| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cone from cone or cylinder |
| Difficulty | Standard +0.8 This is a multi-part centre of mass problem requiring composite body techniques with a removed cone, followed by equilibrium analysis with suspension. While the method is standard for M3 (using volume ratios and the formula for cone COM), it requires careful coordinate setup, algebraic manipulation with multiple parameters (4r, 4h, 3r, 3h), and part (b) adds geometric reasoning with trigonometry. More demanding than routine COM questions but follows established M3 techniques. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Mass: \(\frac{1}{3}\pi \times 16r^2 \times 4h\) \ | \(\frac{1}{3}\pi \times 9r^2 \times 3h\) \ | \(\frac{1}{3}\pi \times 37r^2 h\) |
| (Ratio: 64 \ | 27 \ | 37) |
| Dist: \(h\) \ | \(h + \frac{3}{4}h\) \ | \(\bar{x}\) |
| \(64h - 27 \times \frac{7}{4}h = 37\bar{x}\) | M1A1ft | M1: Use mass ratio and distances to form moments equation with 3 terms; A1ft: Correct equation following their mass ratio and distances |
| \(\bar{x} = \frac{67}{148}h\) accept \(0.45(270\ldots)h\) or better | A1 (5) | Correct distance from \(O\). Exact or min 2 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\tan\theta = \frac{h - \bar{x}}{3r} = \frac{1 - \frac{67}{148}}{3}\) | M1, A1ft | M1: Use \(\bar{x}\) to form expression for \(\tan\theta\); \(h - \bar{x}\) needed but fraction can be either way up. A1ft: Fraction either way up and \(h = r\) used |
| \(\theta = \tan^{-1}\left(\frac{148-67}{3 \times 148}\right)\) \((= 10.34\ldots)\) | ||
| Required angle \(= \tan^{-1}\left(\frac{r}{3r}\right) - \theta = 8.096\ldots°\) | M1A1 (4) | M1: Complete method to obtain required angle; A1: Correct size of angle, 2 sf min. Radians accepted: 0.14 or better |
| Accept 8.1 or better | NB: If solid hung from point on rim of base give M1A0M0A0 | |
| ALT: \(\tan\theta = \frac{3r}{h-\bar{x}} = \frac{3}{1 - \frac{67}{148}}\) | M1A1 | |
| Required angle \(= \theta - \tan^{-1}(3) = 8.096\ldots°\) | M1A1 |
# Question 2(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Mass: $\frac{1}{3}\pi \times 16r^2 \times 4h$ \| $\frac{1}{3}\pi \times 9r^2 \times 3h$ \| $\frac{1}{3}\pi \times 37r^2 h$ | B1 | Correct ratio of masses, any equivalent. Mark the ratio, not formulae |
| (Ratio: 64 \| 27 \| 37) | | |
| Dist: $h$ \| $h + \frac{3}{4}h$ \| $\bar{x}$ | B1 | Correct distances from $O$ or any other point |
| $64h - 27 \times \frac{7}{4}h = 37\bar{x}$ | M1A1ft | M1: Use mass ratio and distances to form moments equation with 3 terms; A1ft: Correct equation following their mass ratio and distances |
| $\bar{x} = \frac{67}{148}h$ accept $0.45(270\ldots)h$ or better | A1 (5) | Correct distance from $O$. Exact or min 2 sf |
---
# Question 2(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\tan\theta = \frac{h - \bar{x}}{3r} = \frac{1 - \frac{67}{148}}{3}$ | M1, A1ft | M1: Use $\bar{x}$ to form expression for $\tan\theta$; $h - \bar{x}$ needed but fraction can be either way up. A1ft: Fraction either way up and $h = r$ used |
| $\theta = \tan^{-1}\left(\frac{148-67}{3 \times 148}\right)$ $(= 10.34\ldots)$ | | |
| Required angle $= \tan^{-1}\left(\frac{r}{3r}\right) - \theta = 8.096\ldots°$ | M1A1 (4) | M1: Complete method to obtain required angle; A1: Correct size of angle, 2 sf min. Radians accepted: 0.14 or better |
| Accept 8.1 or better | | NB: If solid hung from point on rim of base give M1A0M0A0 |
| **ALT:** $\tan\theta = \frac{3r}{h-\bar{x}} = \frac{3}{1 - \frac{67}{148}}$ | M1A1 | |
| Required angle $= \theta - \tan^{-1}(3) = 8.096\ldots°$ | M1A1 | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a67e3644-13fa-4196-a2ef-ea1e26f5726c-04_264_438_269_753}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform solid right circular cone $R$, with vertex $V$, has base radius $4 r$ and height $4 h$. A right circular cone $S$, also with vertex $V$ and the same axis of symmetry as $R$, has base radius $3 r$ and height $3 h$. The cone $S$ is cut away from the cone $R$ leaving a solid $T$. The centre of the larger plane face of $T$ is $O$. Figure 1 shows the solid $T$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance from $O$ to the centre of mass of $T$.
The point $A$ lies on the circumference of the smaller plane face of $T$. The solid is freely suspended from $A$ and hangs in equilibrium. Given that $h = r$
\item find the size of the angle between $O A$ and the downward vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q2 [9]}}