| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Challenging +1.2 This is a standard M3 non-uniform body equilibrium problem requiring center of mass calculation for composite cones and moment equilibrium about the suspension point. While it involves multiple steps (finding individual centers of mass, combining them, then applying equilibrium condition with the 30° angle), the techniques are routine for M3 students and the algebraic manipulation, though somewhat involved, follows a predictable path to the given answer. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
**Main Solution:**
Distance of centre of mass from O at angle $30°$
M1A1
Ratio of masses: $M$, $kM$, $(1+k)M$
Distances from O: $1$, $k$, $\frac{1+k}{1}$
M1A1A1ft
$M(O) = \frac{1}{4}h + \frac{k}{4}kh - \frac{r}{3}$
$\frac{k^2 h}{4} - \frac{r}{3}(1+k) = 0$
A1 [6]
$k = \frac{4r}{h} - 1$
**Alt 1: By moments about A**
M1A1, M1A1
$\frac{1}{4}kMg kh\cos 30° = rsin 30°$, $Mg \frac{1}{4}h\cos 30° = rsin 30°$
$h\cos 30° - k^2 - 1 = \frac{4r\sin 30°}{k-1}$
A1ft
$k - 1 = \frac{4r}{h}\tan 30°$
$k = \frac{4r}{h} - 1$
A1
**Alt 2: Find $x$ first**
M1 A1
$M(0) = \frac{1}{4}h + \frac{k^2h}{4}(1+k)x$
A1
$x = \frac{h(k-1)}{4}$ or equivalent
M1
$\tan 30° = \frac{x}{r}$
A1ft
$\frac{h(k-1)}{4} = \frac{1}{\tan 30°}$
$k = \frac{4r}{h} - 1$
A1
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{00388805-5d60-4327-a10e-c0df74a0cb75-09_403_790_210_577}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a uniform solid spindle which is made by joining together the circular faces of two right circular cones. The common circular face has radius $r$ and centre $O$. The smaller cone has height $h$ and the larger cone has height $k h$. The point $A$ lies on the circumference of the common circular face. The spindle is suspended from $A$ and hangs freely in equilibrium with $A O$ at an angle of $30 ^ { \circ }$ to the vertical.
Show that $k = \frac { 4 r } { h \sqrt { 3 } } + 1$
\hfill \mbox{\textit{Edexcel M3 2015 Q5 [6]}}