| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass with given condition |
| Difficulty | Challenging +1.2 Part (a) is a standard M3 centre of mass derivation using disc integration with the result given, requiring setup and algebraic manipulation. Part (b) requires applying the equilibrium condition (centre of mass at the curved surface) and combining centres of mass of composite bodies, which is a multi-step problem but follows standard M3 techniques without requiring novel insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \((\pi\rho)\int_0^r xy^2\,dx = (\pi\rho)\int_0^r x(r^2 - x^2)\,dx\) | M1 | Use of \((\pi\rho)\int_0^r xy^2\,dx\) with \(y^2 = r^2 - x^2\), attempt integration. Limits not needed |
| \(= (\pi\rho)\left[\frac{1}{2}x^2r^2 - \frac{x^4}{4}\right]_0^r\) | A1 | Correct integration, limits not needed |
| \(= (\pi\rho)\frac{r^4}{4}\) | A1 | Sub correct upper limit |
| \(\frac{2\pi\rho r^3}{3}\bar{x} = \pi\rho\int xy^2\,dx\) | M1 | Use of \(V\rho\bar{x} = \pi\rho\int xy^2\,dx\) with their result, where \(V\) is volume of hemisphere |
| \(\bar{x} = \frac{\pi\rho r^4}{4} \div \frac{2\pi\rho r^3}{3} = \frac{3}{8}r\) | A1* (5) | \(\bar{x} = \frac{3}{8}r\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Mass: Hemisphere \(\frac{2}{3}\pi r^3\), Cone \(\frac{1}{3}\pi kr^3\) | B1 | Correct mass ratio. Total mass not needed |
| Distance of CoM from centre of common plane: Hemisphere \(\frac{3}{8}r\), Cone \(\frac{1}{4}kr\) | B1 | Correct distances for both. Both can be positive or one negative. Alternatives: from vertex of cone (H) \(kr + \frac{3}{8}r\), (C) \(\frac{3}{4}kr\); from peak of hemisphere (H) \(\frac{5}{8}r\), (C) \(r + \frac{1}{4}kr\) |
| \(\frac{2}{3} \times \frac{3}{8}r = \frac{k}{3} \times \frac{1}{4}kr\) | M1A1ft | Form dimensionally correct moments equation with correct \(\bar{x}\) depending on where moments taken. A1ft correct equation, follow through masses and distances |
| \(k^2 = 3\), \(k = \sqrt{3}\) | A1 (5) | Correct exact result |
# Question 5:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $(\pi\rho)\int_0^r xy^2\,dx = (\pi\rho)\int_0^r x(r^2 - x^2)\,dx$ | M1 | Use of $(\pi\rho)\int_0^r xy^2\,dx$ with $y^2 = r^2 - x^2$, attempt integration. Limits not needed |
| $= (\pi\rho)\left[\frac{1}{2}x^2r^2 - \frac{x^4}{4}\right]_0^r$ | A1 | Correct integration, limits not needed |
| $= (\pi\rho)\frac{r^4}{4}$ | A1 | Sub correct upper limit |
| $\frac{2\pi\rho r^3}{3}\bar{x} = \pi\rho\int xy^2\,dx$ | M1 | Use of $V\rho\bar{x} = \pi\rho\int xy^2\,dx$ with their result, where $V$ is volume of hemisphere |
| $\bar{x} = \frac{\pi\rho r^4}{4} \div \frac{2\pi\rho r^3}{3} = \frac{3}{8}r$ | A1* (5) | $\bar{x} = \frac{3}{8}r$ |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Mass: Hemisphere $\frac{2}{3}\pi r^3$, Cone $\frac{1}{3}\pi kr^3$ | B1 | Correct mass ratio. Total mass not needed |
| Distance of CoM from centre of common plane: Hemisphere $\frac{3}{8}r$, Cone $\frac{1}{4}kr$ | B1 | Correct distances for both. Both can be positive or one negative. Alternatives: from vertex of cone (H) $kr + \frac{3}{8}r$, (C) $\frac{3}{4}kr$; from peak of hemisphere (H) $\frac{5}{8}r$, (C) $r + \frac{1}{4}kr$ |
| $\frac{2}{3} \times \frac{3}{8}r = \frac{k}{3} \times \frac{1}{4}kr$ | M1A1ft | Form dimensionally correct moments equation with correct $\bar{x}$ depending on where moments taken. A1ft correct equation, follow through masses and distances |
| $k^2 = 3$, $k = \sqrt{3}$ | A1 (5) | Correct exact result |
---\begin{enumerate}
\item (a) Use algebraic integration to show that the centre of mass of a uniform solid hemisphere of radius $r$ is at a distance $\frac { 3 } { 8 } r$ from the centre of its plane face.\\[0pt]
[You may assume that the volume of a sphere of radius $r$ is $\frac { 4 } { 3 } \pi r ^ { 3 }$ ]
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e837bb9-4ada-4f0f-8b21-2730611335f2-16_355_574_571_749}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform solid hemisphere of radius $r$ is joined to a uniform solid right circular cone made of the same material to form a toy. The cone has base radius $r$ and height $k r$. The centre of the base of the cone is $O$. The plane face of the cone coincides with the plane face of the hemisphere, as shown in Figure 3.
The toy can rest in equilibrium on a horizontal plane with any point of the curved surface of the hemisphere in contact with the plane.\\
(b) Find the exact value of $k$
\hfill \mbox{\textit{Edexcel M3 2022 Q5 [10]}}