| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Challenging +1.2 This is a standard M3/Further Mechanics centre of mass question requiring application of standard formulas for solids of revolution (volume and centroid integrals), followed by a straightforward equilibrium application. The integrals involve polynomial functions which are routine to evaluate, and part (b) is a direct application of equilibrium geometry. While it requires multiple steps and knowledge of specific formulas, it follows a well-established template without requiring novel insight. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x \times \int_0^l \rho y^2 \, dx = \int_0^l \rho x y^2 \, dx\) \(\therefore x \times \int_0^l y^2 \, dx = \int_0^l x y^2 \, dx\) | B1 | |
| \(\int_0^l y^2 \, dx = \int_0^1 (x^2 + 1)^2 \, dx = \int_0^1 (x^4 + 2x^2 + 1) \, dx\) | M1 A1 | |
| \(= [\frac{1}{5}x^5 + \frac{2}{3}x^3 + x]_0^1 = \frac{28}{15}\) | M1 A1 | |
| \(\int_0^l x y^2 \, dx = \int_0^1 x(x^2 + 1)^2 \, dx = \int_0^1 (x^5 + 2x^3 + x) \, dx\) | M1 | |
| \(= [\frac{1}{6}x^6 + \frac{1}{2}x^4 + \frac{1}{2}x^2]_0^1 = \frac{7}{6}\) | M1 A1 | |
| \(\therefore \bar{x} = \frac{7}{6} + \frac{28}{15} = \frac{5}{8}\), \(\bar{y} = 0\) by symmetry \(\therefore\) chords are \((\frac{5}{8}, 0)\) | M1 A1 | |
| (b) \(\tan \alpha = \frac{1 - \frac{5}{8}}{2} = \frac{3}{16}\) | M1 A1 | |
| \(\therefore \alpha = 11°\) (nearest degree) | A1 | (13) |
(a) $x \times \int_0^l \rho y^2 \, dx = \int_0^l \rho x y^2 \, dx$ $\therefore x \times \int_0^l y^2 \, dx = \int_0^l x y^2 \, dx$ | B1 |
$\int_0^l y^2 \, dx = \int_0^1 (x^2 + 1)^2 \, dx = \int_0^1 (x^4 + 2x^2 + 1) \, dx$ | M1 A1 |
$= [\frac{1}{5}x^5 + \frac{2}{3}x^3 + x]_0^1 = \frac{28}{15}$ | M1 A1 |
$\int_0^l x y^2 \, dx = \int_0^1 x(x^2 + 1)^2 \, dx = \int_0^1 (x^5 + 2x^3 + x) \, dx$ | M1 |
$= [\frac{1}{6}x^6 + \frac{1}{2}x^4 + \frac{1}{2}x^2]_0^1 = \frac{7}{6}$ | M1 A1 |
$\therefore \bar{x} = \frac{7}{6} + \frac{28}{15} = \frac{5}{8}$, $\bar{y} = 0$ by symmetry $\therefore$ chords are $(\frac{5}{8}, 0)$ | M1 A1 |
(b) $\tan \alpha = \frac{1 - \frac{5}{8}}{2} = \frac{3}{16}$ | M1 A1 |
$\therefore \alpha = 11°$ (nearest degree) | A1 | (13)6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{00776cc0-0214-4029-8ef1-c1cba89f4b87-4_455_540_201_660}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Figure 3 shows part of the curve $y = x ^ { 2 } + 1$. The shaded region enclosed by the curve, the coordinate axes and the line $x = 1$ is rotated through $360 ^ { \circ }$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the centre of mass of the solid obtained.
The solid is suspended from a point on its larger circular rim and hangs in equilibrium.
\item Find, correct to the nearest degree, the acute angle which the plane surfaces of the solid make with the vertical.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [13]}}