| B1 | 'masses' in correct ratios: \(0.04 : 0.2h\) | |
| B1 | Correct use of '\(h\)' and '\(h/2\)' | |
| A1 | All correct | |
| \(Y = \frac{5h^2 + 2h}{2 + 10h}\) | E1 | Convincingly shown |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let the lowest point of contact of the cylinder with the plane be A; on point of tipping G is vertically above A | B1 | May be shown on a diagram |
| Angle AGO is \(\alpha\) | B1 | May be shown on a diagram; can be implied by subsequent work |
| \(\tan\alpha = \frac{0.1}{Y} = \frac{2}{3}\) | M1 | Allow reciprocal of RHS |
| \(0.3 = 2 \times \frac{2h + 5h^2}{2 + 10h}\) | A1 | |
| \(50h^2 + 5h - 3 = 0\) AG | A1 | cwo and convincingly shown |
| Either: \((5h-1)(10h+3) = 0\) | M1 | Clear evidence of 2 roots |
| (only positive root is) \(h = 0.2\) | E1 | No need to comment on the negative root |
| or: \(50 \times (0.2)^2 + 5 \times 0.2 - 3 = 2 + 1 - 3 = 0\) | B1 | |
| And this is the only positive root | E1 | Need statement but no need to show this |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\sin\beta = 0.6;\ \cos\beta = 0.8)\); c.w. moments about 'furthest' point of the base (through which acts the NR) | M1 | Both forces present in a moments equation |
| \(T\cos\beta \times 0.5 - T\sin\beta \times 0.2 - 42 \times 0.1 = 0\) | M1 | Attempt to find moment of force of magnitude \(T\) in horizontal and vertical components |
| A1 | Correct distances | |
| A1 | Correct equation; numerical values of cos/sin do not need to be substituted | |
| \((T(0.4 - 0.12) = 4.2)\) | ||
| \(T = 15\) | A1 | cao |
| [5] |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let centre of mass be at G; G is on CO by symmetry | B1 | |
| Let $CG = Y$ and curved surface density be $\sigma$ | | Accept $\sigma$ taken to be 1 without comment |
| $\left(\pi(0.1)^2 \times 4\sigma + 2\pi \times 0.1 \times h \times \sigma\right)Y = \pi(0.1)^2 \times 4\sigma \times h + 2\pi \times 0.1 \times h \times \sigma \times \frac{h}{2}$ | M1 | Complete method |
| | B1 | 'masses' in correct ratios: $0.04 : 0.2h$ |
| | B1 | Correct use of '$h$' and '$h/2$' |
| | A1 | All correct |
| $Y = \frac{5h^2 + 2h}{2 + 10h}$ | E1 | Convincingly shown |
| **[6]** | | |
---
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let the lowest point of contact of the cylinder with the plane be A; on point of tipping G is vertically above A | B1 | May be shown on a diagram |
| Angle AGO is $\alpha$ | B1 | May be shown on a diagram; can be implied by subsequent work |
| $\tan\alpha = \frac{0.1}{Y} = \frac{2}{3}$ | M1 | Allow reciprocal of RHS |
| $0.3 = 2 \times \frac{2h + 5h^2}{2 + 10h}$ | A1 | |
| $50h^2 + 5h - 3 = 0$ **AG** | A1 | cwo and convincingly shown |
| Either: $(5h-1)(10h+3) = 0$ | M1 | Clear evidence of 2 roots |
| (only positive root is) $h = 0.2$ | E1 | No need to comment on the negative root |
| or: $50 \times (0.2)^2 + 5 \times 0.2 - 3 = 2 + 1 - 3 = 0$ | B1 | |
| And this is the only positive root | E1 | Need statement but no need to show this |
| **[7]** | | |
---
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\sin\beta = 0.6;\ \cos\beta = 0.8)$; c.w. moments about 'furthest' point of the base (through which acts the NR) | M1 | Both forces present in a moments equation |
| $T\cos\beta \times 0.5 - T\sin\beta \times 0.2 - 42 \times 0.1 = 0$ | M1 | Attempt to find moment of force of magnitude $T$ in horizontal and vertical components |
| | A1 | Correct distances |
| | A1 | Correct equation; numerical values of cos/sin do not need to be substituted |
| $(T(0.4 - 0.12) = 4.2)$ | | |
| $T = 15$ | A1 | cao |
| **[5]** | | |4 Fig. 4.1 shows a hollow circular cylinder open at one end and closed at the other. The radius of the cylinder is 0.1 m and its height is $h \mathrm {~m} . \mathrm { O }$ and C are points on the axis of symmetry at the centres of the open and closed ends, respectively. The thin material used for the closed end has four times the density of the thin material used for the curved surface.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-5_366_656_443_717}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}
Cylinders of this type are made with different values of $h$.\\
(i) Show that the centres of mass of these cylinders are on the line OC at a distance $\frac { 5 h ^ { 2 } + 2 h } { 2 + 10 h } \mathrm {~m}$ from O .
Fig. 4.2 shows one of the cylinders placed with its open end on a slope inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 2 } { 3 }$. The cylinder does not slip but is on the point of tipping.\\
(ii) Show that $50 h ^ { 2 } + 5 h - 3 = 0$ and hence that $h = 0.2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-5_383_497_1178_1402}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
Fig. 4.3 shows another of the cylinders that has weight 42 N and $h = 0.5$. This cylinder has its open end on a rough horizontal plane. A force of magnitude $T \mathrm {~N}$ is applied to a point P on the circumference of the closed end. This force is at an angle $\beta$ with the horizontal such that $\tan \beta = \frac { 3 } { 4 }$ and the force is in the vertical plane containing $\mathrm { O } , \mathrm { C }$ and P . The cylinder does not slip but is on the point of tipping.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8fb49c8b-92e5-49e5-9a3a-e8391c82d9a1-5_451_679_1955_685}
\captionsetup{labelformat=empty}
\caption{Fig. 4.3}
\end{center}
\end{figure}
(iii) Calculate $T$.
\hfill \mbox{\textit{OCR MEI M2 2016 Q4 [18]}}